I used to think group theory lectures were badly done but actually topology is much, much harder and usually the lectures are not good. These are fantastic!

Beautiful lecture, using Hausdorff property. Thank you. I hope you make more playlists!!!

Very good lecture on point set topology! Very clear and helpful to give motivations of definitions!

Also the only convergent sequences in a discrete space are the eventually constant sequences. That's why these spaces are not interesting from the point of view of analysis, as well as non Hausdorff spaces. Great videos, keep it up. I hope your channel grows up!

Thank you so much sir

@mariusfurter

3 жыл бұрын

You're very welcome!

In the last proposition, why does a1 necessarily have to lie in U_p? Why does the new the neighborhood of p disjoint from a0 have to intersect A inside of U?

18:32 "why is this?"... because picking any point p ∈ X \ {f0}, it is proven that p is in the interior of X \ {f0} (as in, there's a neighborhood of p completely contained in X \ {f0})... which means the interior of X \ {f0} is equal to X \ {f0}... which means X \ {f0} is open;

For the last proposition, how do we know/prove that these sets are infinite?

@mariusfurter

Жыл бұрын

The argument works without needing to know in advance that A is infinite. We can always find a new point in it using the combination of the limit point and Hausdorff definitions. You are right that as a consequence A (and hence X) must be infinite. But this just means that in a finite Hausdorff space you can't have any limit points. In fact, it is a consequence of the fact that every finite subset of Hausdorff spaces is closed that the only Hausdorff topology on finite spaces is the discrete topology (can you see why?). Sorry for the long response time. I was on a boat without internet.

Hi Marius. Thanks for these very interesting videos. I think there's a problem with the last proposition: let X={a,b,c,d} and suppose that T is the power set of X, i.e. every subset of X is open. Then T is the discrete topology on X. So (X,T) is Hausdorff, by your second example. Alternatively place a,b,c and d at the corners of a square and use Manhattan distance as a metric. Then we have a metric space, Hausdorff by your first example. But X is finite, so how can any limit point contain infinitely many points of any set? In the illustration of the last proposition, you have placed p outside A. It's a big assumption that this is possible in finite spaces. Am I missing something?

@mariusfurter

Жыл бұрын

Hi, you are very welcome for the videos! Thanks for engaging with the material. This proposition seems to be causing some confusion (see the similar comment by Ji Tae Park below). The reason there is no problem is that finite Hausdorff spaces can't have limit points because the topology on a finite Hausdorff space is necessarily discrete. For instance, we know that finite subsets of a Hausdorff space are closed. Now take any subset. Its complement is finite and hence closed. Thus every subset is open. One could probably come up with a reasonably simple direct argument to show that points are open based on the Hausdorff condition by unioning and intersecting separating neighborhoods (of which there will always be finitely many). By definition, any neighborhood of a limit point p of a set A contains an element of A other than p. But since we are now in the setting of the discrete topology, {p} is a neighborhood of p that does not contain any element other than p. In particular it does not contain any element of A other than p. So there are no limit points. Therefore you are right that the proof should not work in finite spaces, but this is simply because there are no limit points there for the argument to apply to.

@sgut1947

Жыл бұрын

Thank you Marius. I think my confusion arose from not spotting the fact that (unless I'm still mistaken?) the limit of a sequence may possibly not be a limit point, if the sequence is eventually constant.

I really love your playslist :) I have a question which torments me in the example with the constant sequence 2. If you add the open space (1,3), 3 is still in the neighborhood of 2? What I did not understand ? I thank you for your help :)

@mariusfurter

Жыл бұрын

I am not sure which example you are referring to. Can you post the time-stamp? The open set {1,3} would not be a neighborhood of 2, since it does not contain the point 2. By definition a neighborhood of x is an open set containing x.

@awazin4031

Жыл бұрын

@@mariusfurter thanks for your answer :) This is the example to with the constant sequence of 2 kzread.info/dash/bejne/fIGto5iAeqfNhJc.html. As you said, this sequence has two limits 2 and 3. If you add an open set {1,3}, you said that the sequence has this time only 2 in limit. But 3 is still in the neighborhood of 2 since it is contained in {1,2,3}. I know I'm wrong but I don't see what I'm missing.

@mariusfurter

Жыл бұрын

Thanks for clarifying. For x to be a limit, we need the sequence to eventually lie in any neighborhood of x. Since the sequence is constant, this condition simplifies to 2 lying in every neighborhood of x. If the only neighborhood of 3 is the entire space {1,2,3}, then 3 satisfies the condition for being a limit. If we additionally add {1,3}, then this would be a neighborhood of 3 that does not contain 2, hence 3 stops being a limit. The fact that {1,2,3} is still a neighborhood of 3 that still contains 2 does not matter since the condition is that all neighborhoods of 3 must contain 2. In a less trivial example consider the sequence 1/n for n=1,2,3,... in R with the usual topology. This sequence has limit 0 since for every open interval around 0 the sequence eventually lies in that interval. However, every other point in R also has neighborhoods with this property. For instance 1 has a neighborhood (-1,2) for which the sequence eventually lies in it (in fact this neighborhood contains the entire sequence). The reason 1 is not a limit of 1/n is because this property does not hold for every neighborhood of 1, for instance (1/2,3/2).

@awazin4031

Жыл бұрын

@@mariusfurter thx ! Topology is fascinating but there are many subtleties that are not always easy to understand :) I continue to watch your videos with great pleasure :)

Is there such a thing as a limit of a set? Isn't it always a limit of a sequence?

@mariusfurter

3 жыл бұрын

You are right. Limits are defined for sequences. I have also used the concept of a limit point of a set in this series: A point x is a limit point of a set A, if every open neighborhood of x contains a point of A, other than x itself. The two notions are related, but they don't coincide in general. For example, if y is a limit of a sequence of distinct elements of a set A, then y is also a limit point of A. (Proof: For any neighborhood N of y, the sequence eventually lies in N, so N contains a point of A other than y). On the other hand, in sufficiently nice spaces, any limit point of A is the limit of a sequence in A. This occurs, for example, in metric spaces.

I'm confused about the beginning example in lecture 7 that 2 is also a limit point of set {2,2,2,2...}. According to the definition of limit point in lecture 3, a point p is a limit point of a set if any open set containing p contains points of the set other than the point p itself. There are two open sets that contain 2, which are {1,2,3} and {1,2}, and neither them meets the requirement of the definition, then how could 2 be a limit point of the constant set {2,2,2,2,...}?

@mariusfurter

4 ай бұрын

You might be confusing "limit point" with "limit". Not all limits are limit points. It is true that 2 is not a limit point of the set {2,2,...} = {2}. It is however a limit of the constant sequence (2,2,2,...) since every neighborhood of 2 must include 2, and hence all of the sequence elements. In the example it is also the case that 1 is a limit of the sequence. In fact 1 is also a limit point of the set {2}.

@shengzheyang74

4 ай бұрын

After some searching work, I now know that the definition of 'limit point of a set' is not the same as 'limit of sequences'. I thought them were totally same not being two distinct definitions. Than you very much!@@mariusfurter