# How to prove that these two topologies are equal when X is discrete?

I have an exercise, but first two definitions:

topology of pointwise convergence:

Given a point x of the set X and an open set U of the space Y, let:

$S(x,U)=\{f | f \in Y^X \text{ and } f(x) \in U\}$. The sets $S(x,U)$ are a subbasis for a topology on $Y^X$, which is called the topology of pointwise convergence.

topology of compact convergence:

Let $(Y,d)$ be a metric space; let X be a topological space. Given an element $f \in Y^X$, a compact subspace C of X, and a number $\epsilon > 0$, let $B_C(f,\epsilon)$ denote the set of all those elements g of $Y^X$ for which:

$\sup\{d(f(x),g(x)) | x \in C\}<\epsilon$.

The sets $B_c(f,e)$ for a basis for a topology on $Y^X$. It is called the topology of compact convergence.

a. Show that as the topologies are concerned pointwise convergence $\subset$ uniform convergence.

b. If X is discrete show that the topologies are equal.

my attempt and questions:

a. Assume we have $S(x,U)$. Let $f \in S(x,U)$, then $f(x) \in U$, since U open, $B_Y(f(x),\epsilon)\subset U$. Since $\{x\}$ is compact, we have $B_{\{x\}}(f,\epsilon)\subset S(x,U)$, hence the result follows, since in for any subbasis element and any function in that subbasis element, there is an open set in the topology of compect convergence, that contains the function, and is contained in the subbasis element. Hence the topology of uniform convergence is finer. Is this correct?

b. This is the one I really struggle with. Assume that X is discrete. Then any set is open, and hence, any set is closed. Assume that I have the open ball in the compact convergence topology: $B_C(f,\epsilon)$. Let $g \in B_c(f,\epsilon)$. I need to find a basis element in the pointwise convergence topology, containing g, and this basis element must be contained in $B_C(f,\epsilon)$. Can you guys please help me here?, I do not know how to continue.

• (a) seems quite correct. Commented Nov 2, 2015 at 7:00

Re (b): If $X$ is discrete, $C \subseteq X$ is compact iff it's finite. ($\Leftarrow$: obvious; $\Rightarrow: \{\{c\}\mid c \in C\}$ is an open cover of $C$, which has a finite subcover.)
Say $C = \{c_0, \dots, c_{n-1}\}$. For $\varepsilon > 0$, $$B_C(f,\varepsilon) = \{g\in Y^X \mid (\forall i<n)\, d(f(c_i),g(c_i))<\varepsilon\}\text{.}$$ For $i<n$, let $U_i = \{y\in Y\mid d(y,f(c_i))<\varepsilon\}$ be the $\varepsilon$ neighborhood of $f(c_i)$ in $Y$, and let $S_i = S(c_i, U_i)$, so $S_i = \{g \in Y^X\mid d(g(c_i),f(c_i)) < \varepsilon\}$. Clearly, $$\bigcap_{i<n}S_i = B_C(f,\varepsilon)\text{.}$$ is a basis element in the pointwise convergence topology.
• Thank you very much, it was a good answer! :) But I think you mean $Y^X$, not $X^Y$, but this is just a typo. Commented Nov 2, 2015 at 13:52
• I do mean $Y^X$, thanks. Originally I had "$g\colon X\to Y$" but then thought, oh I should use your notation... but it seems I only halfway through the conversions. Fixed. Commented Nov 2, 2015 at 16:32