Continuous Image of Compact Space is Compact/Corollary 3/Proof 1
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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
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:
- $\map \inf {f \sqbrk S} \in f \sqbrk S$
$\blacksquare$
Sources
- 1953: Walter Rudin: Principles of Mathematical Analysis ... (previous) ... (next): $4.16$
- 1975: W.A. Sutherland: Introduction to Metric and Topological Spaces ... (previous) ... (next): $5$: Compact spaces: $5.5$: Continuous maps on compact spaces: Corollary $5.5.4$