# Pointwise convergence topology.

I would like to know if the following statement is true or false.

Let $X$ be a compact Hausdorff space and an infinite set. If $X$ has a topology strictly weaker than the discrete topology, then the compact convergence topology is strictly finer than the pointwise convergence topology.

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For clarity, let me do the case where $X$ is metrizable first. Choose a metric $d$ on $X$ and a limit point $x$ (such a point exists because $X$ is not discrete). Choose a sequence $x_{n} \to x$ with $x_{n} \neq x$. Using the metric, it is easy to construct a continuous function $0 \leq f_{n} \leq 1$ such that $f_{n}(x_n) = 1$ and $f_{n}(y) = 0$ for all $y$ with $d(y,x_{n}) \geq \frac{1}{2}d(x_{n},x)$. Let us check that $f_{n} \to 0$ pointwise: Since $f_{n}(x) = 0$ for all $n$, we need only consider the case $y \neq x$. For $n$ large enough we have $d(x,x_{n}) \leq \frac{1}{2} d(x,y)$ so that $d(x_{n},y) \geq \frac{1}{2} d(x_n,x)$ by the triangle inequality. It follows that for such $n$ we have $f_{n}(y) = 0$, so the sequence $(f_n)$ converges indeed pointwise to zero. As $f_{n}(x_n) = 1$, the sequence $(f_n)$ cannot converge uniformly and none of its subsequences or subnets can.
Since $X$ is not discrete, it has a limit point $x$. Choose a neighborhood filter $\mathcal{U}$ of $x$ and for $U \in \mathcal{U}$ choose $x_{U} \in U \smallsetminus \{x\}$. Then the net $(x_{U})_{U \in \mathcal{U}}$ converges to $x$. By Urysohn's lemma, we may choose a continuous function $0 \leq f_{U} \leq 1$ such that $f_{U}(x) = 0$, $f_{U}(x_{U}) = 1$ and with support inside $U$ (Exercise!). Clearly, $f_{U} \to 0$ pointwise but not uniformly.