Continuous Image of Compact Space is Compact

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 $$T_2$$.

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 $$f \left({T_1}\right)$$.

Proof of Corollary 1
Follows directly from the above, the definition of topological property and homeomorphism.

Proof of Corollary 2
Follows from the above and Compact Subspace of Metric Space is Bounded.

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.