Closure of Non-Empty Bounded Subset of Metric Space is Bounded

Theorem

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

Let $S \subseteq A$ be bounded in $M$.

Then:

$\map \cl S$ is also bounded in $M$.

where $\map \cl S$ denotes the closure of $S$ in $M$.

Proof

By definition of bounded:

$S$ is bounded if and only if:

$\exists K \in \R: \forall x, y \in M': \map {d_B} {x, y} \le K$

That is, such that $S$ has a diameter $K$.

From Diameter of Closure of Subset is Diameter of Subset, if $S$ has a diameter $K$, then so does $\map \cl S$.

That is, $\map \cl S$ is also bounded.

$\blacksquare$

Sources

• 1975: W.A. Sutherland: Introduction to Metric and Topological Spaces ... (previous) ... (next): $3$: Continuity generalized: topological spaces: Exercise $3.9: 32$