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

On page 186 of Munkres' Topology

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

Uniform topology is defined as topology induced by uniform metric $p$ which is stated as follows.

For any two points $a$, $b$ in $\mathbb{R}$, $$\bar{d}(a,b) = \text{min}\{|a-b|,1\}$$ For any two points $x$, $y$ in $[0,1]^{\omega}$ $$x = \{x_i:i<\omega\}$$ $$y=\{y_j:j<\omega\}$$ $$p(x,y) = \text{sup}\{\bar{d}(x_i,y_i):i<\omega\}$$

Is there some method to gain intuition on infinite product of topological space in various topologies? Could we visualize it as the finite product case?

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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.
Thank you for your answer. Allow me to take the liberty to guess "$C$ compact" in the first line is a typo, which should be "$C$ hausdorff" as stated in Munkres' Topology, Theorem 29.2. If $C$ is compact, it's automatically local compact. – Metta World Peace Feb 3 '13 at 15:26
What you say is correct of course, but part of a chain of conclusions leading to a contradiction: if l.c., then for any compact set there is a closed (and so compact) set within it; but the second part shows that such a set cannot be compact. So, unrolling to the start, the space is not l.c. (at $0$). – gnometorule Feb 3 '13 at 16:17
I see. It's me who misunderstood your statement. Your formulation of local compactness is different from the one I'm referring to. I thought $C$ is the underlying topological space that we try to define local compactness on. Thank you very much. – Metta World Peace Feb 3 '13 at 16:28