Subset of Hilbert Sequence Space with Non-Empty Interior is not 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 $\R$.

Let $H$ be a subset of $\ell^2$ whose interior is non-empty.

Then $H$ is not compact in $\ell^2$.

Proof

Let $x \in H^\circ$, where $H^\circ$ denotes the interior of $H$.

By definition, $H^\circ$ is an open set of $\ell^2$ containing $x$.

Again by definition, $H$ is a neighborhood of $x$.

But from Point in Hilbert Sequence Space has no Compact Neighborhood, $x$ has no compact neighborhood in $\ell^2$.

Thus $H$ cannot be compact in $\ell^2$.

$\blacksquare$

Sources

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