# Continuous Image of Compact Space is Compact/Corollary 3

## Corollary to Continuous Image of Compact Space is Compact

Let $S$ be a compact topological space.

Let $f: S \to \R$ be a continuous real-valued function.

Then $f$ attains its bounds on $S$.

## Proof 1

By Continuous Image of Compact Space is Compact: Corollary 2, $f \sqbrk S$ is bounded.

By Supremum of Bounded Above Set of Reals is in Closure:

- $\map \sup {f \sqbrk S} \in \map \cl {f \sqbrk S}$

and by Infimum of Bounded Below Set of Reals is in Closure:

- $\map \inf {f \sqbrk S} \in \map \cl {f \sqbrk S}$

From Continuous Image of Compact Space is Compact, $f \sqbrk S$ is compact in $\R$.

From Non-Closed Set of Real Numbers is not Compact, it follows from the Rule of Transposition that $f \sqbrk S$ is closed in $\R$.

From Closed Set equals its Closure:

- $f \sqbrk S = \map \cl {f \sqbrk S}$

Hence the result that:

- $\map \sup {f \sqbrk S} \in f \sqbrk S$

and:

- $\map \inf {f \sqbrk S} \in f \sqbrk S$

$\blacksquare$

## Proof 2

By Continuous Image of Compact Space is Compact, $f \left({S}\right)$ is compact.

From Compact Metric Space is Complete and Compact Metric Space is Totally Bounded, $f \left({S}\right)$ is complete and totally bounded.

A Totally Bounded Metric Space is Bounded.

Hence both the supremum and the infimum of $f \left({S}\right)$ exist in $\R$.

Because $f \left({S}\right)$ is complete:

- $\sup f \left({S}\right) \in f \left({S}\right)$

and:

- $\inf f \left({S}\right) \in f \left({S}\right)$

$\blacksquare$