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23:04
One las thing If $A\ U$ and $A$ is non empty then $G= \{B\subset U: A\subset B\}\cup\{\varnothing\}$ is clearly a topology. But the exercises says what is the smallest base for the topology. mmm I think that are just the proper subset containing A ...
Hi @DanielF @Jose
Hi @TedS.
HI @TedShifrin how's it going?
Hi! I feel quite happy today. I think I am going to get better and better.
Grrrreat :D
r9m
r9m
23:24
@Chris'ssis didn't see it :O (missed it)
@Chris'ssis AAAAA !!! screams and drops unconscious
@r9m Have you seen it before?
@r9m lol
r9m
r9m
@Chris'ssis nope :|
@r9m OK :-)))
I need some sleep ...
r9m
r9m
okay ,, :)
ow .. nice !! with the $x_i = e^{-t_i}$ change of variables it becomes an Arnold's form of multiple integral !! cool :D
@DanielFischer Are you there?
23:34
hello, @Anthony
@MikeMiller !!!!
@Anthony Not for long. It's more or less bedtime.
I was wondering if something was a requirement, I have a proof for something, but it doesn't use all the things I was given...
Hi @Mike, how's things?
@DanielFischer It's okay, I'll harass Mike <3
@MikeMiller the problem said Prove that if X and Y are compact Hausdorff spaces and if f is a function from X to Y
whose graph is closed in X × Y , then f is continuous.
23:35
Things is good, @DanielFischer
We didn't use Hausfdorffness of Y.
Sounds problematic, @Anthony
Does that not work?
Show me your proof.
We took some closed set in Y, aiming to show it's preimage was closed.
23:37
Keep going.
@Anthony You need the Hausdorffness of $Y$ for the other direction, if $f$ is continuous then its graph is closed.
@MikeMiller I'm slow hold on
@DanielFischer Thanks for correcting me, I didn't know if it was necessary or not.
So the graph restricted to that set, $C$ and its preimage is closed, because it's the intersection of $(X,Y)$ and $(f^{-1}(C),C)$.
God I can't type.
Funny, now that I hear graphs I imagine -graphs-, not functions..
23:39
Why is its preimage closed, @Anthony?
Oh god I really can't type
Hold on
the intersection of those sets is closed.
That's all I'm saying.
Why is the second set closed?
Oh I wrote the wrong set.
Sorry sorry sorry.
(X,Y) and (X,C)
I object to your notation, too. You should write $X \times Y$, not $(X,Y)$, which took me some time to decipher.
Sorry sorry sorry.
23:43
I think a found other mistake in the notes. This says: Let $\varnothing \subset A_1 \subset A_2 \.ldots A $ where $A$ is the union and the inclusions are proper. SO $T=\{A_ns, A, \varnothing and X\}$ is clearly a topology.
OK, @Anthony, so you're just saying it's the intersection of $X \times C$ and the graph, right? By $(X,Y)$ you mean the graph of the function?
BUt in one point this says Let $Y\subset X$ then Y is compact iff on of the following statements holds. 1 $Y\subset A_N$ for some N of exists $x\in Y \setminus A$
So we have $X \times Y \cap X \times C$. This is closed? And in a compact space, so it's compact. Moreover, the projection of this set onto $X$ is continuous for the product topology, and continuous maps take compact sets to compact sets, so $f^{-1}(C)$ is compact. Since it's compact and Hausdorff, it's closed. So the preimage of closed is closed, and our function is continuous.
But I think is not true because if we let the sets $\varnothing \subset \{o\}\subset \{0,2\}\ldots$ the even numbers
So yeah, $X \times Y$ just means the whole space. You should have just called the graph $G$ or something; $(X,Y)$ doesn't make much sense. Other than notation, you're golden.
23:46
lol
Thanks @MikeMiller.
the return does not work for $Y=2\mathbb{N}\cup \{0\}$ which can be cover by the $A_n$s and the $X$ this satisfies the second condition but does not have a finite subcover.
@MikeMiller Wait my friend just read you only need compactness of $Y$ and closedness of the graph, is this true?
probably
lol
Oh god.
I think a found other mistake in the notes. This says: Let $\varnothing \subset A_1 \subset A_2 \.ldots A $ where $A$ is the union and the inclusions are proper. SO $T=\{A_ns, A, \varnothing and X\}$ is clearly a topology. But in the notes this says Let $Y\subset X$ then Y is compact iff on of the following statements holds. 1 $Y\subset A_N$ for some N of exists $x\in Y \setminus A$. One part is really trivial, but the return does not work
for $Y=2\mathbb{N}\cup \{0\}$ which can be cover by the $A_n$s and the $X$ this satisfies the second condition but does not have a finite subcover.
Assuming $X=\mathbb{N}$ andlet the sets $\varnothing \subset \{o\}\subset \{0,2\}\ldots$ the even numbers
23:50
@JoseAntonio But that's not contained in one of the $A_N$, nor does it contain an element of $\mathbb{N}\setminus A$.
ups instead of zero has to be $Y=2\mathbb{N}\cup \{1\}$
@Anthony Sounds very dubious. What for do you think one only needs compactness of $Y$ and closedness of the graph?
@JoseAntonio That's not contained in $A$, so every cover of that must contain $X$.
ok I see
THanks I see the point.
@DanielFischer Nevermind, I meant Hausdorff too. The closed graph theorem?
@Anthony Generally, the projections from a product to the factors are not closed. And if $X$ is not quasicompact, then you only know that $X\times C$ is closed in $X\times Y$, and hence that $X\times C \cap \Gamma(f)$ is closed, not that it is quasicompact. So the quasicompactness of $X$ and that every quasicompact subset of $X$ is closed, are also important.

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