How to show that $[0,1]^{\omega}$ is not locally compact in the uniform topology?

Solution 1:

Use the equivalent definition of local compactness: if $x \in U \subset C$, $U$ open, $C$ compact, then there must be a ball $B_{\epsilon}(x)$ whose closure $\bar{B}$ is contained in $U$. Apply this to $x = 0$. Then the closure of such a ball is compact as it is a closed subset of a compact set in a metric space. But it is not: look at the set $A = \{ x_i = (0, \dots, 0, \epsilon, 0, \dots ), i \in \mathbb{N} \} \subset \bar{B}$, with the $\epsilon$ in position $i$. It has no limit point $x$ in $A$ ($x$ cannot contain a coordinate in $(0, \epsilon)$ or a small enough ball around $x$ will not contain any $y \in A$; and in the other case, it is at distance $0$ or $\epsilon$ from any point in $A$ - think about that!). So $A$ is not limit-point compact, which in metric spaces is equivalent to being compact. Contradiction. 

Solution 2:

Let U be a nbd of 0. By a slight modification in Munkres 29.2, there exist a basic open set $V = [0, \epsilon)^{\omega}$ of zero such that $V \subset \overline{V} \subset U$, where $\overline{V}$ is compact.

We can show that $\overline{V} = [0,\epsilon]^{\omega}$, which is not compact as $ \{ U_n = [0,\epsilon \frac{(n-1)}{n})^{\omega} \}_{n \ge 0}$ is an open cover which doesn't admit a finite subcover for $\overline V$. Hence $[0, 1]^{\omega}$ is not locally compact at 0.