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Let $F(X, Y)$ be the set of all functions from $X$ to $Y$.

Prove that $Y$ is Hausdorff if and only if $F(X, Y)$ is Hausdorff under the compact-open topology.(Although compact-open topology is generally defined on $C(X, Y)$, the set of all continuous functions from $X$ to $Y$, I would assume that it can be defined in the same way as $C(X, Y)$.)

I already proved 'only if' part, but I'm stuck in 'if' part.

Here is my attempt:

Let $y_1, y_2 \in Y$ and $x \in X$, then there are two functions $f_1, f_2 \in F(X, Y)$ such that $f_1 (x)=y_1, f_2 (x) = y_2$. Since $F(X, Y)$ is Hausdorff, there are open $X_1, X_2$ such that $f_1 \in X_1, f_2 \in X_2, X_1 \cap X_2=\emptyset$. Also, let $Y_1 = X_1 \cap S(x, V_1), Y_2 =X_2\cap S(x, V_2)$, where $V_1, V_2$ are neighborhoods of $y_1, y_2$, respectively, and $S(x, V_i) = \{f \in F(X, Y): F(x)\in V_i\}$ for each $i$. Then it is obvious that $f_1 \in Y_1, f_2 \in Y_2, Y_1 \cap Y_2 =\emptyset$.

I am stuck here and can't find ways to proceed. What should I do?

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    You need to assume $X$ is nonempty for this to be true. – 2017-02-08

2 Answers 2

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Let $y,z\in Y.$ Then, choose $f,g\in F(X,Y)$ such that $f(X)=\left \{ y \right \}$ and $g(X)=\left \{ z \right \}.$ Then, as $F(X,Y)$ is Hausdorff, there are compact sets $C_i, K_j\in P(X)$ and $U_i,V_i\in \tau_Y\ $ with $1\le i\le n$ and $1\le j\le m$ such that $f\in \bigcap_i S(C_i,U_i):=S_f;\ g\in \bigcap_j S(C_j,V_j):=S_g$ and $S_f\cap S_g=\emptyset. $

Now, take $\tilde U=\cap_i U_i$ and $\tilde V=\cap_j V_j.$ These are open and non-empty because the former contains $y$ and the latter contains $z$. Summarizing now, we have $y\in \tilde U$ and $z\in \tilde V\ $ and $\tilde U\cap \tilde V=\emptyset,\ $ and so $Y$ is Hausdorff, as claimed.

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    It would be better to use two disjoint basis elements rather than subbasis elements($S(C_f, U), S(C_g, V)$) because two elements of subbasis in Hausdorff space need not be disjoint. – 2017-02-09
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    Anyway, changing the subbasis elements to basis elements and arguing as the same, I solved it. Thanks. – 2017-02-09
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    I revise the last assertion of first comment as: 'Hausdorff implies that we can always find two disjoint "basis" elements containing each point, but does not imply that we can always find such "subbasis" elements." One of the counterexamples is $\mathcal S = \{ (a, \infty), (-\infty, b)|a, b \in \mathbb R\}$, which is a subbasis of standard topology in $\mathbb R$. For a bounded open set, we cannot find a subbasis element contained in it. – 2017-02-09
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    Yes you are right. I fixed it. Thanks.... – 2017-02-09
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A general way to look at this problem: For any $y \in Y$, let $c_y$ be the function in $F(X,Y)$ that is constantly $y$. This is always a continuous function (regardless of topology). Define $i: Y \rightarrow F(X,Y) ,i(y) =c_y$.

Also, whatever the (non-empty) compact set $C$, $S(C, U) \cap i[Y] = i[U]$ (every constant function that maps $C$ into $U$ has value in $U$, and if its value is in $U$, it maps $C$ into $U$ as well). Also $i^{-1}[C(\{x\},U)] = U$, so $i$ is continuous.

This implies that $i$ embeds $Y$ into $F(X,Y)$ as a subspace ($i$ is open and a continuous bijection between $Y$ and $i[Y]$), and Hausdorffness is hereditary.

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    very nice indeed. – 2017-02-09
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    @chilangoincomprendido Glad you like it! – 2017-02-09