Closed and Bounded Subspace is not necessarily Compact

Theorem

Let $M = \struct {A, d}$ be a metric space.

Let $H \subseteq A$ be a subset of $A$.

Let $M_H = \struct {H, d_H}$ be the subspace of $M$ induced by $d$.

Let $H$ be closed and bounded.

Then it is not necessarily the case that $M_H$ is compact.

Proof

Proof by Counterexample:

Let $A = \openint 0 1$ be the open unit interval.

From Open Real Interval is not Compact, $\openint 0 1$ is not a compact space.

Let $H = \openint 0 1$.

Then $H \subseteq A$.

From Underlying Set of Topological Space is Clopen, $\openint 0 1$ is both closed and open in $\openint 0 1$.

Thus $H$ is closed and bounded but not compact.

$\blacksquare$

Sources

• 1975: W.A. Sutherland: Introduction to Metric and Topological Spaces ... (previous) ... (next): $5$: Compact spaces: $5.7$: Compact Subspaces of $\R^n$: Remark $5.7.2$