Continuous Function on Compact Space is Bounded

From ProofWiki
Jump to: navigation, search


Let $\left({X, \tau}\right)$ be a topological space.

Let $\left(Y,\left\|\cdot\right\|\right)$ be a normed vector space.

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

Then $f$ is bounded.


Suppose that $f$ is not bounded.

Let $A_n=\left\{x\in X:\left\|f\left(x\right)\right\|<n\right\}$ for every $n\in\mathbb N$.

Then each $A_n$ is open, since $A_n=\left\|f\right\|^{-1}\left(\left(-n,n\right)\right)$.

Moreover, $X=\bigcup A_n$.

Since $X$ is compact, there are $n_1,n_2,\dots,n_m\in\mathbb N$ such that $X=\bigcup A_{n_k}$.

However, since $f$ is not bounded, there is $x\in X$ such that $\left\|f\left(x\right)\right\|\geq\max\left\{n_1,n_2,\dots,n_m\right\}$, which is a contradiction.