**Theorem**: Let $\left(X, \mathcal{T}_X\right)$ and $\left(Y, \mathcal{T}_Y\right)$ be [[Topology|topological spaces]]. If $f: X \rightarrow Y$ is [[Continuity|continuous]] and $X$ is [[Compact topological spaces|compact]] then $f(X)$ is compact. **Proof**: Let $\mathcal{U}$ be an open cover of $f(X)$. Then, assuming $f$ is continuous, the sets $f^{-1}(U), U \in \mathcal{U}$, are open, and form a cover of $X$. Since $X$ is compact, there is a finite subcover, $\left\{f^{-1}\left(U_1\right), \ldots, f^{-1}\left(U_n\right)\right\}$, of $X$. Now, every $y \in f(X)$ is given by $y=f(x)$ for some $x \in X$. We have $x \in f^{-1}\left(U_j\right)$ for some $j$, so $y=f(x) \in U_j$. So $\left\{U_1, \ldots, U_n\right\}$ is an open cover of $f(X)$. **Remark**: This shows that compactness is a topological property: if $X$ is compact and $f: X \rightarrow Y$ is homeomorphism then $Y=f(X)$ and so $Y$ is compact.