Thank you, Dr., for opening the closed notations on the part of the topology (open sets). Is the Borel sigma-algebra also related to proof 3 at 16:10? Kindly acknowledge.
At 11:50 you said logically that the intersection of the infinitely many open sets reduce to the zero set , which is closed. At the same time at the beginning at 9:30 you claimed without proof (axiom?) that the Empty set denoted as Zero and S are open. I am confused at this point...can you please help me see where my confusion comes from and how it can be remedied?
Concerning the ending Ex2: Doesn't Q have nothing but interior points if you consider with it a topology defined only within rational numbers? I mean if you consider Q with the topology induced by the Euclidean metric of the reals, as I take was done here, then surely Q will have no interior points, it all depends on which topology you choose to use.
I had the same question about Interior Point Example 2. If we limit ourselves to considering only rationals then any ball B around x∈Q° would contain all the elements of Q within the bounds of B. Are interior points strictly defined to be Reals?
@@Sahanie Generally surely no, but I believe the topology over the reals induced by the Euclidean metric was implicitly assumed in the second example at the end. Considering the discrete topology (power set) over just the rational numbers would make each rational number an interior point.
Hmm this is really interesting. One can clearly see that the proof of 3 for a finite intersection of open sets is not valid for infinite amount of intersections. 18:21 . This is because there would be a infinite amount of r's (r1,r2,...) so instead of a "min" one must use a "inf". Of course, an infimum of a set of positive real numbers can be zero so the proof is not valid because you are using a strict inequality. However, simply saying that a proof does not work for a specific statement is not a proof in its own right. For this, you can consider the counterexample which you gave at around 11:13. Hopefully whatever I said made sense. Thank you Dr. P for the clear explanations as always.
I had to think a bunch about what happens with metric spaces where there's a lower bound on non-zero distances from some points, or, more generally, where there are distances that don't occur. I think you get open balls that are the same as closed balls and individual points being open, which is weird but doesn't cause any problems.
Yes, no problem: look up "discrete topology". Another useful one for your topology toolkit is the "trivial topology". Both are good for testing proposed theorems and finding counter examples. There is a definition of what makes something a metric though and that definition garantees that you get a proper topology out of it. But not every definition of "distance" is a metric.
Okay, I'm not this good at math but I'm going to give this a try. If x is an element of (a,b) and I want to know if I have an open set. Then I define the limits first of (a,b) say, a=1, b=100, x1=10, can I lead myself to believe that A≤x≤B is defining an open set, and AB or A>x
The thing i understood is that boundary point in open sets are not achieved that' s why the small ball can't cross the subset without breaking continuity ... ball is not defined on boundary of subset that why its open subsets ... am i in right track or not if not please tell me where do i wrong
Even technically you are correct. The "boundary" is defined as the closure (smallest containing closed set) minus the interior (there are other equivalent ways to define it as well). See e.g. en.wikipedia.org/wiki/Boundary_(topology). But a set is open if and only if it is equal to its interior. So if a set does not contain any of its boundary points, it is no larger than its interior and that means that it is equal to its interior and therefore open. And if it is open then it is equal to its interior, which is excluded from its boundary, so it does not contain any of its boundary points. I guess that is exactly what you are saying. E.g. the closure of Q (as a subset of R) is R, so the boundary is R (and Q is not open in R). The closure of Q as a subset of Q is Q, so that the boundary is empty (Q\Q) (and Q is open in Q). Happy to be proven wrong and see a counter example.
Note that while this introduction video to open sets is understandably based on a metric (or norm if you like), that still defines a topology and the more abstract topological considerations and facts apply equally well to it.
@@alexfekken7599 If we consider the normal open-ball-topology induced by the Euclidean metric over the reals, then for the semi-open interval E = [0,1), its boundary {0,1} is not included in E but the set E is also not an open set with respect to the space's topology. EDIT: I think the answer to this depends on whether we mean "E does not include its (entire) boundary" or "E does not include any points that belong to its boundary"
Open sets just don’t make sense to me like how do you go outside of the set in form of a+ or- r and still be inside of that set ? Like if a if I am talking about the open set (9-13) how would 8.999999 still be part of it? (With r being 0.000001)
We’re saying there is an r such that this exists, not that for every r this is true. And your 8.999 argument assumes that 9 is in your set which it isn’t
@@amadouz6475 because no matter how much small 'r' u take like 0.00000001, 9-r is still outside this interval i think dr peyam got ur point its not like definition of limit we don't need every r>0 we just need to check if there exists such r I am also learning if I am wrong then please correct me
The points at most r away from x is a closed ball. The open ball around x is the points less than r away from x. Your symbolic statements were all correct. Your verbal ones were flawed.
Ah I wish it was said before, it would have put me out of my misery...I saw the contradiction between what was said at 9:30 and then 11:50 when Dr Peyam derives the converging zero set to be closed. Your comment makes it clear. Thanks!
Correct: you need to specify the enclosing set and its topology (just like you need to specify domains and codomains when you talk about functions). Q as a subset of itself with the usual topology is open (and closed). As a subset of R with the usual toplogy it is not open (nor is it closed). (0,1) as a subset of R with the usual toplogy is open. As a subset of C with the usual toplogy it is not open (nor is it closed).
@@Happy_Abe You define it that way. You act like S is the biggest set lets say, and you wanna know which subsets of S are open. This then coincides with the idea of openness where balls are always contained, if S is the biggest set, any ball is automatically in S so S is open.
@@Happy_Abe This S is the set that's part of the metric space you're working in. That is, if you're working in a metric space (S, d), that S is open in that metric space. On the other hand, any particular set is not open in some other metric space. For example, R is not open in the metric space (C, d(x,y)=|x-y|) (that is, the metric space you'd expect for the complex numbers), because 0 is in R, and, for any r>0, |(0+(r/2)i) - 0| < r but 0+(r/2)i is not in R.
Dr. P always starts with "Thanks for watching" and I wonder: How does he KNOW???? Freaky!
Really enjoyed this one, thanks for your very clear explanation of these foundations.
Thank you so much for this video. It is clear in every detail. Great way to introduce topology.
Such a treat to get a taste of topology on Thanksgiving day! :D Happy Thanksgiving!!
Thanks so much for clear graph explanation ! Really makes such abstract topic comes into sense !
I wish I had seen this video earlier. Good job, doc!
Thank you, Dr., for opening the closed notations on the part of the topology (open sets).
Is the Borel sigma-algebra also related to proof 3 at 16:10?
Kindly acknowledge.
21:10 interior design for topology
Oh yeah, now, we have a taste of powe... topology
At 11:50 you said logically that the intersection of the infinitely many open sets reduce to the zero set , which is closed. At the same time at the beginning at 9:30 you claimed without proof (axiom?) that the Empty set denoted as Zero and S are open. I am confused at this point...can you please help me see where my confusion comes from and how it can be remedied?
A set can be both open and closed, so the empty set is open but it’s also closed because it’s complement is S which is open
Thanks a lot for this amazing video
Again. Amazing
What do you think about bartle - elements of real analysis?
No context opening this at 0:55 is a riot
Concerning the ending Ex2: Doesn't Q have nothing but interior points if you consider with it a topology defined only within rational numbers? I mean if you consider Q with the topology induced by the Euclidean metric of the reals, as I take was done here, then surely Q will have no interior points, it all depends on which topology you choose to use.
I had the same question about Interior Point Example 2. If we limit ourselves to considering only rationals then any ball B around x∈Q° would contain all the elements of Q within the bounds of B. Are interior points strictly defined to be Reals?
@@Sahanie Generally surely no, but I believe the topology over the reals induced by the Euclidean metric was implicitly assumed in the second example at the end. Considering the discrete topology (power set) over just the rational numbers would make each rational number an interior point.
Thanks. It finally clicked with me.
Hmm this is really interesting. One can clearly see that the proof of 3 for a finite intersection of open sets is not valid for infinite amount of intersections. 18:21 . This is because there would be a infinite amount of r's (r1,r2,...) so instead of a "min" one must use a "inf". Of course, an infimum of a set of positive real numbers can be zero so the proof is not valid because you are using a strict inequality. However, simply saying that a proof does not work for a specific statement is not a proof in its own right. For this, you can consider the counterexample which you gave at around 11:13. Hopefully whatever I said made sense. Thank you Dr. P for the clear explanations as always.
Cool. Thank you very much.
I had to think a bunch about what happens with metric spaces where there's a lower bound on non-zero distances from some points, or, more generally, where there are distances that don't occur. I think you get open balls that are the same as closed balls and individual points being open, which is weird but doesn't cause any problems.
Yes, no problem: look up "discrete topology". Another useful one for your topology toolkit is the "trivial topology". Both are good for testing proposed theorems and finding counter examples.
There is a definition of what makes something a metric though and that definition garantees that you get a proper topology out of it. But not every definition of "distance" is a metric.
Question: 3:43 In that definition, would the 2nd IF be IFF?
Yes. this is true for every definition . 'If' is commonly used for 'Iff' in definitions
6:33 this could be fleshed out using triangle inequality!
Excellent!!!!!!
Okay, I'm not this good at math but I'm going to give this a try. If x is an element of (a,b) and I want to know if I have an open set. Then I define the limits first of (a,b) say, a=1, b=100, x1=10, can I lead myself to believe that A≤x≤B is defining an open set, and AB or A>x
A TASTE of Topology on Thanksgiving. I see what you did there!
Are there any good theorems about when a closed set is indeed not open?
Cool that you’re doing this as I’m getting started on Topology by Munkres!
The thing i understood is that boundary point in open sets are not achieved that' s why the small ball can't cross the subset without breaking continuity ... ball is not defined on boundary of subset that why its open subsets ... am i in right track or not if not please tell me where do i wrong
Can we say in layman’s terms that a set E is open if the “boundary” of E is not included in E?
Not true
@@drpeyam Thank you
Even technically you are correct. The "boundary" is defined as the closure (smallest containing closed set) minus the interior (there are other equivalent ways to define it as well). See e.g. en.wikipedia.org/wiki/Boundary_(topology).
But a set is open if and only if it is equal to its interior. So if a set does not contain any of its boundary points, it is no larger than its interior and that means that it is equal to its interior and therefore open. And if it is open then it is equal to its interior, which is excluded from its boundary, so it does not contain any of its boundary points. I guess that is exactly what you are saying.
E.g. the closure of Q (as a subset of R) is R, so the boundary is R (and Q is not open in R). The closure of Q as a subset of Q is Q, so that the boundary is empty (Q\Q) (and Q is open in Q).
Happy to be proven wrong and see a counter example.
Note that while this introduction video to open sets is understandably based on a metric (or norm if you like), that still defines a topology and the more abstract topological considerations and facts apply equally well to it.
@@alexfekken7599 Yes that is what I meant. Thank you very much for your explanation. My comment was pretty vague 😅
@@alexfekken7599 If we consider the normal open-ball-topology induced by the Euclidean metric over the reals, then for the semi-open interval E = [0,1), its boundary {0,1} is not included in E but the set E is also not an open set with respect to the space's topology.
EDIT: I think the answer to this depends on whether we mean "E does not include its (entire) boundary" or "E does not include any points that belong to its boundary"
Hello Dr Peyam, may you make a clip about Einstein's general theory of relativity in mathematic ? That will be very nice, thank you so much.
Oh no 🌚
At 9:30, what is S?
The metric space. Also what is the 33 in your name?
@@drpeyam It's my roll number in school 😆
Why zero set is not open?
It’s open
Open sets just don’t make sense to me like how do you go outside of the set in form of a+ or- r and still be inside of that set ? Like if a if I am talking about the open set (9-13) how would 8.999999 still be part of it? (With r being 0.000001)
We’re saying there is an r such that this exists, not that for every r this is true. And your 8.999 argument assumes that 9 is in your set which it isn’t
@@drpeyam thank you for your reply, but could you please explain to me why 9 wouldn’t be part of my set?
@@amadouz6475 because no matter how much small 'r' u take like 0.00000001, 9-r is still outside this interval
i think dr peyam got ur point its not like definition of limit we don't need every r>0 we just need to check if there exists such r
I am also learning if I am wrong then please correct me
The points at most r away from x is a closed ball. The open ball around x is the points less than r away from x. Your symbolic statements were all correct. Your verbal ones were flawed.
PLZ DO MORE TOPOLOGY MY TOPOLOGY COMPREHESIVE IS IN FEBRUARY
Check out my playlist
@@drpeyam chi Migi baba there’s not topology playlist
Metric spaces playlist
@@drpeyam I mean like on connectedness compactnesss path connected topological invariants..
Literally that playlist
Thanks soo much sir, l have understood(johnson omutoko)
I thought, that empty set is open set & close set, and the topolog space the same.
That's true but he hadn't defined closed sets yet so he's leaving it for another video
Ah I wish it was said before, it would have put me out of my misery...I saw the contradiction between what was said at 9:30 and then 11:50 when Dr Peyam derives the converging zero set to be closed. Your comment makes it clear. Thanks!
Isn't it the case that Q is an open set, just not an open subset of the real numbers?
Correct: you need to specify the enclosing set and its topology (just like you need to specify domains and codomains when you talk about functions).
Q as a subset of itself with the usual topology is open (and closed). As a subset of R with the usual toplogy it is not open (nor is it closed).
(0,1) as a subset of R with the usual toplogy is open. As a subset of C with the usual toplogy it is not open (nor is it closed).
You said empty said and S is open, did you mean R instead of S?
He means S as in a general set S. This is so that you can define topology for all kinda sets. But yes you could take S to be ℝ
@@helloitsme7553 why is S, a general set open?
@@Happy_Abe You define it that way. You act like S is the biggest set lets say, and you wanna know which subsets of S are open. This then coincides with the idea of openness where balls are always contained, if S is the biggest set, any ball is automatically in S so S is open.
@@Happy_Abe This S is the set that's part of the metric space you're working in. That is, if you're working in a metric space (S, d), that S is open in that metric space. On the other hand, any particular set is not open in some other metric space. For example, R is not open in the metric space (C, d(x,y)=|x-y|) (that is, the metric space you'd expect for the complex numbers), because 0 is in R, and, for any r>0, |(0+(r/2)i) - 0| < r but 0+(r/2)i is not in R.
But you defined the open sets in terms of its underlying metric which is indeed a special case since not all topological spaces are metrizable :c
Can you give an example where this "wiggle room" is useful in solving problems? Thanks for the great videos. I'll skip the 🏀 jokes
I hate topology so much ;( I don’t understand anything ;(
It’s an acquired taste, but check out the playlist