# Munkres proof that $\mathbb{R}^{ \omega }$ is not locally compact (clarification needed)

From section 29 of Munkres' topology. The bar on top of the 'B' indicates that it is the closure of B. The example above assumes that $\mathbb{R}^{\omega}$ is in the product topology with respect to the standard topology of $\mathbb{R}$.

My understanding: To derive a contradiction for the local compactness of $\mathbb{R}$, the proof uses the fact that a closed set of a compact space is compact. Furthermore, if X is a compact space in Y and Y is a compact space in Z, then X is a compact space in Z. If we show that X (in this case, the closure of B) is not compact in Z ($\mathbb{R}^{\omega}$), then we have proven that $\mathbb{R}^{\omega}$ is not locally compact. This part follows from the fact that $\mathbb{R}^{\omega}$ is not compact.

My logic gap: The proof that none of $\mathbb{R}^{\omega}$'s basis elements is contained in a compact subspace seems to rely on the assumption that, if B lies in compact subspace C, then the closure of B also lies in C. To me, this seems to be the only way the above logic can be applied to arrive at the subsequent contradiction, but I can't see any obvious way to prove this.

Am I thinking about this wrong? Any proofs or guiding hints would be much appreciated.

• Compact subsets in euclidean space are closed, and the closure is the smallest closed set above the set. – Randall Jul 14 '18 at 3:22
• $\mathbb{R}^{\omega}$ is not Euclidean. However, Compact subsets are closed in any Hausdorff space, and so Randall's comment "works". – David Hartley Jul 14 '18 at 7:17
• @DavidHartley You’re right. I was arguing in a coordinate slot but that’s not the right way. – Randall Jul 14 '18 at 12:43

As noted in the comments, the closure of $B$ is the intersection of all closed sets containing $B$; since $\mathbb R^\omega$ is Hausdorff, the compact set $C$ is closed; hence the closure of $B$ is contained in it.