Exercise 29.4

Proof. Suppose $X={[0,1]}^{\omega }$ is locally compact, and in particular at $0$. Then, there exists $C$ compact that contains a neighborhood $U\ni 0$. There exists $X\cap B_{\overline{\rho }}(0,𝜖)\subseteq U$; we see that ${\{0,𝜖∕3\}}^{\omega }\subseteq X\cap B_{\overline{\rho }}(0,𝜖)$. ${\{0,𝜖∕3\}}^{\omega }$ is closed since

in the product topology by Theorem $19.5$, which is finer than the uniform topology, and so it is compact by Theorem $26.2$ since it is a closed subset of $C$ compact, i.e., limit point compact by Theorem $28.2$.

We claim this is a contradiction. Consider $x\in X$, and the ball $X\cap B_{\overline{\rho }}(x,𝜖∕9)$. Note that the distance between any two distinct points of ${\{0,𝜖∕3\}}^{\omega }$ is $𝜖∕3$, and so since the diameter of $X\cap B_{\overline{\rho }}(x,𝜖∕9)$ is at most $2𝜖∕9$, $X\cap B_{\overline{\rho }}(x,𝜖∕9)$ contains at most one point of ${\{0,𝜖∕3\}}^{\omega }$. Thus, ${\{0,𝜖∕3\}}^{\omega }$ contains no limit points, and so is not limit point compact, a contradiction. □