Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

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

Let $(X,d)$ be a compact metric space and let $S= \{f \in C(X):\|f\|\le 1\}$ be the closed unit ball of $C(X)$. Show that if $X$ is an infinite set then $S$ will not be compact.

share|cite|improve this question
Thanks Michael ,please let me know is this work for the metric space also? – Athar Raheel Ahmad Oct 4 '12 at 22:36
The space of continuous functions on a compact metric space with the sup-norm is an infinite dimensional normed space. Finite dimensional subspaces are closed. You can use this to find a countable family of disjoint open balls with the same radius in the unit ball. By covering the rest with very small balls, you get an open cover without a countable subcover. – Michael Greinecker Oct 4 '12 at 22:41
hmmm it seems to be use full,can you explain a bit more so that i can easily do my work on it? – Athar Raheel Ahmad Oct 5 '12 at 16:40

Let $a$ be an accumulation point of $X$. Let $f_n:X\to\mathbb{R}$ be given by $$f_n(x)=\max\big\{1-n~|a-x|,0\big\}.$$

You can verify that $\|f_n\|=1$ for all $n$. If the sequence would have a convergent subsequence, it would converge to a discontinuous function.

share|cite|improve this answer
which kind of subsequence could I construct so that it could converge to a discontinuous function – Athar Raheel Ahmad Oct 5 '12 at 18:24
@AtharRaheelAhmad Every subsequence would converge to a function that has the value $1$ at $a$ and $0$ everywhere else. That $a$ is an accumulation point guarantees that such a fnction cannot be continuous and that the sequence cannot be eventually constant. – Michael Greinecker Oct 5 '12 at 23:14

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.