Whether fine topology and uniform topology on C(X,Y) coincide , when metric on Y is bounded

Question

Whether fine topology and uniform topology on C(X,Y) coincide , when metric on Y is bounded

Answer 1

Edit 1:

The fine topology is equal to the uniform topology if, and only if, $X$ is pseudocompact (here, we will suppose that $X$ is Hausdorff and the Tietze extension theorem is valid for $X$).

Proof: Suppose that $X$ is pseoudocompact space. Denote by $\tau_d$ de metric topology and $\tau_f$ the fine topoloy on $C(X,Y)$.

Let $B(f,\epsilon)$ a neighborhood of $f$ in $C(X,Y)$ with respect to the fine topology. As $\epsilon$ is a positive function on the pseoudocompact space $X$, there exists $m>0$ such that $\epsilon(x)\geq m$ for all $x\in X$ (as $X$ is pseoudocompact, then the function $\frac{1}{\epsilon}$ is a positive bounded function and we can obtain such $m$). Thus, the set $B=\{g\in C(X,Y)/\sup_{x\in X}d(f(x),g(x))\leq m\}$ is a neighborhood of $f$ with respect to the metric topology and we have $B\subset B(f,\epsilon)$. Follows that $\tau_f\subset \tau_d$. As always $\tau_d\subset \tau_f$, we conclude that $\tau_d=\tau_f$.

Now, suppose that $\tau_f=\tau_d$ and $X$ is not pseoudocompact.

(1) Suppose that there exists a infinite discret subset $F=\{x_n\} $ of $ X$ such that we can extend continuos function. Then, we can extend on $X$ the function $\epsilon:F\rightarrow \mathbb{R}^+$ given by $\epsilon(x)=\frac{1}{n}$. Since that $Y$ contains a non trivial path, exists a non isoleted point $y_0\in Y$. Thus, for each $n$, we can take $y_n\in B(y_0,\frac{1}{n})$ with $y_n\neq y_0$. Let $h_n\in C(X,Y)$ be the constant function $h_n=y_n$, $n\geq 0$. Now, we will show that $B(h_0,\epsilon)$ is not open in the metric topology. Let $\delta$ any positive real number. Given $n_0$ with $\frac{1}{n_0}<\delta$, we have that $h_{n_0}\in B(h_0,\delta)$, since that $d(h_{n_0}(x),h_0(x))=d(y_0,y_{n_0})<\frac{1}{n_0}<\delta$. Now, given $k$ with $d(y_0,y_{n_0})>\frac{1}{k}$, we have that $ d(h_0(x_k),h_{n_0}(x_k))=d(y_0,y_{n_0})>\frac{1}{k}=\epsilon(x_k)$. Follows that $B(h_0,\delta)\nsubseteq B(h_0,\epsilon)$. As $B(h_0,\delta)$ is a arbitrary neighborhood of $h_0$ with respect the metric topology, we concludes that $\tau_d\neq \tau_f$, a contradiction. Thus, if $\tau_f=\tau_d$ and $Y$ has a non trivial path, then $X$ is pseoudocompact.

(2) The construction of the set $F$. Let $\{U_n\}$ a open cover of $X$ that not admits finite subcover (if every contable open cover of $X$ admit a finite subcover, given continuos function $f:X\rightarrow \mathbb{R}$, as $X\subset \cup f^{-1}(-n,n)$, then $X=f^{-1}(-n_0,n_0)$ for some $n_0$. Follows that $f$ is bounded and, therefore, $X$ is pesoudocompact). Define $V=\cup_{i=1}^nU_n$. Then, $\{V_n\}$ is also a open cover of $X$ that not admit finite subcover. Take $x_1\in V_1$, and for each $n\geq 2$ take $x_n\in V_n-V_{n-1}$.

(2.1) The set $F$ has not limite points outside $F$. Indeed, if $x_{n_k}\rightarrow x$, exists $n_0$ such that $x\in V_{n_0}$. Thus, for large $k$ we have $x_{n_k}\in V_{n_0}$. By construction of $F$, $x_n\notin V_{n_0}$ for large $n$. Then, the sequence $x_{n_k}$ is eventually constant. Therefore, $F$ is closed in $X$.

(2.2) $F$ as subspace of $X$ has the discret topology. Indeed, we have that $\{x_n\}=(V_n -\{x_1,...,x_{n-1}\})\cap F$ is open in $F$ (as $X$ is Hausdorff, $A=\{x_1,...,x_{n-1}\}$ is closed in $X$. Thus, $V_n-A=V_n\cap(X-A)$ is open in $X$). Therefore, any function $\epsilon:F\rightarrow \mathbb{R}^+$ is continuos.

(2.3) As the Tietze extension theorem is valid for $X$ by hypothesis and $F$ is closed in $X$, the continuos function $\epsilon:R\rightarrow \mathbb{R} $ can be extended to a continuos function $X\rightarrow \mathbb{R}^+$. The proof is completed.

For general spaces $X$ I don't know if this characterization is valid.