Continuous Image of Compact Space is Compact

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Theorem

Let $T_1$ and $T_2$ be topological spaces.

Let $f: T_1 \to T_2$ be a continuous mapping.


If $T_1$ is compact then so is its image $f \sqbrk {T_1}$ under $f$.


That is, compactness is a continuous invariant.


Corollary 1

Compactness is a topological property.


Corollary 2

A continuous mapping from a compact topological space to a metric space is bounded.


Corollary 3

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

Suppose $\mathcal U$ is an open cover of $f \sqbrk {T_1}$ by sets open in $T_2$.

Because $f$ is continuous, it follows that $f^{-1} \sqbrk U$ is open in $T_1$ for all $U \in \mathcal U$.

The set $\set {f^{-1} \sqbrk U: U \in \mathcal U}$ is an open cover of $T_1$, because for any $x \in T_1$, it follows that $\map f x$ must be in some $U \in \mathcal U$.

Because $T_1$ is compact, it has a finite subcover $\set {f^{-1} \sqbrk {U_1}, f^{-1} \sqbrk {U_2}, \ldots, f^{-1} \sqbrk {U_r} }$.

It follows that $\set {U_1, U_2, \ldots, U_r}$ is a finite subcover of $f \sqbrk {T_1}$.

$\blacksquare$


Sources