Continuous Function on Compact Space is Bounded

Theorem
Let $\struct {X, \tau}$ be a compact topological space.

Let $\struct {Y, \norm {\, \cdot \, } }$ be a normed vector space.

Let $f: X \to Y$ be continuous.

Then $f$ is bounded.

Proof
$f$ is not bounded.

Let $A_n = \set {x \in X: \norm {\map f x} < n}$ for every $n \in \N$.

Let $\map {B_n}{ \mathbf 0 }$ denote the open ball with radius $n$ and center $\mathbf 0$.

Then each $A_n$ is open, since $A_n = \norm f^{-1} \sqbrk { \map {B_n}{ \mathbf 0 } }$, i.e., the preimage of $\map {B_n}{ \mathbf 0 }$ under $x \mapsto \norm{f(x)}$.

Moreover, $\ds X = \bigcup_{ n \mathop \in \N } A_n$.

By definition of compactness, there exist $m \in \N$ and $n_1, n_2, \dots, n_m \in \N$ such that $\ds X = \bigcup_{ k \mathop \in \set {1, \ldots, m} } A_{n_k}$.

However, since $f$ is not bounded, there exists $x \in X$ such that $\norm {\map f X} \ge \max \set {n_1, n_2, \dots, n_m}$, which is a contradiction.