Continuous Image of Compact Space is Compact/Corollary 3/Proof 1

Proof
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:
 * $\map \inf {f \sqbrk S} \in f \sqbrk S$