"First countable" does not only require that each point can be provided with a countable set of nested neighborhoods, as stated at 3:40; it requires in addition that these nested neighborhoods approach the point more than any arbitrary neighborhood of it; en.wikipedia.org/wiki/First-countable_space
Hi! At 36:06, you discuss the continuity of the function f(x) = x from the euclidean topology to the closed interval topology, and at 36:56 you explain it cannot be continuous because the half-open interval in T_CI (inversely) maps to a half-open interval in T_E. But wouldn't the preimage of that closed interval in T_CI actually exclude the point 0 in T_E, thus making it an open set in T_E?
32:18 Yes, I think it's pretty easy to understand how if each open set of Y maps back to an open set of X, then any point x in X must have a neighborhood that's a subset of any neighborhood of y in Y. But what's not so clear to me is how we can assume the converse of this relationship must be true. That is, how can we show that if every point x has a neighborhood that's a subset of any neighborhood of y, then it necessarily follows that all of Y's open sets must map backwards to open sets of X? Thanks for making these videos.
Maybe the second definition (which seems to me to be the definition of continuity at a point: en.wikipedia.org/wiki/Continuous_function#Continuity_at_a_point) is not correctly stated in the video (at 31:02 it should continue: "and so for any open subset of V"). So, let Ux be some open neighborhood of x in J_x and Vy some open neighborhood of y in J_y; then the second definition should state that for any y in Y, for ANY open subset Sy of Vy there is Ux in J_x that maps to Sy; if you consider that Vy itself is one such Sy, you'll conclude that there is some Ux that maps to Vy; this amounts to the first definition.
Around 16:14: You can build any open set (a,b) say you take [b_n, b_n + 1/m) so that the b_n's enumerate the rationals. The problem is you can't get all "open" sets of the form [a,b). In the case just suggested you can get (sqrt(2), 2), but you can't get [sqrt(2), 2).
Totally agree with you. This part was really confusing, and as a matter of fact, his explanation is wrong. It is not second-countable because the number of [b, a) intervals is aleph1, and you can't construct "[b," part with an infinite union you have to have it in your base, hence your base can't be countable.
at 8:44 how is the nested interval second countable? I can, say specify an interval from 0 to 1/4... it can be constructed from the basis? it seems that the I can use the first element in the base to cover the above interval.... I don't see how the second countability even come into the picture? I am a bit confused...
The interval you specified is not an open set! Second Countability involves making any OPEN set from the union of a countable number of open sets. The nested interval topology is particularly simple in the regard: the entire topology is countable and the topology is its own base!
at 16:28 you explained with the concept of completeness that T_LL is not C-2, but I doubt if it is the same for T_E? if we consider b_i of an interval I_b, where i is an element of Q, we can alway find a gap in between two intervals(because not complete). Wouldn't that imply there exists an interval that is not fully retrievable with the bases of a given collection? we'd be able to fill up all the holes if the base element I_b is an uncountable collection. wait, please confirm with me if the following assumption is true. if we were to consider an interval I, we can find inf{S}=x. Then if we say inf{S}∉S, it is an open set. we can define x'=x+e where e->0. we can always find some rational number p/q such that p/q
One short question around 29:34. if we have the pre-image of an open set in X under the inverse f^-1 then the function is continuous. Q)Should that the pre-image cover all X? If we take an arbitrary union of all pre-image on X mapped by f^-1, then should the arbitrary union has to cover X or is it ok to cover a subset of X? I guess if homeomorphic, it should cover all X and vice versa but not sure if that is one special case.
There is no requirement that this process can cover X with open sets however. For example, f might not be surjective but it still can be continuous. Homeomorphisms are examples of continuous functions that do indeed cover both the domain and the range of the mapping.
Thx. So can I wrap up this way, if we can guarantee that f^-1(u) exists for all u in Y, we know that at least some subset of X generates all output value in Y, which satisfies the definition of function (1 input corresponds to 1 output) I really appreciate you for all your help:)
Jay Lee I would say the inverse only needs to map those open sets in Y which are in the range of f. In order to have an inverse f must be 1-1 from its domain into its range. Neither the domain nor the range need cover X or Y.
At 42:00, why are you taking in the open interval with green boundaries? While the open intervals, basis in cofinite topology are one to one in euclidean topology?
It has been a while since I reviewed this, and I am not quite sure what you are asking,,,, but I agree that the confinite topology is 1-1 with the Euclidean topology, but the Euclidean topology has MORE open sets. That is, it is *finer*. So the mapping is invective but not surjective (not ‘onto”). Does this help?
you are calling arbitrarily chosen open interval in cofinite topology as an open interval. I guess that open intervals in cofinite topology are subsets left after omitting the points.
Tursinbay Oteev I shouldn’t be, if that is what you are hearing. The cofinite topology’s open sets are what is left after removing a finite number of points.
I'm not sure the explanation at 15:30 (of why the LL topology is not 2C) is correct. We can choose our base as all [a, b) with a, b rational. This base can be used to construct any open interval (in a similar way to how the open intervals can be used in the Euclidean topology): choose [a1, b), [a2, b), ... where the a_n converge to a. What this (or any countable base) cannot do, however, is generate all the half-open intervals. See also math.stackexchange.com/questions/1135993/mathbbr-with-the-lower-limit-topology-is-not-second-countable As before, thanks for the excellent series of videos! I'm finding them very instructive.
I’m sold! Of course you are correct! I love the proof at the link you provided! I wish I could add annotations to the video! I think what I will do is put your observation in the comment section! Thank you for paying such close attention and your kind remarks!
@@XylyXylyX My pleasure! I find that unless I try to work through the proofs myself, I quickly forget the material. For anyone reading, the basic idea is this: if you want to converge to a half-open interval [x, a), you must include at least one set whose minimum is x. This is unlike the Euclidean case, where none of the sets converging to (x, a) needs to include x. Since there are uncountably many such x, you can't do it with a countable base. There's one thing I'm unsure of in this proof: proofwiki.org/wiki/Sorgenfrey_Line_is_not_Second-Countable. It's true that Bx must have _infimum_ of x, but it seems to me that it must also be the _minimum_ . Not sure why they wouldn't mention this stronger condition, even if the weaker one will do.
Aditya Prasad I also noticed that this proof used a theorem about separability that I did not cover. I should do a little mini correction lecture about this. I see your point about infimum/minimum. I will try to get to the bottom of that too.
This is extremely useful! I was also confused about the proof in the video about 2nd countability of LL topology. I hope I can vote this comment up and let others see this.
as said from others your original proof about LL topology is not 2-C actually is not correct; we need to take in account half-open intervals instead: from [a,b) with a,b rationals we are able to build a succession of rationals having as limit a given real number however we cannot "reach" it, hence we cannot build any [a,b) with a,b real numbers.
Thank you for your detour through Manifoldland. It's very helpul to have this helicopter view of how pure mathematicians set the concepts of continuity, compactenss, etc on a solid set-theoretical basis. But I have one question about an issue you only skirted in your presentation. You showed us how "continuity "works only in the case of f(x)=x. The question is whether our pure mathematician friends could not dream up of some function that would preserve continuity in some of the cases you discussed. Or does it suffice to show that if it doesn't work for for the identity function it's not continuous?
Continuity is a property of *functions* and some functions will be continuous and others won't. Indeed, some functions do exist between these spaces that are continuous. In those cases such functions might be somewhat contrived, but they exist. I chose f(X)=X because it was easy to demonstrate. Continuity of a function is not the same as homeomorphism between two different topological spaces. In that case I must either find *any* function to prove two spaces are homeomorphic, or prove that no such function exists. That is, if I suspect two spaces are homeomorphic but I have not found the correct mapping, I need to worry that some other mathematician may find it. Not so for continuity, which is easier to check function-by-function.
Take f(X) = constant (ie a single point). This function is always continuous regardless of the topologies involved as the inverse image is either the empty set or X.
I'm a bit confused on the Lower Limit topology and countability. From 12:00 and 17:00, if open intervals are also part of the topology, then can't they be their own bases, instead of requiring infinite (uncountable?) numbers of half-open intervals as bases? Wouldn't this make the Lower Limit topology second countable? I think my understanding of bases might be wrong here ....
Each time someone asks about this material I have to remind myself of the fundamental definitions! You are definitely thinking about this thoroughly and well, which is the key to ultimately getting all to make sense. First, you noted without realizing it maybe, that the LL topology is *finer* than the standard topology: it has more open sets. Not only does it have the half open interval, but it also has the open intervals too, and all the associated unions and intersections. All of these open sets can be constructed from the basis of half-open intervals. A basis MUST be able to make *all* open sets via union and finite intersection. The question is can we identify a *countable* basis? That is, can we find a *countable* number of half open intervals that can make *any* open set in the topology. It is not good enough to find a countable collection of half-open intervals to build a particular open set, it must be one countable set that will build EVERY open set in the topology. The argument that this can not be done is a bit hard to see at first, but upon reflection it slowly becomes “obvious.” Say we have found such a countable basis and call that countable basis “C”. Then that basis can build the open set [x, x+a), which is obviously open. But note that [x, x+a) *may not be in our countable collection* so it is NOT in C, but it can be made with a countable number of members of C. This means that C must contain a collection of half-open intervals that can be “unioned” together with a lower limit of x. THat is not hard to do for any given x, but .... Then take another open set [y, y+a). Now it must be that C contains a collection of half-open intervals with a lower limit of y. Repeat this argument over and over and you will see that C has the impossible job of having to have a collection of half-open intervals that limit to every possible real number. This means it must have a cardinality equal to the real numbers, which is uncountable!
@@XylyXylyX Thanks for your explanation, it will take me a while to digest ... Anyway I genuinely and pleasantly surprised that you still remember what you did almost 3 years back and are still answering questions about them. Thanks to the recent black hole hooha, I am doing both your Tensor and Manifold series concurrently so that I can get to your General Relativity series, and hopefully Lie Groups and Algebras after that. It's very hard going, especially the Manifold ones, but your videos are truly helpful and enjoyable. Thanks again and keep up the good work!
ngozumpa The more I think about your question, the better it is. A full understanding of why the LL topology is second countable is a great exercise to understand the concept.
couldnt you create a nested seies of open sets in the cofinite topology just by choosing 1 point, then 1 more and 1 more and so on? the nth open set would be determined by the 1st n choices of points. each consecutive open set would be contained in its predecessor.
I was trying to use the "neighborhood" method of determining continuity in the example you provided for the identity function from the euclidean topology to the lower limit topology. Any open neighborhood V on the LL topology looks like [a, b) or (a, b). Going back to the euclidean topology, we can always find a neighborhood U (a+delta, b-delta), whose image is a subset of V. So for every V in the LL topology, we can always find an open neighborhood U in the euclidean topology that maps to a subset of V. Thus the identity function is continuous on the LL topology. I must be missing something, as clearly from your example at 39:00 the identity function is not continuous on the LL topology, but I don't see what I'm doing wrong.
At 14:00, if your sequence approaching L corresponds to half open intervals, closed on left side, how can the open set resulting from the limiting process be *open* on both sides? ISTM, that the result would be half open, closed on left side.
Soon thereafter you claim that LL topology not second countable because you found a nested countable sequence which skips over some point P. But the definition of second countable just demands that some countable union exists which contains every open set; not that every such countable union must contain some open set.
Frank Bennett For your first question I will respond "Socratic-ally": If the left side is closed, then what is the lowest member of the set? That is, there should be a highest lower bound to the infinite union that is also a member of the set. If you think about this, you will realize there is no such point!
I see the answer to my second question. The definition requires that some countable base will cover every open set. But there is no such base for the LL topology.
Frank Bennett The definition of second countability is that the topology has a countable basis. My little informal demonstration showed that any nominee countable basis will not be able to construct arbitrary open sets. I probably should had done a more rigorous proof of this point, however.
In discussing the 1-C of Cofinite topology, starting from the original open set that contains P but has 2 more points excluded, if we continue to exclude more points, then the new open sets are subsets of the old, thus forming a nested series. Where am I wrong?
proofwiki.org/wiki/Uncountable_Finite_Complement_Space_is_not_First-Countable is a proof of this point. However, you are correct in noting that *my* hand-waving proof was not good enuf! Good observation! Thanks.
You don't get nested intervals because you can only omit a countable subset from your original interval and the set difference of nested intervals is more than countable
Thomas Impelluso THe discrete topology asserts that every subset of X is a member of the topology of X. If X is the real like, say, then any closed interval is a subset of X and is therefore a member of the discrete topology. "Closed" has two meanings here. 1) a closed set is the complement of an open set and 2) a closed intervals contains its end points. "Closed intervals" are "closed sets" if their complements are open. In the discrete topology the complement of ANY set is still a subset of X and therefore a member of the topology and therefore "open".
I am not sure I am convinced that the justification around 11:00 can be used to demonstrate that the cofinite topology is not first-countable. The claim, as I understand it, is that given an open set S_0 that contains the point A, but excludes points B, and C, there is no open set S_1 that is nested (a proper subset of S_0) which also contains A. But isn't the open set which excludes points B, C, and D a proper subset of S_0 which contains the point A? To demonstrate that the cofinite topology is not first-countable, I believe you must show that for any finite sequence S_0, S_1, S_2 ... of nested neighborhoods of A, there exists a neighborhood of A which contains none of the neighborhoods in the sequence. Or am I missing something here?
This bothered me also. Here what Wolfram had to say via ChatGPT: Yes, the cofinite topology on an open interval of the real line is first countable. In the cofinite topology, the open sets are defined as either the entire space itself or any set whose complement (relative to the space) is finite. This means that the only closed sets are the whole space and finite sets. To demonstrate that this space is first countable, we need to show that each point in the space has a countable local basis. For any point \( x \) in our open interval (let's denote this interval as \( I \)), we can consider the collection of all open sets that contain \( x \) but exclude only a finite number of points. Specifically, for each finite subset \( F \) of \( I \) that does not contain \( x \), we can define an open set \( U_F = I \setminus F \) that contains \( x \) and whose complement is \( F \), which is finite by construction. This collection of sets \( \{ U_F \} \) forms a local basis at \( x \). Since there are countably many finite subsets of \( I \) (even though \( I \) itself may be uncountable), this local basis is countable. Therefore, the cofinite topology on an open interval of the real line is first countable. This property holds regardless of the specific open interval of the real line considered, as long as we're working within the cofinite topology.
Hi, I do not understand your explanation about 2nd countability of half-open intervals. I do not see the reason why I could not construct any open interval. But I think all half-open intervals cannot be created because of the union of [a_n,b) =(a,b), where a_n is decreasing sequence.
I think this is explained at 12:07. I belive the point is that you have to be able to choose a single countable basis that creates all the open sets simultaneously. But for any single countable basis you can always choose an open set (c,d) which is an element of the lower limit topology where c is say between b_n and b_n+1. You can't create this set unless you pick a new basis.
@@stber321 Same problem for the closed interval topology. If your argument is correct, not all random open set (a, b) can be constructed by a chosen countable base of [-1, bn) and (an, 1]
No. Should I? If a space is second countable it has a countable base. If it has a countable base, then it has a countable base at each point. Second C -> First C.
I think in the definition of a homeomorphism you forgot to mention that f has to be a bijection. However, I guess it can be automatically implied depending on how you define what a function is.
It has become traditional. There is a way of inverting all the definitions so that closed sets take center stage. I do not know why the subject has leaned on open set, but Ihave seen all the reversed definitions. It often comes as an exercise, btw. For example: define a topology in terms of closed sets.
Hi. At first, thank you for those very interesting lectures. But there is something I'm not sure to well understand: - at 18'30, you sais that any open interval can be built as the limit of the union of baie elemnt of JLL. It coul be an infinite union but, since the set of indices is N, it' a countable union. - howerver at 23'19, you said exactly the contrary: you said that the open interval (in red) can not be constructed on an countable basis. So, where is the mistake ?
I think the earlier statement is correct. As for the latter one, we can let "b" in the set of rational number (a countable base), then any open set (l,a) can be reached by a limit of [b,a). However, for open sets like [c,a) with c an irrational number, then [c,a) cannot be reached by union of [b,a) (only (c,a) can be reached in this way). Even if you change the base to be [cb, a) to cover [c,a), you can always find another number (for example c*pi) that cannot be reached. Hence, TLL is not second countable.
i have a question about the cofinite topology countability. you said it's not 2-c because we can't build a set in Tcf using a finite number of open sets. if i take a set in which point A is omitted, i can construct it by the union of 2 sets omitting the same point A from each of them and a second point C from set 1 and point B from set 2. what's wrong with this? And i can't think of any set in this topology that can't be built following the same logic.
Great question! I had to think about this a bit, but I think I know the issue. First, the co-finite topology is not first countable, and it can be shown that ALL second countable spaces are first countable. It is common in topology for interesting properties to NOT be inherited, but in this case first- countability is inherited from second-countability. But to address your thought experiment: Pick a nominee countable base for the co-finite topology of a line segment. Can you find an open set of the co-finite line segment that can NOT be constructed from that nominee basis? I think you can! That is, it is not enough to build a countable basis that can construct a given an open set. You need to create a countable basis that can build ANY open set! Let me know if that helps.
At about 11:24 I state the rule when I said (paraphrase) "It is not second countable because we can not find a countable base that can make *any* open set." But you are right to think about it. The distinction between your thought-test and the definition is important.
+XylyXylyX So i must find one base set that can build any open set. in the example i gave i have to choose another base set specific to each open set and that's why it doesn't match the rule. But this raises another question for me. let's consider the case of euclidian topology, if i choose a base set i can always find a smaller set so i have to readjust the size of my initial set base accordingly to be able to construct the given set so i can't find 1 unique base set that can construct any set. please help!!
+John Ron I'm not absolutely sure I understand you new question, but maybe this will help: The standard base of the Euclidean topology is the set of open balls. We can make a smaller, countable, base by considering only the open balls who's centers lie at rational coordinates and with diameters that are rational numbers. From that countable base you have enough open sets to make any open set via the union operation of members of the base. The base of all "rational" open balls is unique, and countable. In order for a topological space to be second countable, you only need to find ONE countable base. Not every element of the base will contribute to the construction of any given open set. I think that gets at what you are struggling with?
BINGO! the phrase "the base of all rational open balls is unique" ended my struggle! i thought we should pick a base of rational numbers with precised diameter for it to be unique but now it's crystal clear! Sorry for being pain in the ass but i just need one more clarification about the language plz. when u were talking about differentiable manifold u said the transition function belongs to C(infinity) which means it can be "derived" infinite times. And when u introduced velocity u said the function f which goes from the manifold to R belongs to C(infinity). But f can't have derivatives so how it is a C(infinity)? what am i missing?
Doesn't the logic for the lower limit topology also apply to the euclidean topology, in the sense that in order to make the base of the euclidean topology countable you have to restrict the bounds of the intervalls to rational number. Now if you want to construct a open set whose bounds are real , you have to use the same limit process as in the lower limit topology ?
I am a little confused regarding why the lower limit topology is not second countable. For the example you provided around minute 15, couldn't I similarly argue that the Euclidean topology number line is NOT second countable because I cannot construct the following arbitrary open set: 0(----------------*(---------)*-----------------)1 using this following basis: 0(---*(-)*---------------------------------------)1 What I mean by this is...aren't you just choosing some arbitrary basis that prevents you from forming the set? -thanks
No for lower limit topology we can create that open intervel. For example if we take all [bn,an) such that bn and an are rational numbers. But I think wat we can't create from this is [b,a) in which b is an irrational number and a is rational or irrational. So this is also not the basis set.
Really great videos which explain the intuition for topology, but for me it is a bit frustrating that you also don't give the "rigorous" definition of some of the concept like the first and second countable or the topological spaces etc. I know you said in the first video that you are going to give "semi-rigorous" definitions, but it is at least for me to "follow" the concepts. I am speaking for myself only, and this is only my opinion, anyway keep up the good work that you do. And if I could ask, what book do you follow, or you would recommend?
Yeah, that is the real balance when making these videos. I am counting on the fact that viewers are either reviewing things they already know or supplementing rigorous study with the lectures. Sometimes I make these videos, ironically, in order to provide some sort of "rigor" to physics students who get a cursory exposure to the deeper math, but in mathematical terms, even my most "rigorous" exposition is hand-waving.
This is an old video, but I'm a bit confused by your proof that the lower-limit topology isn't 2nd countable... Why are you allowed to order your lower limits? For example, the positive rational numbers are countable, but I don't think it's possible to produce an *ordered* counting: in between 0 and any rational number is going to be more rational numbers. Any countable list of rational numbers relies on tricks where you end up zig-zagging between large and fine structures, and you'll *eventually* get to that rational between the two... I'm sure the conclusion is right, I just think it requires some more delicate gloves than what you're using here, unless I'm missing something.
It has been a while. I suspect the answer to your question lies in the fact that any infinite and bounded sequence has a convergent subsequence. I am not sure if ordering is as important as that the subsequence converges. I don't have time now to dig back into this, but I would like to. Please understand that my lectures are a *survey* of the subject. You are obviously interested in the many many important details and clearly you will have to engage in a far more formal study, which it seems you are doing. Let me know if you find your answer! Thank you for watching and good luck with your studies.
@@XylyXylyX Thanks for replying! and I totally understand on the survey bit, I'm mostly reviewing since it's been a few years since I finished my masters thesis on this stuff and I don't want it all to fade :P
Regarding second countability for the plane (or higher dimensions) with the usual topology, ISTM that this is an exceedingly difficult property to prove existent, even for simple shapes like circles and squares. Easy to show for the real line with usual topology, but not for higher dimensions.
Frank Bennett With topology, it is impossible to cite the complexity of any give demonstration for any given topological space because there are so many possible spaces. It is probably true that proofs so second countability may be hard for many many spaces, but some spaces will have easy proofs and it is probably not possible to categorize topological spaces based on difficulty of proofs. There is a terrific book called "Counterexamples in Topology" that is a must read if you want to get a feel for the wide range of possible topological spaces.
For a circle in the plane with usual topology, second countability is easy to show. Just take the union of concentric circles with rational radii converging to the radius of the original circle considered as open set in usual topology. Doesn't matter if the original radius rational or irrational. For anything with a more complex shapes, not easy to prove IMO.
One issue with the above comment re: a circular open set in plane with usual topology. If the radius is irrational, one should be able to define a set of rational radii converging to the length of the irrational radius, because the rationals are dense in the reals. But I don't see how to do this for an arbitrary irrational number (representing the length of the original radius) because there is no way to WRITE or express that irrational number. What's the solution? TIA.
Frank Bennett it only need be possible in principal. That is, say the radius is the irrational number x. We know we can construct a sequence of rational numbers that converge to x. That is all that is required in the abstract universe of topology.
The lower limit topology is not second-countable (like you said), but the proof you presented is incorrect. For the countable basis you use, `[a_n, b_m)`, it's not true that you can reorder them such that `a_{n+1} < a_n`; as a counterexample, take `a_n` to be the nth rational number in (0, 1) (which we can do because the rationals are countable), no such order exists. Indeed if you use `a_n` = the nth rational number in (0, 1) and `b_m` = the mth rational number in (0, 1) you can use this set to construct all (l, r) where (l \in R, r \in R), using basically the same construction that proves that the euclidean line is second-countable. What fails is that you can't use this set to construct [l, r), hence this set is not a basis.
I don't know how you will take it .... well to some degree I do you, want to make knowledge available free to all ........... but your videos should be paid contents on Udemy or coursera etc. This is beyond university classroom level stuff.
"First countable" does not only require that each point can be provided with a countable set of nested neighborhoods, as stated at 3:40; it requires in addition that these nested neighborhoods approach the point more than any arbitrary neighborhood of it; en.wikipedia.org/wiki/First-countable_space
Hi! At 36:06, you discuss the continuity of the function f(x) = x from the euclidean topology to the closed interval topology, and at 36:56 you explain it cannot be continuous because the half-open interval in T_CI (inversely) maps to a half-open interval in T_E.
But wouldn't the preimage of that closed interval in T_CI actually exclude the point 0 in T_E, thus making it an open set in T_E?
I agree. 0 and 1 have empty preimages. All open sets in T_CI have open pre-images.
32:18 Yes, I think it's pretty easy to understand how if each open set of Y maps back to an open set of X, then any point x in X must have a neighborhood that's a subset of any neighborhood of y in Y.
But what's not so clear to me is how we can assume the converse of this relationship must be true. That is, how can we show that if every point x has a neighborhood that's a subset of any neighborhood of y, then it necessarily follows that all of Y's open sets must map backwards to open sets of X?
Thanks for making these videos.
Maybe the second definition (which seems to me to be the definition of continuity at a point: en.wikipedia.org/wiki/Continuous_function#Continuity_at_a_point) is not correctly stated in the video (at 31:02 it should continue: "and so for any open subset of V"). So, let Ux be some open neighborhood of x in J_x and Vy some open neighborhood of y in J_y; then the second definition should state that for any y in Y, for ANY open subset Sy of Vy there is Ux in J_x that maps to Sy; if you consider that Vy itself is one such Sy, you'll conclude that there is some Ux that maps to Vy; this amounts to the first definition.
Around 16:14: You can build any open set (a,b) say you take [b_n, b_n + 1/m) so that the b_n's enumerate the rationals. The problem is you can't get all "open" sets of the form [a,b). In the case just suggested you can get (sqrt(2), 2), but you can't get [sqrt(2), 2).
Totally agree with you. This part was really confusing, and as a matter of fact, his explanation is wrong. It is not second-countable because the number of [b, a) intervals is aleph1, and you can't construct "[b," part with an infinite union you have to have it in your base, hence your base can't be countable.
at 8:44 how is the nested interval second countable? I can, say specify an interval from 0 to 1/4... it can be constructed from the basis? it seems that the I can use the first element in the base to cover the above interval.... I don't see how the second countability even come into the picture? I am a bit confused...
The interval you specified is not an open set! Second Countability involves making any OPEN set from the union of a countable number of open sets. The nested interval topology is particularly simple in the regard: the entire topology is countable and the topology is its own base!
This indeed wasn't clear. I hope this is an example, not the central part of manifold theory...
Wait wat is the difference between a nested set and an subset? At 11:15 the second set is a subset of the first open set.
at 16:28 you explained with the concept of completeness that T_LL is not C-2, but I doubt if it is the same for T_E? if we consider b_i of an interval I_b, where i is an element of Q, we can alway find a gap in between two intervals(because not complete). Wouldn't that imply there exists an interval that is not fully retrievable with the bases of a given collection? we'd be able to fill up all the holes if the base element I_b is an uncountable collection.
wait, please confirm with me if the following assumption is true.
if we were to consider an interval I, we can find inf{S}=x. Then if we say inf{S}∉S, it is an open set.
we can define x'=x+e where e->0. we can always find some rational number p/q such that p/q
The proof at 23:00, why doesn't it work with the usual topology of open intervals?
One short question around 29:34. if we have the pre-image of an open set in X under the inverse f^-1 then the function is continuous. Q)Should that the pre-image cover all X? If we take an arbitrary union of all pre-image on X mapped by f^-1, then should the arbitrary union has to cover X or is it ok to cover a subset of X? I guess if homeomorphic, it should cover all X and vice versa but not sure if that is one special case.
There is no requirement that this process can cover X with open sets however. For example, f might not be surjective but it still can be continuous. Homeomorphisms are examples of continuous functions that do indeed cover both the domain and the range of the mapping.
Thx. So can I wrap up this way, if we can guarantee that f^-1(u) exists for all u in Y, we know that at least some subset of X generates all output value in Y, which satisfies the definition of function (1 input corresponds to 1 output)
I really appreciate you for all your help:)
Jay Lee I would say the inverse only needs to map those open sets in Y which are in the range of f. In order to have an inverse f must be 1-1 from its domain into its range. Neither the domain nor the range need cover X or Y.
I learn a lot from you. Have a great day!
At 42:00, why are you taking in the open interval with green boundaries? While the open intervals, basis in cofinite topology are one to one in euclidean topology?
It has been a while since I reviewed this, and I am not quite sure what you are asking,,,, but I agree that the confinite topology is 1-1 with the Euclidean topology, but the Euclidean topology has MORE open sets. That is, it is *finer*. So the mapping is invective but not surjective (not ‘onto”). Does this help?
you are calling arbitrarily chosen open interval in cofinite topology as an open interval. I guess that open intervals in cofinite topology are subsets left after omitting the points.
Tursinbay Oteev I shouldn’t be, if that is what you are hearing. The cofinite topology’s open sets are what is left after removing a finite number of points.
I'm not sure the explanation at 15:30 (of why the LL topology is not 2C) is correct. We can choose our base as all [a, b) with a, b rational. This base can be used to construct any open interval (in a similar way to how the open intervals can be used in the Euclidean topology): choose [a1, b), [a2, b), ... where the a_n converge to a. What this (or any countable base) cannot do, however, is generate all the half-open intervals. See also math.stackexchange.com/questions/1135993/mathbbr-with-the-lower-limit-topology-is-not-second-countable
As before, thanks for the excellent series of videos! I'm finding them very instructive.
I’m sold! Of course you are correct! I love the proof at the link you provided! I wish I could add annotations to the video! I think what I will do is put your observation in the comment section! Thank you for paying such close attention and your kind remarks!
@@XylyXylyX My pleasure! I find that unless I try to work through the proofs myself, I quickly forget the material.
For anyone reading, the basic idea is this: if you want to converge to a half-open interval [x, a), you must include at least one set whose minimum is x. This is unlike the Euclidean case, where none of the sets converging to (x, a) needs to include x. Since there are uncountably many such x, you can't do it with a countable base.
There's one thing I'm unsure of in this proof: proofwiki.org/wiki/Sorgenfrey_Line_is_not_Second-Countable. It's true that Bx must have _infimum_ of x, but it seems to me that it must also be the _minimum_ . Not sure why they wouldn't mention this stronger condition, even if the weaker one will do.
Aditya Prasad I also noticed that this proof used a theorem about separability that I did not cover. I should do a little mini correction lecture about this. I see your point about infimum/minimum. I will try to get to the bottom of that too.
This is extremely useful! I was also confused about the proof in the video about 2nd countability of LL topology. I hope I can vote this comment up and let others see this.
as said from others your original proof about LL topology is not 2-C actually is not correct; we need to take in account half-open intervals instead: from [a,b) with a,b rationals we are able to build a succession of rationals having as limit a given real number however we cannot "reach" it, hence we cannot build any [a,b) with a,b real numbers.
Thank you for your detour through Manifoldland. It's very helpul to have this helicopter view of how pure mathematicians set the concepts of continuity, compactenss, etc on a solid set-theoretical basis. But I have one question about an issue you only skirted in your presentation. You showed us how "continuity "works only in the case of f(x)=x. The question is whether our pure mathematician friends could not dream up of some function that would preserve continuity in some of the cases you discussed. Or does it suffice to show that if it doesn't work for for the identity function it's not continuous?
Continuity is a property of *functions* and some functions will be continuous and others won't. Indeed, some functions do exist between these spaces that are continuous. In those cases such functions might be somewhat contrived, but they exist. I chose f(X)=X because it was easy to demonstrate. Continuity of a function is not the same as homeomorphism between two different topological spaces. In that case I must either find *any* function to prove two spaces are homeomorphic, or prove that no such function exists. That is, if I suspect two spaces are homeomorphic but I have not found the correct mapping, I need to worry that some other mathematician may find it. Not so for continuity, which is easier to check function-by-function.
Take f(X) = constant (ie a single point). This function is always continuous regardless of the topologies involved as the inverse image is either the empty set or X.
I'm a bit confused on the Lower Limit topology and countability. From 12:00 and 17:00, if open intervals are also part of the topology, then can't they be their own bases, instead of requiring infinite (uncountable?) numbers of half-open intervals as bases? Wouldn't this make the Lower Limit topology second countable? I think my understanding of bases might be wrong here ....
Each time someone asks about this material I have to remind myself of the fundamental definitions! You are definitely thinking about this thoroughly and well, which is the key to ultimately getting all to make sense.
First, you noted without realizing it maybe, that the LL topology is *finer* than the standard topology: it has more open sets. Not only does it have the half open interval, but it also has the open intervals too, and all the associated unions and intersections. All of these open sets can be constructed from the basis of half-open intervals.
A basis MUST be able to make *all* open sets via union and finite intersection. The question is can we identify a *countable* basis? That is, can we find a *countable* number of half open intervals that can make *any* open set in the topology. It is not good enough to find a countable collection of half-open intervals to build a particular open set, it must be one countable set that will build EVERY open set in the topology. The argument that this can not be done is a bit hard to see at first, but upon reflection it slowly becomes “obvious.” Say we have found such a countable basis and call that countable basis “C”. Then that basis can build the open set [x, x+a), which is obviously open. But note that [x, x+a) *may not be in our countable collection* so it is NOT in C, but it can be made with a countable number of members of C. This means that C must contain a collection of half-open intervals that can be “unioned” together with a lower limit of x. THat is not hard to do for any given x, but .... Then take another open set [y, y+a). Now it must be that C contains a collection of half-open intervals with a lower limit of y. Repeat this argument over and over and you will see that C has the impossible job of having to have a collection of half-open intervals that limit to every possible real number. This means it must have a cardinality equal to the real numbers, which is uncountable!
@@XylyXylyX Thanks for your explanation, it will take me a while to digest ... Anyway I genuinely and pleasantly surprised that you still remember what you did almost 3 years back and are still answering questions about them. Thanks to the recent black hole hooha, I am doing both your Tensor and Manifold series concurrently so that I can get to your General Relativity series, and hopefully Lie Groups and Algebras after that. It's very hard going, especially the Manifold ones, but your videos are truly helpful and enjoyable. Thanks again and keep up the good work!
ngozumpa The more I think about your question, the better it is. A full understanding of why the LL topology is second countable is a great exercise to understand the concept.
couldnt you create a nested seies of open sets in the cofinite topology just by choosing 1 point, then 1 more and 1 more and so on? the nth open set would be determined by the 1st n choices of points. each consecutive open set would be contained in its predecessor.
I dont think this would be an open set?
I was trying to use the "neighborhood" method of determining continuity in the example you provided for the identity function from the euclidean topology to the lower limit topology. Any open neighborhood V on the LL topology looks like [a, b) or (a, b). Going back to the euclidean topology, we can always find a neighborhood U (a+delta, b-delta), whose image is a subset of V. So for every V in the LL topology, we can always find an open neighborhood U in the euclidean topology that maps to a subset of V. Thus the identity function is continuous on the LL topology.
I must be missing something, as clearly from your example at 39:00 the identity function is not continuous on the LL topology, but I don't see what I'm doing wrong.
I see, thanks. I had misunderstood, not understanding that the n-hood U in X had to be an n-hood of x, the original point.
At 14:00, if your sequence approaching L corresponds to half open intervals, closed on left side, how can the open set resulting from the limiting process be *open* on both sides? ISTM, that the result would be half open, closed on left side.
Soon thereafter you claim that LL topology not second countable because you found a nested countable sequence which skips over some point P. But the definition of second countable just demands that some countable union exists which contains every open set; not that every such countable union must contain some open set.
Frank Bennett For your first question I will respond "Socratic-ally": If the left side is closed, then what is the lowest member of the set? That is, there should be a highest lower bound to the infinite union that is also a member of the set. If you think about this, you will realize there is no such point!
I see the answer to my second question. The definition requires that some countable base will cover every open set. But there is no such base for the LL topology.
Frank Bennett The definition of second countability is that the topology has a countable basis. My little informal demonstration showed that any nominee countable basis will not be able to construct arbitrary open sets. I probably should had done a more rigorous proof of this point, however.
In discussing the 1-C of Cofinite topology, starting from the original open set that contains P but has 2 more points excluded, if we continue to exclude more points, then the new open sets are subsets of the old, thus forming a nested series. Where am I wrong?
proofwiki.org/wiki/Uncountable_Finite_Complement_Space_is_not_First-Countable
is a proof of this point. However, you are correct in noting that *my* hand-waving proof was not good enuf! Good observation! Thanks.
You don't get nested intervals because you can only omit a countable subset from your original interval and the set difference of nested intervals is more than countable
Could you please explain (around 44:30), why is a closed interval OPEN in the discrete topology?
Thomas Impelluso THe discrete topology asserts that every subset of X is a member of the topology of X. If X is the real like, say, then any closed interval is a subset of X and is therefore a member of the discrete topology. "Closed" has two meanings here. 1) a closed set is the complement of an open set and 2) a closed intervals contains its end points. "Closed intervals" are "closed sets" if their complements are open. In the discrete topology the complement of ANY set is still a subset of X and therefore a member of the topology and therefore "open".
I am not sure I am convinced that the justification around 11:00 can be used to demonstrate that the cofinite topology is not first-countable. The claim, as I understand it, is that given an open set S_0 that contains the point A, but excludes points B, and C, there is no open set S_1 that is nested (a proper subset of S_0) which also contains A. But isn't the open set which excludes points B, C, and D a proper subset of S_0 which contains the point A?
To demonstrate that the cofinite topology is not first-countable, I believe you must show that for any finite sequence S_0, S_1, S_2 ... of nested neighborhoods of A, there exists a neighborhood of A which contains none of the neighborhoods in the sequence.
Or am I missing something here?
This bothered me also. Here what Wolfram had to say via ChatGPT: Yes, the cofinite topology on an open interval of the real line is first countable.
In the cofinite topology, the open sets are defined as either the entire space itself or any set whose complement (relative to the space) is finite. This means that the only closed sets are the whole space and finite sets.
To demonstrate that this space is first countable, we need to show that each point in the space has a countable local basis. For any point \( x \) in our open interval (let's denote this interval as \( I \)), we can consider the collection of all open sets that contain \( x \) but exclude only a finite number of points. Specifically, for each finite subset \( F \) of \( I \) that does not contain \( x \), we can define an open set \( U_F = I \setminus F \) that contains \( x \) and whose complement is \( F \), which is finite by construction.
This collection of sets \( \{ U_F \} \) forms a local basis at \( x \). Since there are countably many finite subsets of \( I \) (even though \( I \) itself may be uncountable), this local basis is countable. Therefore, the cofinite topology on an open interval of the real line is first countable. This property holds regardless of the specific open interval of the real line considered, as long as we're working within the cofinite topology.
The set of all integers is not called I but Z and I think this is a universal convention.
I want to thank you for this lecture series. May I ask, what your background is? How do you know so much maths and physics?
Kind regards
Hi, I do not understand your explanation about 2nd countability of half-open intervals. I do not see the reason why I could not construct any open interval. But I think all half-open intervals cannot be created because of the union of [a_n,b) =(a,b), where a_n is decreasing sequence.
I think this is explained at 12:07. I belive the point is that you have to be able to choose a single countable basis that creates all the open sets simultaneously. But for any single countable basis you can always choose an open set (c,d) which is an element of the lower limit topology where c is say between b_n and b_n+1. You can't create this set unless you pick a new basis.
@@stber321 Same problem for the closed interval topology. If your argument is correct, not all random open set (a, b) can be constructed by a chosen countable base of [-1, bn) and (an, 1]
6:27 When you say second-countability implies first-countability, you're excluding the discrete topology?
No. Should I? If a space is second countable it has a countable base. If it has a countable base, then it has a countable base at each point. Second C -> First C.
@@XylyXylyX Sorry for the confusion.
I think in the definition of a homeomorphism you forgot to mention that f has to be a bijection. However, I guess it can be automatically implied depending on how you define what a function is.
incredible videos!!!!!
Why the lower limit topology is not countable is not at all clear. Can you find a countable basis for the euclidean topology?
why we always talk in term of open sets, not closed ones when we talk about topology?
It has become traditional. There is a way of inverting all the definitions so that closed sets take center stage. I do not know why the subject has leaned on open set, but Ihave seen all the reversed definitions. It often comes as an exercise, btw. For example: define a topology in terms of closed sets.
Hi. At first, thank you for those very interesting lectures.
But there is something I'm not sure to well understand:
- at 18'30, you sais that any open interval can be built as the limit of the union of baie elemnt of JLL. It coul be an infinite union but, since the set of indices is N, it' a countable union.
- howerver at 23'19, you said exactly the contrary: you said that the open interval (in red) can not be constructed on an countable basis.
So, where is the mistake ?
I think the earlier statement is correct. As for the latter one, we can let "b" in the set of rational number (a countable base), then any open set (l,a) can be reached by a limit of [b,a). However, for open sets like [c,a) with c an irrational number, then [c,a) cannot be reached by union of [b,a) (only (c,a) can be reached in this way). Even if you change the base to be [cb, a) to cover [c,a), you can always find another number (for example c*pi) that cannot be reached. Hence, TLL is not second countable.
i have a question about the cofinite topology countability.
you said it's not 2-c because we can't build a set in Tcf using a finite number of open sets.
if i take a set in which point A is omitted, i can construct it by the union of 2 sets omitting the same point A from each of them and a second point C from set 1 and point B from set 2.
what's wrong with this?
And i can't think of any set in this topology that can't be built following the same logic.
Great question! I had to think about this a bit, but I think I know the issue. First, the co-finite topology is not first countable, and it can be shown that ALL second countable spaces are first countable. It is common in topology for interesting properties to NOT be inherited, but in this case first- countability is inherited from second-countability. But to address your thought experiment: Pick a nominee countable base for the co-finite topology of a line segment. Can you find an open set of the co-finite line segment that can NOT be constructed from that nominee basis? I think you can! That is, it is not enough to build a countable basis that can construct a given an open set. You need to create a countable basis that can build ANY open set! Let me know if that helps.
At about 11:24 I state the rule when I said (paraphrase) "It is not second countable because we can not find a countable base that can make *any* open set." But you are right to think about it. The distinction between your thought-test and the definition is important.
+XylyXylyX So i must find one base set that can build any open set.
in the example i gave i have to choose another base set specific to each open set and that's why it doesn't match the rule.
But this raises another question for me.
let's consider the case of euclidian topology, if i choose a base set i can always find a smaller set so i have to readjust the size of my initial set base accordingly to be able to construct the given set so i can't find 1 unique base set that can construct any set. please help!!
+John Ron I'm not absolutely sure I understand you new question, but maybe this will help: The standard base of the Euclidean topology is the set of open balls. We can make a smaller, countable, base by considering only the open balls who's centers lie at rational coordinates and with diameters that are rational numbers. From that countable base you have enough open sets to make any open set via the union operation of members of the base. The base of all "rational" open balls is unique, and countable. In order for a topological space to be second countable, you only need to find ONE countable base. Not every element of the base will contribute to the construction of any given open set. I think that gets at what you are struggling with?
BINGO! the phrase "the base of all rational open balls is unique" ended my struggle! i thought we should pick a base of rational numbers with precised diameter for it to be unique but now it's crystal clear!
Sorry for being pain in the ass but i just need one more clarification about the language plz. when u were talking about differentiable manifold u said the transition function belongs to C(infinity) which means it can be "derived" infinite times. And when u introduced velocity u said the function f which goes from the manifold to R belongs to C(infinity). But f can't have derivatives so how it is a C(infinity)? what am i missing?
Doesn't the logic for the lower limit topology also apply to the euclidean topology, in the sense that in order to make the base of the euclidean topology countable you have to restrict the bounds of the intervalls to rational number. Now if you want to construct a open set whose bounds are real , you have to use the same limit process as in the lower limit topology ?
P. Zhao yes! That is correct.
Doesn't that mean that the Euclidean topology is not 2C?
I am a little confused regarding why the lower limit topology is not second countable. For the example you provided around minute 15, couldn't I similarly argue that the Euclidean topology number line is NOT second countable because I cannot construct the following arbitrary open set:
0(----------------*(---------)*-----------------)1
using this following basis:
0(---*(-)*---------------------------------------)1
What I mean by this is...aren't you just choosing some arbitrary basis that prevents you from forming the set?
-thanks
No for lower limit topology we can create that open intervel. For example if we take all [bn,an) such that bn and an are rational numbers. But I think wat we can't create from this is [b,a) in which b is an irrational number and a is rational or irrational. So this is also not the basis set.
Really great videos which explain the intuition for topology, but for me it is a bit frustrating that you also don't give the "rigorous" definition of some of the concept like the first and second countable or the topological spaces etc. I know you said in the first video that you are going to give "semi-rigorous" definitions, but it is at least for me to "follow" the concepts. I am speaking for myself only, and this is only my opinion, anyway keep up the good work that you do.
And if I could ask, what book do you follow, or you would recommend?
Yeah, that is the real balance when making these videos. I am counting on the fact that viewers are either reviewing things they already know or supplementing rigorous study with the lectures. Sometimes I make these videos, ironically, in order to provide some sort of "rigor" to physics students who get a cursory exposure to the deeper math, but in mathematical terms, even my most "rigorous" exposition is hand-waving.
This is an old video, but I'm a bit confused by your proof that the lower-limit topology isn't 2nd countable... Why are you allowed to order your lower limits? For example, the positive rational numbers are countable, but I don't think it's possible to produce an *ordered* counting: in between 0 and any rational number is going to be more rational numbers. Any countable list of rational numbers relies on tricks where you end up zig-zagging between large and fine structures, and you'll *eventually* get to that rational between the two...
I'm sure the conclusion is right, I just think it requires some more delicate gloves than what you're using here, unless I'm missing something.
It has been a while. I suspect the answer to your question lies in the fact that any infinite and bounded sequence has a convergent subsequence. I am not sure if ordering is as important as that the subsequence converges. I don't have time now to dig back into this, but I would like to. Please understand that my lectures are a *survey* of the subject. You are obviously interested in the many many important details and clearly you will have to engage in a far more formal study, which it seems you are doing. Let me know if you find your answer! Thank you for watching and good luck with your studies.
@@XylyXylyX Thanks for replying! and I totally understand on the survey bit, I'm mostly reviewing since it's been a few years since I finished my masters thesis on this stuff and I don't want it all to fade :P
@@QRebound Learning, forgeting, re-learning, re-forgetting, re-relearning, re-reforgeting..... sigh.... that is why I want to be a robot.
Regarding second countability for the plane (or higher dimensions) with the usual topology, ISTM that this is an exceedingly difficult property to prove existent, even for simple shapes like circles and squares. Easy to show for the real line with usual topology, but not for higher dimensions.
Frank Bennett With topology, it is impossible to cite the complexity of any give demonstration for any given topological space because there are so many possible spaces. It is probably true that proofs so second countability may be hard for many many spaces, but some spaces will have easy proofs and it is probably not possible to categorize topological spaces based on difficulty of proofs. There is a terrific book called "Counterexamples in Topology" that is a must read if you want to get a feel for the wide range of possible topological spaces.
For a circle in the plane with usual topology, second countability is easy to show. Just take the union of concentric circles with rational radii converging to the radius of the original circle considered as open set in usual topology. Doesn't matter if the original radius rational or irrational. For anything with a more complex shapes, not easy to prove IMO.
One issue with the above comment re: a circular open set in plane with usual topology. If the radius is irrational, one should be able to define a set of rational radii converging to the length of the irrational radius, because the rationals are dense in the reals. But I don't see how to do this for an arbitrary irrational number (representing the length of the original radius) because there is no way to WRITE or express that irrational number. What's the solution? TIA.
Frank Bennett it only need be possible in principal. That is, say the radius is the irrational number x. We know we can construct a sequence of rational numbers that converge to x. That is all that is required in the abstract universe of topology.
Opps replace "can construct" with "there exists".... otherwise you will ask me to construct it! My point is for the proof you don't have to.
What is "countable"?
"Countable" means you can put the set in 1 to 1 correspondence with the natural numbers.
Also, a set is countable if you can put all the elements on a line, and "count" up to any of them in finite time.
eres un crack
My concern is that even if you have countably different basis, you cannot necessarily put their lower limit in an order b_{n+1}
The lower limit topology is not second-countable (like you said), but the proof you presented is incorrect.
For the countable basis you use, `[a_n, b_m)`, it's not true that you can reorder them such that `a_{n+1} < a_n`; as a counterexample, take `a_n` to be the nth rational number in (0, 1) (which we can do because the rationals are countable), no such order exists.
Indeed if you use `a_n` = the nth rational number in (0, 1) and `b_m` = the mth rational number in (0, 1) you can use this set to construct all (l, r) where (l \in R, r \in R), using basically the same construction that proves that the euclidean line is second-countable. What fails is that you can't use this set to construct [l, r), hence this set is not a basis.
Thank you for catching this. I will revisit it as soon as I can!
Holey balls!
I don't know how you will take it .... well to some degree I do you, want to make knowledge available free to all ........... but your videos should be paid contents on Udemy or coursera etc. This is beyond university classroom level stuff.
Let the content creators decide man. Watch or don't that is your choice.