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Show that in the compact-open topology, $C(X, Y)$ is Hausdorff if $Y$ is Hausdorff and regular if $Y$ is regular

In the first statement, let $f,g$ be 2 functions in $C(X,Y)$, we need to find 2 disjoint subbasis elements $S(C_{1}, U_{1}), S(C_{2}, U_{2})$ that contains $f,g$. I think we can choose $C_{1} = C_{2}$, and then choose 2 disjoint open sets $U_{1}, U_{2}$, then $S(C_{1}, U_{1})$ and $S(C_{2}, U_{2})$ are disjoint. But how can we choose $C$ ?

In my book, there's a hint: If $\bar{U} \subset V$, then $\overline{S(C,U)} \subset S(C,V)$, but I even can't prove this hint, cause I don't know how to express an element of $\overline{S(C,U)}$.

Can someone help me clarify all these problems? Thanks

YuiTo Cheng
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le duc quang
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1 Answers1

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If $f\ne g$, there is a point $x\in X$ such that $f(x)\ne g(x)$. Let $U_1$ and $U_2$ be disjoint open nbhds of $f(x)$ and $g(x)$, respectively, and let $C=\{x\}$.

As you probably realized, the hint is for the second part of the question and does most of the work, so the problem really is to prove the hint. Suppose that $f\in\operatorname{cl}S(C,U)$. This means that for any compact $K\subseteq X$ and open set $W$ in $Y$ such that $f[K]\subseteq W$ there is a $g\in C(X,Y)$ such that $g[K]\subseteq W$ and $g[C]\subseteq U$. You want to use this to show that $f[C]\subseteq V$. Suppose that $f[C]\nsubseteq V$; then there is some $x\in C$ such that $f(x)\notin V$. Recall that by hypothesis $\operatorname{cl}U\subseteq V$, so $f(x)\notin\operatorname{cl}U$. Let $K=\{x\}$ and $W=Y\setminus\operatorname{cl}U$; what can you conclude?

Brian M. Scott
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  • So great. Your choice of compact space is too good. A finite set is compact. Thanks so much Brian! – le duc quang Oct 28 '13 at 15:55
  • @leducquang: You’re welcome! – Brian M. Scott Oct 28 '13 at 16:01
  • @BrianM.Scott I'm not sure I understand your answer fully. Are you saying that if $K={x}$ and $W=Y\setminus \mathrm{cl} U$, then there exists a $g$ such that $g(x)\subseteq Y\setminus\mathrm{cl} U$ and $g(C)\subseteq U$, which is a contradiction since $x\in C$? – Orca_1 Dec 01 '17 at 22:15
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    @BrianM.Scott Also, I'm not sure how the hint helps answer the question. Why showing the hint is equivalent to showing that $C(X,Y)$ is regular whenever $Y$ is regular? I understand that being regular means that a point can be separated from a closed set by an open set. – Orca_1 Dec 01 '17 at 22:17
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    @Orca_1 - This is due to an alternative characterization of a regular space which says that a space $X$ is regular if and only if for any $x \in X$ and any neighborhood $V$ of $x$, there is a neighborhood $U$ of $x$ such that $\bar U \subseteq V$. – user1236 Dec 26 '17 at 20:42
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    @user1236 Now that makes much more sense. Do you happen to know if there's a similar characterization of a normal space? – Orca_1 Dec 28 '17 at 09:06
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    Yup. Is the same thing but instead of a point, we are talking about a closed set. – user1236 Dec 28 '17 at 21:21
  • @user1236 Got it. This is very helpful. Thank you. – Orca_1 Dec 29 '17 at 04:48