Відео

This lecture is infnitely beautiful, Brilliant presentation.

Davis Susan Martinez Jennifer Rodriguez Nancy

Didn't follow the argument about continuity that well. I was kind of expecting a measure to be introduced but I may be too stuck in my metric space ways.

Moore James Lopez Laura Wilson Anthony

Thanks for the videos, helped me a lot.

12:35 I was trying to prove the lemma by my self, wondering why I should take a finite union of open sets U_p_i (actually a union was also an open set). Then I thought about the intersection of sets V_p_i, which is open only if finite and everything became clear.

55:15 one should show that P is closed.

29:57 the symbol ought to be \in and not =.

28:48 here one starts from two manifolds with boundaries (which are not manifolds) and eventually ends up with a manifold; impressive.

22:30 this is the "Vel" symbol, the "Wedge" symbol is the Vel turned upside down.

4:08 here is said that a point in A can have multiple values in f(A), how is that possible if f is a function?

You forgot to show that the topology is unique in the porosition that starts at 20:20. Great videos!💪👍🖖

10:41 in the definition of subset topology there's written V subset of X, but I reckon you meant V element of the topology of X.

I was thinking: since topological manifold with boundaries are not manifolds, one cannot define an atlas on them. If so, it would be composed of coordinate charts made out of different coordinate domains. Does this make any sense?

20:20. Quick and easy explanation. Just one question: where the uniqueness of topology is proven? Moreover, in the converse implication where the hypothesis of unique topology is used? Thanks in advance.

Great lecture! I've been following it very closely and had one doubt. Would be grateful if someone could help. At 37:41, you prove that f(|V) ^(-1) (U) i.e. the purple set is open because it is the pre-image of U (an open set) under a restrictive function f which you assumed to be continuous. I'm assuming you draw this conclusion from the proposition (c) proved at 24:32 If so, doesn't this require f^(-1) (U) to be an open set first (as mentioned in the proposition) ? Since it is not an iff statement, we can't draw any reverse conclusions. Thanks

16:27 I reckon you meant one point Hausdorff space is closed.

This is a fascinating subject! Are you going to do the next in the series on inference? I sure hope so.

16:20, if we defined U=f^{-1}(V) by (*) U is open, then f^{-1}(V) is open; why writing f^{-1}(V)=f^{-1}(f(f^{-1}(V)))=...?

You talk about balls in a general topological space (X, T), could you define a ball supposing the topology is not induced by a metric? I head your answer at 10:16, now it's clear.

Each connected component is only closed need not be open??

Finally an exciting stuff

1:00

Wouldn´t be better to say that (X,T) is a topological space rather than just X?

That's a great video. Just a thing that bothered me, in the omega example. Youve drawn a set u that does not contain [0]=[1] on the unit circle. But for some reason its inverse under omega contained both the points 1 and 0. But it's my first time studying topology, and im kind of suprised that i can comprehend all of that without having to rewing every two minutes like most complicated topics. So from lack of experience in other learning sources i can't really judge but i would say you are doing a great job!

0:50, Your T looks like S-bar, and your P also looks like S-bar, or possibly a T but with more of a Disney font. Impressive…

34:15, does anyone else hear Kermit the Frog talking about Epsilons?

Thanks for the series mate, I finally understood that I need to selfstudy more intense)

That was a tough one, thanks) I've got a question about real analysis: which course/book/source would you recommend for it? I just don't have this one at uni but I studied complex analysis so I am little dummy at this one)

How is N an open set? It can be neither open nor close depending on the topology chosen right

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;

Hi mate, I didn't get the thing about 'yota-s': is this about sequences of subspace S?

One thing; I think the topological definition might be redundant; The (i) axiom is redundant if we consider intersections and unions over no sets; Intersection over (no sets) = Universe set = X Union over (no sets) = Empty set (ii) and (iii) imply (i)

Wondering if these morphing spaces are used by orian to track down the power of money counter accumulation on the black markets.

These are some of the most thorough and clear videos (on any topic) I've seen so far. I'm amazed this one only has 267 views after three years.

I have followed your lectures carefully until now but you didn't explain the meaning of COUNTABLE

I started learning this subject because I thought it was about donuts, turns out its just set theory on steroids, much like abstract algebra. Now I understand the importance of set theory as a foundation for these fields

Cool!

I think G in the last example is not continuous because the sphere is open and the cube is not. You should change the max to be < 1

thank you for this.

For the final example about the connected set that is not path connected, I was having trouble figuring out the detail in the extra notes about the existence of the minimum value a. I managed to figured out a proof which might help others understand. The claim is that there is a minimum a such that g(a)=(0,y). Notice that any value t which satisfies g(t)=(0,y) equivalently satisfies (fg)(t)=0, where f is the projection function along the first coordinate. Hence the set of such t is exactly the set of zeroes for the composite function fg. We know the projection function is continuous and g is continuous by assumption, so fg is also continuous. An important fact to know is that the zeroes of a continuous real function form a closed set. This set is a subset of [0,1], so it is bounded below. This and the fact that it’s closed implies that the set of zeroes contains its infimum. We take this infimum to be the value of a.

I don’t think the wedge sum example quite works because we don’t know if the singleton {y0} is closed in Y. If it is closed, we can make the wedge sum, but I don’t think this works for a general topology Y.

love your videos! 1 question, on the basis criterion proof , <= direction, why is it not enough to use the definition of open set that every point in the set has a neighbourhood contained in the set? with B_p as the neighbourhood?

Awesome video. I just have one nitpick. Whenever you talk about open sets in the product topology, like in 15:50, you write the open sets as unions of open sets in the component spaces. I don’t think it’s necessary to do all that given that each union would be an open set in each component space already. All you really need is that U=U1xU2.

For a less intuitive but simpler version we can take the identity function from a space to itself, but with two different topologies; the topology in the domain being strictly finer than the one in the codomain. For example X is a space with at least two points, the discrete topology in the domain, the trivial topology in the codomain. This example is the same, we can change the topology in the interval so that it becomes homeomorphic to the circle, but it won't be the subspace topology in R^2!

Great set of videos. Thanks.

At 32:40 you should not assume that the points are not colinear.

the textbook I'm using says the open cover only needs to contain the space X. Not that it needs to equal it.

Wonderful explanations Thx

Thank you professor ❤