This is the best video for the subject, along with the whole playlist, thank you!

@mariusfurter

2 жыл бұрын

I'm happy you are enjoying them!

Thank you so much for making these wonderful videos! You've been saving countless lives.

You just saved my life thank you

As clear as the first video in the series. Excellent! Thank you! I look forward to the rest.

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)

@35:36, *B contained in A. Great series.

Wonderful explanations Thx

I'm loving this video series! What program do you use to make them? I'd like to use it and the color theme on the classes I give!

@mariusfurter

2 жыл бұрын

I'm happy you are enjoying the videos. I am using goodnotes on ipad. It has some colors by default but I had to set up the pastel chalk colors myself by creating custom colors. I believe I copied the rgb values of some theme I found online or used a color picker. In particular the white is slightly off-white. The gray background is the dark theme in the app. There are also nice screen sharing options that hide the controls.

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

15:06 I think you mean, with correction, "and this shows that the complement of the yellow set (which is the red set) is open, so the yellow set is closed."

the closure of open unit ball is include in the closed unit ball . not always equal

@bookl5578

Жыл бұрын

could you tell me why

@ktayeb6774

Жыл бұрын

@@bookl5578 take d(x,y)=1 if x different of y and d(x,y)=1 if x=y. The open unit ball of radius 1 around any point x is the singleton set {x}. Its closure is also the singleton set. However, the closed unit ball of radius 1 is everything

@ktayeb6774

Жыл бұрын

the closure of the open unit ball is contained in the closed unit ball but not always equal. This can not happen in normed spaces.

27:15 Part of this is tricky to prove. Suppose by way of contradiction that a nbdh B of y is completely contained in X\A, so B contains no points of A. Since B is open, X\B is closed. Now A ⊆ X\B and since Ā is the smallest closed set containing A, we have that Ā ⊆ X\B. This means B ⊆ X\ Ā = Ext A, hence y ∈ Ext A, which contradicts y ∈ ∂A.

@matheusjahnke8643

20 күн бұрын

∂A=X\(Å ∪ Ext A)=(X \ Å) ∩ (X \ Ext A) = (X \ Å) ∩ Ā = Ā \ Å Which means if y is in the boundary of A.... it is equivalent to being on the closure of A, which means (c) applies, while not being in the interior which means the equivalent statements in (a) are false; (a) and (c) may be used to prove (b)

I might have misunderstood, but I have a question: on a set X is possible to define different topologies, therefore there are different topological spaces (X,T). Now, since closed sets are defined to be complements of open sets, which are by definition elements of a topology, does this mean that a set which is closed for some topology T1, might not be for another T2? So shouldn't we specify to which topology/ topological space (intended as couple set/topology) we are referring to when we say that some set S is closed? For example we could write: Let (X,T) be a topological space, then S is closed with respect to the specific topology T on X (?)

@mariusfurter

2 жыл бұрын

You are absolutely right! The closed sets depend on the topology we put on X. As a simple example consider (X,T1) and (X,T2) where T1 = {X, \empyset} and T2 = Powerset(X). Then in (X,T1) only {X, \emptyset} are closed, while in (X,T2) every set is closed. If we wanted to be super precise, we would always write a topological space X as a pair (X,T), where X is a set and T a topology on X. However, we generally abbreviate this to "Let X be a topological space" for brevity. So we leave T implicit and assume we are making all our claims relative to that implicit T. We could expand the definition of closed set I gave in the video to be more precise in the following way: Let (X,T) be a topological space. A set C \sub X is closed in (X,T) iff its complement is open in (X,T). I know that it can be difficult in the beginning when people abbreviate things, so thanks for the question.

@Evan-ne5bu

2 жыл бұрын

@@mariusfurter Thank you for be very clear!

Hello, I was wondering, why is it that the topology is closed under finite intersection and not under arbitrary intersections. Is it because it could go to the Empty set which is Open and Closed at the same time ? But then it would imply ,that it wouldn't work for the arbitrary unions either since X could be in the the result of the union, so I suppose it isn't due to the empty set ,but then why is that ? In any case thank you for sharing your knowledge.

@mariusfurter

Жыл бұрын

Topological spaces should generalize the way open sets behave in euclidean space R^n. There it is not the case that arbitrary intersections of open sets are open. For instance, consider the sequence of open intervals (0, 1+1/n) for n in the natural numbers. If you take the infinite intersection of these types of intervals over all natural numbers you get the interval (0,1] which is not open in the standard topology on R. So if we would put the requirement that arbitrary intersections of opens should be open, R with its standard topology would not be a topological space.

23:08 you said compliment of the exterior even though it should just be the exterior

@mariusfurter

2 жыл бұрын

It's possible I misspoke, but I can't find the instance you are referring to. The misunderstanding might also be a consequence of the fact that bracketing is ambiguous, i.e X-(ext(A) u int(A)) sounds the same as (X-ext(A)) u int(A). In any case, what is written in the video seems correct, so orient by that.

What is the definition of a set open in relation to another set ? I saw this in a book and i still cannot clear my mind. Obs.: there was a mistake a typo in a part y = { ......}. y was contained or belonged to that set as set came as an element of a set an in the context it was not a set so it it would not be equal to a set. I should have marked the slide. Some mathematicians can be very hard on you for the slightest mistakes. Nobody is soft in this field of knowledge and you know that. A small mistake can kill you. Best wishes.

@mariusfurter

Жыл бұрын

Suppose A is a subset of a topological space X. Then one says that a subset U of A is relatively open in A, if it is open in the subspace topology on A, i.e there exists some open set V in X such that U is the intersection of A and V. You can find more info on this in the video on the subspace topology.

@monicamir

Жыл бұрын

@@mariusfurter Thank you very much for your attention. I was studying sigma-algebra and Lebesgue measure. We had much difficulty at that time. The professor was German. A brilliant person, but hard to understand. Some time ago I decided to return and finish what I started. Best wishes