Continuous Functions Preserve Compactness
# Statement
Let $K, Y$ be Topological Spaces and suppose $K$ is Compact. Let $f: K \to Y$ be a Continuous Function. Then $f(K)$ is Compact.
# Proof
Let ${V_{i}}{i \in I}$ be a Open Cover of $f(K)$. Then, because $f$ is continuous, ${f^{-1}(V{i})}{i \in I}$ is a collection of Open sets in $K$. Furthermore, because ${V{i}}{i \in I}$ is an Open Cover of $f(K)$, $$K \subset f^{-1}(f(K)) \subset f^{-1}(\bigcup\limits{i \in I} V_{i}) = \bigcup\limits_{i \in I}f^{-1}(V_{i}),$$ so ${f^{-1}(V_{i})}{i \in I}$ is an Open Cover of $K$. Since $K$ is Compact, we can reduce to a finite Open Subcover of $K$, ${f^{-1}(V{i_{j}})}{j \in [n]}$ for some $n \in \mathbb{N}$. Then $$f(K)) \subset f ( \bigcup\limits{j \in [n]} f^{-1} (V_{i_{j}}) )= \bigcup\limits_{j \in [n]}f(f^{-1}(V_{i_{j}})) = \bigcup\limits_{j \in [n]} V_{i_{j}}.$$ Since ${V_{i}}_{i \in I}$ was an arbitrary Open Cover, and we reduced it to finite Open Subcover of $f(K)$, we see that $f(K)$ is Compact. $\blacksquare$