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

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Aiming for a contradiction, suppose $f$ is not bounded.

Let $g : X \to \R$ be defined by $\map g x = \norm {\map f x}$.

From Norm is Continuous and Composite of Continuous Mappings is Continuous, it follows that $g$ is continuous.

For all $n \in \N$, set $A_n := g^{-1} \sqbrk{ \map {B_n}{ 0 } }$, where $\map {B_n}{ 0 }$ denotes the open ball in $\R$ with radius $n$ and center $0$.

From Open Ball is Open Set in Normed Vector Space and the definition of continuity, it follows that all $A_n$ are open in $X$.

For all $x \in X$, we have $x \in A_n$ for all $n \ge \norm{ \map f x }$.

It follows that $\ds X \subseteq \bigcup_{ n \mathop \in \N } A_n$.

By definition of open cover, $\set {A_n}_{n \in \N}$ is an open cover of $X$.

As $X$ is compact, there exist $m \in \N$ and $n_1, \dots, n_m \in \N$ such that $\set {A_{n_k} }_{k \mathop \in \set{ 1, \ldots , m } }$ is a finite subcover of $X$.

It follows that $\ds X \subseteq \bigcup_{ k \mathop \in \set{ 1, \ldots , m } } A_{n_k}$.

As $f$ is not bounded, there exists $x_0 \in X$ such that $\norm {\map f {x_0} } > \max \set {n_1, \ldots, n_m}$.

It follows that $x_0 \notin \ds \bigcup_{ k \mathop \in \set{ 1, \ldots , m } } A_{n_k}$ , which is a contradiction.

$\blacksquare$