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
Compactness is a topological property.

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