Continuous Image of Compact Space is Compact/Corollary 3/Proof 2
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Corollary to Continuous Image of Compact Space is Compact
Let $S$ be a compact topological space.
Then $f$ attains its bounds on $S$.
Because $f \sqbrk S$ is complete:
- $\sup f \sqbrk S \in f \sqbrk S$
- $\inf f \sqbrk S \in f \sqbrk S$