🐵 A Cute Topology Proof on Connectedness

A quick video on a nice topology proof I wrote years ago. How would you prove this claim? Comment below!

Music:

"Rainbow in the Dark" as performed by Corey Taylor
Originally by Dio

"Blue Wednesday" - All That Jazz

"Lateralus" as performed by Sakis Strigas
Originally by Tool

My current equipment for making videos (affiliate links):

[samhollins]

Hah! Who else recognizes the symbol he signed his proof with? 🎸

[davidfenoll]

Being unfamiliar with topology, couldn't help but to burst a laugh when you stated that A is not clopen

[MathManMcGreal]

feeling cute might do some topology today idk

[benthayermath]

I had a reserved, Polish professor as well for complex analysis. Really went the extra mile in giving a proof of every result. I once waved to him outside of class when I saw him walking home, but he just gave me a strange look. Interesting guy

[cinvhetin]

I did it exactly backwards compared to your proof:
Suppose A is both open and closed. Then since A is closed, closure(A)=A and since A is open, closure(X\A) = X\A. Therefore the intersection of the two closures is empty and so - by assumption - A is either empty or the whole space. Hence the only subsets, which are both open and closed, are empty or the entire space and so X is connected.

[EpicMathTime]

Be sure to follow me there if you want to see my videos a little earlier than scheduled!

[alkankondo]

NICE JOB!! I was afraid I wasn't going to understand any of this, but nope! It was pretty easy to follow all the way through (albeit with me having to pause-and-ponder a handful of times). It was a great balance of being both concise and easy-to-understand -- the mark of an excellent, optimal proof. No wonder your professor gave you kudos for it!


(By the way, this is coming from someone who has never taken a proper topology course, but has taken a course that had point-set topology in it, which was a couple years ago.)

[fountainovaphilosopher]

I particularly don't understand topology, but nevertheless this was a *nice* video to watch.
It's fair to say, you're among my favorite personalities from the youtube part of the math community. Hail

[johannesoertel]

It is allways true that boundary(S)=closure(S) intersect with closure(M\S). Therefore the statement is equivalent to boundary(S) is nonempty for all S not equal to M or the empty set. This however is just another way of saying that the only subsets of M that are both open and closed are M itself and the empty set. And thats one definition of a connected Topological space. So we see that the statement does not only imply that (M, T) is connected but is actually equivalent to it.

[saschabaer]

My attempt: Let X be a space with that property. If X is disconnected, then there exist disjoint open sets A, B such that X is their union. But then A and B are also closed, as each is the complement of an open set. So the closure of A is A and same for B. This gives us a contradiction: the intersection of A and B must be empty (by assumption), but also nonempty (as the intersection of closure(A) and closure(B) is nonempty, but closure(A) = A and same for B)

[Jocularious]

"Clopen", um excuse me wtf...

[Jcarr]

Suppose X = A \/ B, A and B open and mutually disjoint. Then A^c = B is closed, and B^c = A is closed. So A^c's closure is B. The closure of a closed set is the set itself, so invoking the property above, we have A closure /\ B closure = A /\ B is non-empty, contradicting the hypothesis that A and B were disjoint.

[arkapointer]

Good stuff. Keep this channel going. Love from India.

[EssentialsOfMath]

Suppose X is disconnected. Then X = U (u) V, where U and V are disjoint open sets. Then V = X/U, U = X/V. Then U and V are also closed. By assumption, close(U) (n) close(X/U) is non empty, but since U is closed, and X/U = V is closed, this implies U (n) V is non empty, which is a contradiction.

[grouseski]

Your content is so well done keep it up my bro

[estebanmartinez]

Supose X is not connected, then there are two open disjoint sets A, B, none of them empty, such that their union is X.
As non of them is empty, and they are disjoint they must be proper subsets of X.
As A is open, X-A is closed and so we have by hipothesis that cl(A) intersection cl(X-A) =cl(A) intersection X-A =/= emptyset
With this, A cannot be closed cause there is an element of the clousure of A that is not an element of A, but X-A =B which is open, so A is closed and we have a contradiction.
Therefore, X is connected.

Your proof is cuter <3

[tomificationness]

I'd love to see some graduate topology, if you have it.

[adrianparism]

I assumed there are two disjoint open sets whose union is X (another standard definition of connectedness); as each of them is the complement of the other, they are both closed, so equal to their closure; by hypothesis, then, their intersection is nonempty, but that contradicts the assummption that they are disjoint. Hence such two sets don't exist, and X is connected.

[MasterHigure]

I went for the contrapositive (always consider the contrapositive of what you want to prove). The steps are basically yours in reverse: Assume (X, tau) is disconnected. Then there is a proper, non-empty clopen subset A. Both A and its complement are closed, so the intersection given in the problem statement is empty.

[Chek]

I hope this won't be too annoying to read, but here's my take on this.
The premise is equivalent to the negation of condition 2 on an arbitrary subset. Connectedness follows from lemmas 2 and 1.

Definition -- connectedness:
A set is connected iff the set is not the union of any two open disjoint nonempty true subsets.

Definition -- closure:
The set's closure is the intersection of all closed sets which contain the set.

Definition -- edge:
The set's edge is the intersection of the set's closure and the set's complement's closure.

Lemma 1:
Connectedness is equivalent to the following:
No nonempty true subset is clopen. (condition 1)

Proof -- lemma 1:
Let (X, T) be a topological space.

Let X be not connected.
We obtain sets A and B which are open disjoint nonempty true subsets of X and whose union is X.
We see that B is the complement of A and thus A is clopen.
Condition 1 implies connectedness.

Let A be a clopen nonempty true subset of X.
Now A and the A's complement are open disjoint nonempty true subsets of X whose union is X.
Connectedness implies condition 1.

Connectedness is equivalent to condition 1. QED

Lemma 2:
Let A be a subset of a topological space.
The following conditions are equivalent:
A's edge is empty. (condition 2)
A is clopen. (condition 3)

Proof -- lemma 2:
Let (X, T) be a topological space and A be a subset of X.

Let A be clopen.
Now the closures of A and A's complement are themselves and their intersection is empty.
By definition the edge is empty.
Condition 3 implies condition 2.

Let A's edge be empty.
The closures of A and A's complement must necessarily be disjoint.
Let x be an element of A's closure.
Now x cannot be an element of A's complement, since it would then be an element of A's complement's closure.
A's closure is a subset of A and the sets must be identical -- symmetrically A's complement's closure is identical to of A's complement.
A is clopen.

Condition 2 implies condition 3.

Conditions are equivalent. QED