where limits aren't what they seem...
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That "small mistake" seems wrong. 1 and -1 are at a distance of 1, which is larger than 1/2, from 0 according to this f metric
Agreed!
Clearly d_f(1, 0) = | f(1) - f(0) | = | 1 - 0 | = 1, which is not < 1/2, and the same applies to d_f( -1, 0 ). Neither of those two x values are in the open disc. I suspect that whoever edited the video mistakenly thought that 1/2 < x < 3/2 could include x = 1 (and similarly for -3/2 < x < -1/2), but the piecewise definition excludes those.
The problem starts where he uses x>0 which includes numbers outside the domain of [-1,1]. He should have used x€ (0,1]. But then he has to break that into 2 cases because f(1) is not x-1, but rather x. Then would show the d(1)=1 which is not
@@bobh6728 He should have just split into cases according to the piecewise definition of f, namely x € (0,1), x € (-1,0), and x € {-1,0,1}
They're also expressible as quotients of each other by powers of themselves.
And concurrent derivatives of transformations of each other
I think it would have made sense to build the type of metric you're using for your limit into the limit notation, because with the definition at 17:00, there are two different definitions of the limit at work that give different results, but share the same notation. (The top limit uses the f metric, the bottom limit uses the regular absolute value of the difference of the inputs metric)
I agree. For those curious, the first condition in the orange bubble "limit a_n = L" is a statement about what happens in the *top* number line. To answer the question "What is the limit of the sequence a_n" you *have* to be a little more specific. A fully complete "answer" would be a sentence like "The sequence a_n converges to L in the metric space M." Here the top number line and bottom number line are 2 different metric spaces. In the top number line the "metric" is d_f(x,y) and in the bottom number line the metric is abs(x-y). The orange bubble is adding the needed details by saying that "limit a_n = L" *MEANS* "The sequence a_n converges to L in the FIRST metric space, where the metric is d_f." In other words the metric d_f compares a_n to L and gives back regular real numbers which approach 0, hence we say "lim a_n = L in the top metric space / using the metric d_f."
Indeed, and there ways to clarify this:
1. Write "d_f-lim" and "|.|-lim" (Euclidean)
2. Use the "-->" notation and write the metric d_f on top of the arrow
3. Just stick to "d(a_n, L) -> 0"
4. With words (like how @xane256 does it!)
5:53 = "a mathematician's favorite trick, which is to add zero": but when to add and what form of zero? Therein lies the real magic!
The way that I look at things like this is that you are encoding each real number in a unique way and consider the distance between two reals to be the distance between their encoded values.
Yeah, that's the way I look at it, you can do this for general metric spaces as well, so you would define d_f(a,b) = d(f(a),f(b)). Then since f is injective you have a bijection from your space X to the image f(X). And then you're essentially using that association
What's happening here is simply that we are identifying our set with a subspace of R, namely the image of the injective map f. Pulling back the metric by f then simply means that we inherit the metric (hence topology) of said subspace.
Exactly
can't wait for the topology class on MathMajor!!! been excited about it for like a year :D
20:55
Conjecture: if f is also continuous (in the usual sense of Euclidean topology) in addition to being injective, then the metric d_f generates the usual euclidian topology.
Yes, that is true. Since f is continuous any ball is open, which implies that d_f can induce at most the euclidean topology. Now since f has to be monotone as well one can show that balls can be arbitrarily small neighbourhoods of points, which is enough to show that d_f actually induces the euclidean topology.
So basically we’ve taken a metric space (Real numbers with absolute value) and we have “pushed it forward” through an injective function.
This reminds me a lot of the Pushforward Measure in Measure Theory, where one constructs a new measure, as the image of a function.
This brought back fond memories of topology class in grad and undergrad. I remember the opening idea. And he's right all the function has to be is injective and if I remember correctly measurable.
There are no measure spaces here so measurability has nothing to do with it. (you could assume the Lebesgue measure on *R* but measurability still doesn't come into play for d_f being a metric)
What if you had a space/metric where two different points are separated by zero distance? I feel like that could be useful somehow
There is. It's called a pseudo-metric space. However, I think limit and convergence cannot be defined uniquely. Ie, a sequence can converge to two or more limiting values. (This is due to the topology induced not being T2.) Actually, if d(x,y)=0, anything that has a limit value of x also has a limit value of y.
Yes, these are called pseudo-metrics and any pseudo-metric still induces a topology. Their pseudoness (non-positive-definiteness) can be expressed topologically: a pseudo-metric space is a metric space if and only if the induced topology is Hausdorff. Similarly, a pseudo-metrizable space is metrizable if and only if it is Hausdorff.
There is sense in which not the Hausdorff condition (T2) but rather Kolmogorov condition (T0) is more fundamental here, but T2 and T0 are equivalent for pseudo-metrizable spaces.
It is very very very useful in physics. The Minkowski spacetime is an example of this, and the whole theory of special relativity is grounded on it. The metric implies that the distance from any two points located in the worldline of a particle which travels exactly at the speed of light is zero, even if the points are different! This is one of the ways in which we can see that even particles with no mass can have linear momentum. Very neat stuff!
Actually, I got things a little mixed up. Ok, we can define a pseudometric in spacetime, no problem. It is indeed a consequence of the usual structure we impose on it. However, the general setting of relativity requires more structure, namely, a pseudo-Riemannian manifold with other specific properties. This is what enables the whole power of relativity. We can define a pseudometric from it. But, if we begin only with this pseudometric, there will not be enough structure to get the full story. So, I guess that a pseudometric by itself is enough in other settings, but not in relativity!
I think we can just use a function that "flips" 0 and -1, i.e. f(0) = -1, f(-1) = 0 and f(x)=x if x not in {0,-1}. In this case d(1/n, -1) = |f(1/n) - f(-1)| = |1/n - 0| = 1/n -> 0
Pretty sure the thing you mentioned at the end is an iff condition i.e. if you have 2 induced topologies that are different you can construct a sequence that has different limits in the 2 spaces, this because you could choose a sequence approaching a point that's not in the neighborhood for both topologies as a target, but I didn't verify that so could be wrong
What about opposite question, what if I have two metrics that have all limits go to same values, then its mean its same topology?
I'm surprised this actually satisfies the triangle inequality
If you have a set X, and a set Y equipped with a metric d_Y,
and an injective function
f : X -> Y
Well, because X is just a set without any extra structure on it,
you can think of f : X -> Y as if it it a way to see X as just a subset of Y, with some different labels.
Any we have a metric on Y.
So, we get a metric on X.
If you disregard the fact that Y (and X) in the example shown, are both the real numbers,
(because f isn’t required to respect any properties of the real numbers)
then this producing a metric on X, shouldn’t be much of a surprise, I think?
Maybe I’m just too used to things to see it as surprising?
But, I suspect the source of the feeling of surprise, is the expectation that f should respect the properties of the real numbers in some way,
rather than just being “an injective function from some set, to some set, where there is a metric on the second set.”
@@drdca8263 I guess I didn't think of it that way
I was thinking of a 'weird' example being x in reals; f(x) => (x in Q then x, else -x) and take balls at places other then the origin.
Which book did you use for your topology class ?
You are the only person on TH-cam giving quality knowledge in calculus, geometry, analysis, algebra, infinite series, IMO, Putnam etc
You're just awesome :)
Do the p-adic numbers define a metric?
Yes they do: the p-adic order of an integer a is the largest power of p that divides a, i.e. the power of p in the prime factorisation of a (if a≠0), and is denoted by v_p(a). this extends to the p-adic order of rational numbers by v_p(a/b) = v_p(a) - v_p(b). If we then define the p-adic norm by |x|_p = e^(-v_p(x)) then |x|_p is a so-called "absolute value" (specifically a "non-Archimedean absolute value") and induces a metric just like a norm induces metric: by d_p(x, y) = |x - y|_p. The metric space of p-adic numbers has many strange properties, such as every open ball being closed and vice versa, and any point inside a ball is a center of the ball - this means that there can be many points all equidistant from each other, in contrast to the familiar Euclidean metric where only 2 points can be equidistant on a line (or 3 points equidistant in the plane, or 4 points equidistant in space).
"Now, before we look at an example that I cooked up ..."
Nice one
Love your videos. Keep it coming
Is there some notion that it's a limit as n goes to zero from above?
lim n-> 0+ (plus is superscript)?
I think the concept of directional limits kind of (well... you could do other things instead...) requires having an order on the set you are dealing with (in a way that really ought to be compatible with the metric, and he hasn’t prescribed an order to go along with this metric.
Of course, one could define the new ordering in the same way as he defined this new metric, by composing the usual ordering with the function f in both slots.
(... I guess an alternate way of talking about directional limits is to talk about limits along a (continuous) path, where the path determines the direction. I guess that’s probably the more general version of the concept...
...
Hmm... ok, defining a notion of this in arbitrary(?) metric spaces (or more generally, topological spaces) might be interesting,
especially if we don’t want to require that the path be like, a continuous function from [0,1) to our topological space, but rather having a more general domain...
I suppose one way to define it could be, if we have some topological space B with a designated base point b_0, where b_0 is an accumulation point,
then, if we have a function
f : X -> Y
(with X any Y topological spaces, but f not necessarily continuous)
then, if we had a continuous function
i : B -> X
and, then looked at the composition of f with i,
uhh...
ok, this doesn’t seem like quite all the things such a generalization should have... I think we need a notion of two “B-paths” in X, approaching i(b_0) “from the same direction”.
Or, maybe don’t need that, at this point, and can define that *on top* of the definition of the limit of f of x as x approaches i(b_0) in the direction “i”, defining what it means for two “directions” to be the same, later?
Hm. Idk.)
Nice idea and very good presentation!
How do you know that there is only 1 limit?
metric spaces are hausdorff
The important concept for me in all this is: it challenges naive notion of performing metrics limited to a nice domain of a function.
Why should metrics not be performed on image/co-domain of a function or its inverse? 🙂
I hope this means you’ll make more topology videos
I don't think what you get at the end is weird. You just flipped how you quantify "closeness" where a big value means close in contrary to what one "usually" gets i.e. 0
Right?
When you say “does not need to be well-defined”, I’m guessing you meant to say “well-behaved”, and this was just a slip of the tongue?
pretty sure he misspoke, based on what he is saying before and after. good catch!
Hi,
You may as well prove that the limit is 1 by taking the green segment instead of the magenta one, no?
It's a common mistake to have d(x,y)>=0 as an axiom of a metric. You can show that it must be always positive.
"Mistake" isn't the right word - you probably mean 'it's a common misconception that d(x,y)>=0 is needed as an axiom of a metric'.
is there some sense in which taxicab metrics and max metrics are dual to each other, while the euclidian metric is dual to itself?
Yes, the same sense in which -1 is dual to 1 and 0 is dual to itself :), although, in a more general setting, they are simply points of a continuum: just like, for an analogy, Hyperbolic vs Euclidean vs Elliptic spaces are points of a continuum parametrized by the curvature of space...
@@jpdiegidio i was hoping for something like that but if it's simple additive inversion, it's not jumping out at me. after all, i can't get a taxicab metric by writing min(Δx, Δy), do i perhaps have to rotate by 45° or something?
@@KD-onegaishimasu the Euclidean metric is the l_2 metric d_2((x_1, ..., x_n), (y_1, ..., y_n)) := (Σ_i |x_i - y_i|^2)^(1/2), the taxicab metric is the l_1 metric d_1((x_1, ..., x_n), (y_1, ..., y_n)) := (Σ_i |x_i - y_i|^1)^(1/1), and the maximum metric is the "l_∞ metric" "d_∞((x_1, ..., x_n), (y_1, ..., y_n)) = (Σ_i |x_i - y_i|^∞)^(1/∞)": to see this last one, these l_p metrics are induced by the l_p norms |x_1, ..., x_n|_p := (Σ_i |x_i|^p)^(1/p) in the usual way as d_p( *x* , *y* ) = | *x* - *y* |, and the reason why letting p approach infinity gives |x_1, ..., x_n|_∞ = "(Σ_i |x_i|^∞)^(1/∞)" = max(|x_1|, ..., |x_n|) is because the coordinate x_j of greatest absolute value will swamp all the other terms in Σ_i |x_i|^p (because p is huge), so Σ_i |x_i|^p ≈ |x_j|^p and so (Σ_i |x_i|^p)^(1/p) ≈ |x_j| = max(|x_1|, ..., |x_n|).
So, using Michael's notation: d_E = d_2, d_T = d_1, and d_M = d_∞. Therefore it is not an additive inversion as jpdiegidio says - the parameter values are 1, ∞, 2, not -1, 1, 0.
@@schweinmachtbree1013 I haven't said anywhere that it's an *additive inversion*, I have simply given a couple of (not random) analogies of points that more generally belong to a continuum... But you have taken the time to find and give the actual formulas: thanks for that.
P.S. In fact I did suggest more than just that, for how at an intuitive level: but indeed in the space of l_p spaces, as parameterized by p, which is, by definition of p-norm, a real number >= 1, it is the case that l_1 is "dual" to l_oo and l_2 is "dual" to itself, though in the sense of a *multiplicative inversion relative to p-1* (I am not sure about the right terminology here)... no?
injectivity isn't 1:1 though? bijective is, which is injective + surjective
There is some confusing terminology that is sometimes used. The difference is _one-to-one_ vs _one-to-one correspondence._
One-to-one means the same thing as injective.
One-to-one correspondence means the same thing as bijective.
Does anyone know if there are any videos that can clearly explain what is and how we can compute (1+ε) factor approximation?
Notice than as n goes to infinity this 1/n term is going to zero (18:00) That's all😂
C'est lim f(1/n) =-1 et non lim 1/n??.
7:05
You said “this function does not have to be well-defined”, and this provides an example of the kind of thing that can happen with bad terminology. I’ve always objected to the way the term “well-defined” is typically used, because it is confusing and is unnecessary - it in some sense violates Ockham’s Razor. Let me explain: most authors and lectures I’ve known use this term as a synonym for the phrase “is a function”, or equivalently, for the phrase “has the function property”. If one has written a formula that describes a relation R and then proves that this relation satisfies a “test” that is analogous to the vertical line test of elementary algebra and and introductory calculus, then it is common to summarize with something like “this shows that R is a well-defined function”. I’ve always found it rather odd that some instructors will write down a formula as if it defines a function and then claim that we need to determine if that formula actually defined a function however, it is common in textbooks and lecture materials and videos, etc.
In any case it is not common at all to say anything like “let f be a function that is not well-defined”; in fact, I’ve never seen such a thing in 40 years of teaching & research until today. In this case it was given at the beginning that lowercase f is an injective function. Nowhere until this moment in the recording was it hinted that f is a relation that may fail the criteria to be a function. This, for me created some cognitive dissonance: do you really mean that we could use a relation, in place of f, that might not be a function? While I have worked with metric spaces before, this question has never come up for me, so let’s consider it now.
Let R be a binary relation between points of a set X and real numbers with the following property:
If ordered pairs (p,q) and (p,u) are in the relation R, then q=u.
Define a relation d between pairs of points of X and nonnegative real numbers as follows:
Given p, q in X and a nonnegative real number r, the pair ((p,q),r) is in d if and only if there are real numbers t and u such that pRt, qRu, and r=|t-u|.
The problem we want to address is this: Is d a metric on the domain of R?
In order for this to work, d must be a function; that is, it must possess the function property. This would require that if ((p,q),r) and ((p,q),s) are in the relation d, then r=s.
Assume that d does have the above property but that R is not a function, say p is a point of X such that pRs and pRt for distinct real numbers s and t. Then we have the following:
((p,p),0) is in d because pRs and pRs and 0=|s-s|, and
((p,p),|s-t|) is in d because pRs and pRt.
Since 0 is not equal to |s-t|, it follows that d is not a function.
Consequently, if R is not a function, then d is not a metric, because d is not a function.
I’ve posted a question to Quora about the reason people use this phrase:
youtubemathswhyspace.quora.com/?ch=17&oid=3233113&share=396067ef&srid=tbyW&target_type=tribe
I started to write a similar comment, but came to the conclusion that it was a mere slip of the tongue, and that he meant to say “well-behaved”.
I guess he didn’t catch this in editing.
In most cases I’ve seen, “well defined” is usually used in regards to derivatives/integrals, and is used to refer to functions that don’t have discontinuities that would case derivative issues. So, I think he’s using the same definition here, especially with his piecewise function (which I would define to be not well defined)
@@drdca8263
Good point, but now perhaps he will edit it? 😃
In any case, there are some nice related questions. For example, what happens if f is not injective? Well, it seems clear that you then get a psuedo-metric that is not a metric, but I did not write a proof of this. Also, several nice variants can be considered in relation to adding some "well-behavedness" requirements. If f is injective and continuous, some nice things happen. What about when f is injective and piecewise continuous? That's got to be interesting.
But my criticism is more about the terminology as it is usually used: If the term is to be truly of value, it's negation should be also useful, but it isn't, because there is no such thing as a function that is not well-defined.
@@Megumin_Random
This usage also does not make sense. For example, the absolute value function is a funcntion and so is well-defined. It has no derivative at zero, but this does not mean that the absolute value function is not "well-defined" at zero.
The piecewise defined function he uses is well-defined on its domain, namely the interval [-1,1]. For no function is it ever required that it be "well-defined" outside its domain. That would make no sense.
@@writerightmathnation9481 To say that something is “not well-defined” is in some ways (not necessarily in most ways) like saying that something “doesn’t exist”.
It has been said, “existence is not a predicate”.
When we say “Bigfoot does not exist”, we are not saying “There is a thing called ‘Bigfoot’, which has the property of not existing.” That wouldn’t make sense. It can’t both exist and not-exist!
To meaningfully ask whether a predicate is true of something, there must be a thing which we are asking whether the predicate is true of that thing.
That is, the thing must exist.
So, to ask whether a thing exists, cannot be to ask whether a predicate “it exists” is true of it. It is a different kind of thing.
“Does Bigfoot exist”, in a sense, is sorta not a question about Bigfoot, so much as about the concept of Bigfoot. (If you are familiar with the programming language Lisp, perhaps compare the difference between Lisp functions and Lisp macros. “Is tall” is like a Lisp function, while “exists” is perhaps like a Lisp macro. )
One might phrase it as “Does there exist an instantiation of the concept of Bigfoot?” or “Does there exist a thing which is-Bigfoot?”.
The same sort of thing is the case, I think, for the question “Is X well-defined”.
We are not assuming that we have some concrete specific thing X (perhaps a function from the reals to the reals), and we are asking “is this thing well-defined?”.
Rather, we are asking of a potential definition (or perhaps more broadly a concept), “is this a valid definition, in the sense of uniquely specifying a single object / property-for-an-object-to-have?” (or, if a concept, “is there a good(best?) choice of a definition (or class of equivalent definitions) for this concept we have, that would uniquely pick out a specific object/property?”)
For example, in the topic of voting systems, there is a concept of a (I forget the name, but it means a candidate that would win all possible pairwise matchups. Maybe “Condorcet winner”? But that might refer to something else. Idr.)
This property “is a (this type of winner)” is well-defined, and, in elections where there is such a winner, “the (that type of winner) for this election” is well-defined.
But, on the other hand, I would claim that, in the general case, (the concept of) “the will of the people” is “not well-defined” (though, of course, in restricted cases, “the will of the people in context X, regarding the question Y”, may be well-defined, and in many cases this may be sufficient.)
So, in summary, I think “is well-defined” is kinda like “exists”, in that it isn’t a predicate which is applied to the (alleged) object it appears to be applied to, but instead is kind of more like it is applied to the (name or description which potentially refers to an object), where this (name or description) is put in the place that a (name or description of an actual object) would go if the “exists” or “is well-defined” *were* a predicate being applied to an object.
Exploring p-adic numbers would be neat.
The claim for the small mistake was a big mistake.
Determining which sequences have limits using the discrete metric is a silly, if simple, exercise.
There's a similar but more interesting result if you look at whether metric spaces are complete (every Cauchy sequence converges to a value in the metric space). Real numbers: complete. Rational numbers: not complete. Integers: complete!?
Does this definition of a metric mean that the Minkowski metric isn't a metric?
O is not in.
very good !
Isn't this argument tautological? In your calculation of the limit of 1/n as n approaches infinity by the metric definition, you employ that 1/n approaches 0 if n approaches infinity. This is contradictory to your final result that 1/n approaches -1 as n approaches infinity.
The limit in the metric is the classical limit of the distance.
He uses the fact that the limit of 1/n is 0 in usual sens.
It does not contradict that lim 1/n =-1 with df metric
I think you are perhaps using the word “tautological” to mean something other that the standard meaning?
Pretty interesting 😮
For example, the cofinite topology (open sets are the empty set and sets whose complement is finite) on the real numbers has some interesting properties regarding limits of sequences!
surely it is -1/12
Click bait. You are just making up arbitrary rules, but make no mention of those rules in the title card. Surely what you meant to say is that: lim n->∞ f(1/n) = -1, leaving out the f makes it just wrong.
At a glance it actually looks like you produce a lot of good content, this video just doesn't reflect that. I think you should take it down, or at the very least change the title card.
Everything seemed perfectly straight-forward to me. It was mostly based on the principal that the distance between something and itself must be zero hence for different definitions of distance you get different limits when you consider the formula giving the distance between the two points which are postulated to be the same, naturally. That's why the title reads *"where* limits aren't what they seem..." (in places with different metric spaces obviously). Also why the thumbnail reads "this is true [arrow pointing to the mathematical formula, 'the limit of 1/n as n approaches infinity is equal to -1']...from a certain point of view". Obviously he never meant that this was true in conventional mathematics.
Also, these rules aren't really arbitrary. Just the standard rules for dealing with metrics and they're defined that way for good reasons which he explained in the beginning of the video.
I think it's clear from both the title and thumbnail that the standard topology on R is not being used.
I’d like to share a functional equation problem if anyone’s interested
for all real numbers x,y determine all continuous functions such that f(0)f(x+y)>=f(x)f(y)
also f(x)>0 for all real numbers x
I cannot quite prove why the answer is the exponential function
Fairly straightforward. Here are a few hints to help:
1- Observe that if f(z) is a solution then so is g(z) = A * f(z) for some constant A. Hence, if f(0) is not zero then we may instead solve the functional equation g(z+w) >= g(z)g(w) by letting A = 1/f(0).
2- Let g(1) be any arbitrary value, then relate g(n) to g(1) when n is natural, followed up by relating g(q) to g(1) when q is rational.
3- A continuous function is uniquely defined by its values on any dense set. Q is dense in R, hence it suffices to know g for all rationals.
@@Nameless.Individual How can one deduct g(2)=g(1)*g(1) and furthermore g(n+1)=g(n)*g(1) given g(x+y)>=g(x)*g(y) ? Is it possible to treat this inequality as an equation?
I’m probably on the wrong track here or missing some obvious reason why this is true but your help would be much appreciated.
i squared?