# $\Bbb{R}^\omega$ is not Locally Compact

Here is example 2 in chapter 3.29 in Munkres:

The space $\Bbb{R}^\omega$ [defined as $\Bbb{R} \times \Bbb{R} \times \Bbb{R} \times ...$ endowed with the product topology] is not locally compact...For if $B = (a_1,b_1) \times ... \times (a_n,b_n) \times \Bbb{R} \times ...$ were contained in a compact subspace, then its closure $\overline{B}$ would be compact, which it is not.

Would the following be a sufficient way of filling in the details? Recall that every compact subspace of a Hausdorff space ($\Bbb{R}^\omega$ is in fact metrizable) is necessarily closed. Thus this compact space containing $B$ must be closed. Now since the closure of $B$ is the smallest closed set containing $B$, the closure of $B$ must be contained in this compact set also. Again, recall that every closed subspace of a compact subspace is itself compact, making $\overline{B}$ compact too. However, if it were compact, then its image under the continuous map $\pi_{n+1}$, the projection of the $(n+1)$-th factor, would also be compact, but this is a contradiction since the image is $\Bbb{R}$, which is known not to be compact.

• You want to justify that just using a base set B is sufficient. – William Elliot Jul 22 '17 at 21:24

All the ideas are there, but I'd streamline it a bit like this:

Suppose that for some $x \in \mathbb{R}^\omega$ we have that it is contained in a compact set $C$ that contains a neighbourhood $O$ (open, WLOG) of $x$ (his definition). Then there is a basic open set $B$ as above such that $x \in B \subseteq O \subseteq C$. Indeed $C$ is closed as $\mathbb{R}^\omega$ is Hausdorff, so $\overline{B} \subseteq \overline{C} = C$, and in any space, a closed subset of a compact subset is compact, so $\overline{B}$ is compact. Contradiction as its $(n+1)$-th coordinate (in your notation) indeed is $\mathbb{R}$ which is not compact.

So $\mathbb{R}^\omega$ is not locally compact at any point.

Proposition: The product space $\prod X_\alpha$ is locally compact if and only if each $X_\alpha$ is locally compact and $X_\alpha$ is compact for all but finitely many values of $\alpha.$
Proof. Recall that the projections $\pi_\beta:\prod X_\alpha\to X_\beta$ are continuous open maps. Suppose that $\prod X_\alpha$ is locally compact. Let $\mathbf{x} \in \prod X_\alpha.$ Then there exist a compact subspace $C$ of $\prod X_\alpha$ containing a neighborhood of $\mathbf{x}.$ This neighborhood contains a basis element for $\prod X_\alpha$ containing $\mathbf{x},$ say $U=\prod U_\alpha$ where $U_\alpha=X_\alpha$ for all but finitely many indices, say $\alpha_1, \ldots, \alpha_n.$ Let $\beta \neq \alpha_i$ for all $i=1,\ldots,n.$ Then $\pi_\beta(C)$ is compact and contains $\pi_\beta(U)=X_\beta,$ so $\pi_\beta(C)=X_\beta$ and $X_\beta$ is compact. Hence $X_\alpha$ is compact for all but finitely many values of $\alpha.$ Now we prove that $X_{\alpha_i}$ is locally compact for $i=1,\ldots,n.$ Let $x \in X_{\alpha_i}.$ Let $\mathbf{x} \in \prod X_\alpha$ be such that $x_{\alpha_i}=x.$ There is a compact $C$ containing a basis neighborhood $U$ of $\mathbf{x}.$ Then $\pi_{\alpha_i}(C)$ is a compact subspace of $X_{\alpha_i}$ containing the neighborhood $\pi_{\alpha_i}(U)$ of $x.$ So the spaces $X_{\alpha_i}$ are (at least) locally compact.
Now suppose that each $X_\alpha$ is locally compact and $X_\alpha$ is compact for all but finitely $\alpha.$ Let $\mathbf{x} \in \prod X_\alpha.$ For each $\alpha,$ there exists a compact subspace $C_\alpha$ of $X_\alpha$ containing a neighborhood $U_\alpha$ of $x_\alpha.$ By hypothesis, for all but finitely many indices we can assume that $C_\alpha=U_\alpha=X_\alpha.$ By the Tychonoff theorem, $\prod C_\alpha$ is compact, and it contains the neighborhood $\prod U_\alpha$ of $\mathbf{x}.$ Hence $\prod X_\alpha$ is locally compact.
Any neighbourhood $V$ of $x$ is of the form $x+V_0$ where $V_0$ is a neighbourhood of $0$. But any neighbourhood of $0$ contains a closed elementary one $W_0$ which is the product of a hypercube (of finite dimension $d$) and the product of full lines ($\epsilon>0$ and $F\subset \omega$ finite with $d$ elements) $$W_0=(\prod_{i\in F} [-\epsilon,\epsilon])\times \mathbb{R}^{\omega - F}$$ hence contains an infinite dimensional subspace which is not compact.