Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question

Yes, that's true. To see this, it suffices to exhibit a net of continuous functions that converges pointwise to zero but not uniformly.

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.

The general case follows the same idea, replacing the use of the metric by Urysohn's lemma. Here's the construction:

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.

share|cite|improve this answer
By the way: an infinite compact space must have a topology strictly weaker than the discrete topology, since the discrete topology is clearly not compact, so the statement of your question is a bit redundant. – t.b. May 31 '11 at 10:40

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.