# Topology question

• Mar 1st 2010, 01:23 PM
Defunkt
Topology question
Hi, I would appreciate some guiding with this question. Thanks.

Let X be some topological space and Y a compact, Hausdorff topological space. Let $\displaystyle f:X \to Y$ and define $\displaystyle \Gamma_f = \{(x, f(x)) : x \in X\}$.

Show that if $\displaystyle \Gamma_f$ is a closed subset of $\displaystyle X \times Y$ with the product topology, then f is continuous.
• Mar 1st 2010, 01:36 PM
Drexel28
Quote:

Originally Posted by Defunkt
Hi, I would appreciate some guiding with this question. Thanks.

Let X be some topological space and Y a compact, Hausdorff topological space. Let $\displaystyle f:X \to Y$ and define $\displaystyle \Gamma_f = \{(x, f(x)) : x \in X\}$.

Show that if $\displaystyle \Gamma_f$ is a closed subset of $\displaystyle X \times Y$ with the product topology, then f is continuous.

What have you tried?

P.S. $\displaystyle Y$ need not be Hausdorff.
• Mar 1st 2010, 02:01 PM
Drexel28
Theorem: Let $\displaystyle X$ be any space and $\displaystyle Y$ be compact. Then, if $\displaystyle \Gamma_f\subseteq X\times Y$ is closed we have that $\displaystyle f:X\mapsto Y$ is continuous.

Proof:

Lemma: Let $\displaystyle Y$ be compact, then $\displaystyle \pi:X\times Y\mapsto X$ is a closed.
Proof: Let $\displaystyle E\subseteq X\times Y$ be closed. We prove that $\displaystyle \pi(E)$ is closed by proving that it's compliment is open. So, let $\displaystyle x\notin\pi(E)$. Then, it follows that $\displaystyle \left(x\times Y\right)\cap E=\varnothing$, and for each $\displaystyle y\in Y$ $\displaystyle (x,y)$ has a basic open neighborhood $\displaystyle G_x(y)\times F_y\subseteq X\times Y-E$. Since the tube $\displaystyle x\times Y$ is compact (since it's homeomorphic to $\displaystyle Y$) there is a finite subcovering $\displaystyle \left\{U_{y_1},\cdots,U_{y_m}\right\}$. The neighborhood $\displaystyle U=\bigcap_{k=1}^{m}U_x(y_k)$ is clearly disjoint from $\displaystyle \pi(E)$. The conclusion follows. $\displaystyle \blacksquare$

Now, it is easily proved that since $\displaystyle \pi:X\times Y\mapsto X$ is closed so is $\displaystyle \pi\mid \Gamma_f:\Gamma_f\mapsto X$. But, it is clearly a bijective continuous mapping and since it's closed it's also a homeomorphism. The problem follows after considering the following lemma

Lemma: A mapping $\displaystyle f:X\mapsto Y$ is continuous if $\displaystyle \pi\mid \Gamma_f:\Gamma_f\mapsto X$ is a homeomorphism.
Proof: We must merely note that $\displaystyle f=\pi_Y\circ\left(\pi_X\mid \Gamma_f\right)^{-1}$.
• Mar 1st 2010, 02:05 PM
Defunkt
Quote:

Originally Posted by Drexel28
What have you tried?

P.S. $\displaystyle Y$ need not be Hausdorff.

I think I came up with a solution. Tell me if you see anything wrong please.

I want to show that f is continuous, ie. that if $\displaystyle V \subset Y$ is open then $\displaystyle f^{-1}(V)$ is open as well.

Let V be an open subset of Y. Let $\displaystyle U = V^c$, then U is closed, and so is $\displaystyle X \times U$. Since $\displaystyle \Gamma_f$ is closed as well, $\displaystyle \Gamma_f \cap (X \times U)$ is also closed.

From a previous result, we know that the projection $\displaystyle \pi_x:X \times Y \to X$ is a closed mapping when Y is compact. Therefore, $\displaystyle \pi_x(\Gamma_f \cap (X \times U) )$ is closed.

However, note that $\displaystyle \pi_x(\Gamma_f \cap (X \times U) )$ is exactly the set of points in X which f does not map into V, and it is closed, therefore $\displaystyle f^{-1}(V)$ is open and we are done.

I know it is a bit informal, but that is simply the outline.

Thanks again for helping.
• Mar 1st 2010, 02:16 PM
Drexel28
Quote:

Originally Posted by Defunkt
I think I came up with a solution. Tell me if you see anything wrong please.

I want to show that f is continuous, ie. that if $\displaystyle V \subset Y$ is open then $\displaystyle f^{-1}(V)$ is open as well.

Let V be an open subset of Y. Let $\displaystyle U = V^c$, then U is closed, and so is $\displaystyle X \times U$. Since $\displaystyle G$ is closed as well, $\displaystyle G \cap (X \times U)$ is also closed.

From a previous result, we know that the projection $\displaystyle \pi_x:X \times Y \to X$ is a closed mapping when Y is compact. Therefore, $\displaystyle \pi_x(G \cap (X \times U) )$ is closed.

However, note that $\displaystyle \pi_x(G \cap (X \times U) )$ is exactly the set of points in X which f does not map into V, and it is closed, therefore $\displaystyle f^{-1}(V)$ is open and we are done.

I know it is a bit informal, but that is simply the outline.

Thanks again for helping.

I'm having a little trouble following it, but I think you are saying that $\displaystyle \left[f^{-1}(V)\right]'=\pi_X\left(\Gamma_f\cap\left(X\times V'\right)\right)$. If you can prove that I'll believe you.
• Mar 1st 2010, 02:29 PM
Defunkt
Quote:

Originally Posted by Drexel28
I'm having a little trouble following it, but I think you are saying that $\displaystyle \left[f^{-1}(V)\right]'=\pi_X\left(\Gamma_f\cap\left(X\times V'\right)\right)$. If you can prove that I'll believe you.

$\displaystyle \Gamma_f \cap (X \times V') = \{(x, f(x)) : x \in X\} \cap \{(x,y) : y \notin V\} = \{(x, f(x)) : f(x) \notin V\} =$ $\displaystyle \{(x, f(x)) : x \notin f^{-1}(V)\}$

Sorry for being messy, by the way I mistakenly wrote G in my first answer instead of $\displaystyle \Gamma_f$, but I fixed it now.
• Mar 1st 2010, 02:31 PM
Drexel28
Quote:

Originally Posted by Defunkt
$\displaystyle \Gamma_f \cap (X \times V') = \{(x, f(x)) : x \in X\} \cap \{(x,y) : y \notin V\} = \{(x, f(x)) : f(x) \notin V\} =$ $\displaystyle \{(x, f(x)) : x \notin f^{-1}(V)\}$

Sorry for being messy, by the way I mistakenly wrote G in my first answer instead of $\displaystyle \Gamma_f$, but I fixed it now.

Then I would agree!