Prove that if $X$ is locally compact, then $\epsilon$ is continuous (with the product topology on the domain of $\epsilon$). Let $X$,$Y$ be topological spaces and $C(X,Y)$ be the set of continuous maps $X \rightarrow Y$. If $K \subset X$ and $U \subset Y$, define $O(K,U)$ to be the set of $f \in C(X,Y)$ such that $f(K) \subseteq U$.
Let the set $$\beta = \{O(K,U) \mid K \subseteq X \text{ is compact }, U \subseteq Y \text{ is open}\}$$ be a subbasis for a topology on $C(X,Y)$. 
With the notation of above, define
$\epsilon:X \times C(X,Y) \rightarrow Y,\epsilon(x,f):=f(x),\forall x\in X,f \in C(X,Y)$
Prove that if $X$ is locally compact, then $\epsilon$ is continuous (with the product topology on the domain of $\epsilon$).
 A: We assume that a space $X$ is locally compact provided any its point $x$  has a local base $\mathcal B_x$ consisting of compact neighborhoods of $x$. This condition holds provided $X$ is a Hausdorff space and each point $x\in X$ has a compact subspace of $X$ containing a neighborhood of $x$ (see Munkres). 
Now let $x$ be any point of $X$, $f$ be any function of $C(X,Y)$, and $U$ be any open neighborhood of the point $\epsilon(x,f)=f(x)$. Since the map $f$ is continuous and the space $X$ is locally compact, there exists a compact neighborhood $K$ of the point $x$ such that $f(K)\subset U$. In particular, $f\in O(K,U)$. Moreover, let $x’\in K$ be any point and $g\in O(K,U)$ be any function. Then $\epsilon(x’,g)=g(x’)\in g(K)\subset U$.
The claim can fail if we require only that each point $x\in X$ has a compact subspace of $X$ containing a neighborhood of $x$. Indeed, let $X=\Bbb Q$ with an attached point $\infty$, endowed with the following topology. A base of each point $x\in\Bbb Q$ is its base in the usual topology and a base at the point $\infty$ consists of subsets $Z$ of $X$ such that $X\setminus Z$ is a finite subset of $\Bbb Q$. It is easy to see that the space $X$ is compact. Put $Y=X$ and let $f\in C(X,Y)$ be the identity map, that is we have $f(x)=x$ for each $x\in X$. In particular, we have $f(0)=0\in (-1,1)\subset\Bbb Q$. We claim that for each neighborhood $V$ of $0$, each compact subset $K$ of $X$, and each open subset $U$ of $X$ exists a point $x\in V$ and a function $g\in (K,U)$ such that $g(x)\not\in (-1,1)$, that is $\epsilon (V\times (K,U))\not\subset (-1,1)$. Indeed, suppose to the contrary that  $\epsilon (V\times (K,U))\subset (-1,1)$. Since $f\in (K,U)$, we have that $K\subset U\subset\Bbb Q$. Since $K$ is compact, $K$ is a closed subset of the space $\Bbb R$ endowed with the usual topology. Since the set $\Bbb Q$ is dense in $\Bbb R$, a closure $V’$ of the set $V$ in the space $\Bbb R$ is a neigborhood of $0$ in $\Bbb R$. Pick any point $y\in V’\setminus\Bbb Q$. Since $K\subset\Bbb Q$, $y\not\in K$. So there exists an neighborhood $W$ (in $\Bbb R$) of $y$ such that $W\cap K=\varnothing$. Since $y\in V’$, there exists a point $x\in V\cap W$. In particular, $x\not\in K$. So there exists a irrational number $\delta>0$ such that $O_x\cap K=\varnothing$, where $O_x=(x-\delta,x+\delta)$. Pick any point $z\in\Bbb Q\setminus (-1,1)$ such that $O_z\cap  (K\cup O_x)=\varnothing$, where $O_z=(z-\delta,z+\delta)$. Now define the map $g:X\to X$ as follows. For each $t\in X\setminus (O_x\cup O_z)$ put $g(t)=t$, and $g$ swaps intervals $O_x$ and $O_z$. The latter means that $g(t)=t+z-x$ for each $t\in O_x$ and $g(t)=t+x-z$ for each $t\in O_z$. It is easy to check that the map $g$ is contnuous. Since $g(t)=t$ for each $t\in K$, we have $g(K)=K\subset U$. So $g\in (K,U)$. But since $x\in O_x$, $g(x)=z\not\in (-1,1)$, whereas $x\in V$, a contradiction.
