Hilbert Sequence Space is not Sigma-Compact

Theorem

Let $A$ be the set of all real sequences $\sequence {x_i}$ such that the series $\ds \sum_{i \mathop \ge 0} x_i^2$ is convergent.

Let $\ell^2 = \struct {A, d_2}$ be the Hilbert sequence space on $\R$.

Then $\ell^2$ is not $\sigma$-compact.

Proof

By Compact Subset of Hilbert Sequence Space is Nowhere Dense, a compact subset of $\ell^2$ is nowhere dense in $\ell^2$.

We have that Hilbert Sequence Space is Complete Metric Space.

Hence $\ell^2$ is non-meager.

This article, or a section of it, needs explaining. In particular: Find a result that shows a complete metric space is non-meager. S&S seem to imply that it follows. You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by explaining it. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{Explain}} from the code.

It follows that $\ell^2$ is not $\sigma$-compact.

This article, or a section of it, needs explaining. In particular: How? You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by explaining it. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{Explain}} from the code.

$\blacksquare$

Sources

• 1978: Lynn Arthur Steen and J. Arthur Seebach, Jr.: Counterexamples in Topology (2nd ed.) ... (previous) ... (next): Part $\text {II}$: Counterexamples: $36$. Hilbert Space: $4$