Continuous Image of a Compact Space is Compact

Theorem

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

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

If $T_1$ is compact then so is $T_2$.

That is, compactness is a continuous invariant.

Corollary 1

Compactness is a topological property.

Corollary 2

Any continuous mapping from a compact space to a metric space is bounded.

Corollary 3

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

If $S$ is a compact space, then $f$ attains its bounds on $S$.

Proof

Suppose $\mathcal U$ is an open cover of $f \left({T_1}\right)$ by sets open in $T_2$.

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

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

Because $T_1$ is compact, it has a finite subcover $\left\{{f^{-1} \left({U_1}\right), f^{-1} \left({U_2}\right), \ldots, f^{-1} \left({U_r}\right)}\right\}$.

It follows that $\left\{{U_1, U_2, \ldots, U_r}\right\}$ is a finite subcover of $T_2 = f \left({T_1}\right)$.

$\blacksquare$

Proof of Corollary 1

Follows directly from the above, the definition of topological property and homeomorphism.

$\blacksquare$

Proof of Corollary 2

Follows from the above and Compact Subspace of Metric Space is Bounded.

$\blacksquare$

Proof of Corollary 3

By Corollary 2, $f \left({S}\right)$ is bounded.

By Closure of Real Interval, $\sup \left({f \left({S}\right)}\right) \in \operatorname {cl} \left({f \left({S}\right)}\right)$ and $\inf \left({f \left({S}\right)}\right) \in \operatorname {cl} \left({f \left({S}\right)}\right)$.

By the above and Compact Subspace of Hausdorff Space is Closed, $f \left({S}\right)$ is closed in $\R$.

Hence $\sup \left({f \left({S}\right)}\right) \in f \left({S}\right)$ and $\inf \left({f \left({S}\right)}\right) \in f \left({S}\right)$.

Hence the result.

$\blacksquare$

Alternatively, one could reason as follows:

By the main theorem, $f \left({S}\right)$ is compact.

By the Heine-Borel Theorem (General Case), $f \left({S}\right)$ is complete and totally bounded.

A totally bounded metric space is bounded.

Hence the supremum $\sup f \left({S}\right)$ and the infimum $\inf f \left({S}\right)$ exist.

Because $f \left({S}\right)$ is complete, $\sup f \left({S}\right) \in f \left({S}\right)$ and $\inf f \left({S}\right) \in f \left({S}\right)$.

$\blacksquare$

Sources

• Lynn Arthur Steen and J. Arthur Seebach, Jr.: Counterexamples in Topology (1970)... (previous)... (next): $\text{I}: \ \S 1$: Functions