Let $X$ be a compact metric space. Prove that there exists $ \epsilon>0$ such that $\left \|f(x) \right \| \geq \epsilon$ for every $x \in X$.

Question

Let $X$ be a compact metric space and let $f: X \rightarrow \mathbb{R}^{p}$ be a continuous aplication such that $f(x) \neq 0$ for every $x \in X$. Prove that there exists $\epsilon >0$ such that $\left \|f(x) \right \| \geq \epsilon$ for every $x \in X$.

My attempt

I guess that the best way to prove that is by contradiction. So the statement would be "Prove that there exists $\epsilon >0$ such that $\left \|f(x) \right \| < \epsilon$ for every $x \in X$"

But I don't know what else to do. Any suggestion?

Answer 1

Let $m=\inf_{k\in K}\|f(x)\|$. Then there is a minimizing sequence $x_k$ such that $\|f(x_k)\|<m+\frac1k$. Now that sequence contains a convergent subsequence by compactness, and $f$ is continuous, and the value at the limit point is non-zero,...

Answer 2

Since $f$ is continuous and the norm is continuous, so is $||f|| : X \to \mathbb R$.

By a theorem that we know (if you do not, reply in the comments), every function on a compact set attains it's infimum within the set. That is, there exists $x_0 \in X$ such that $\displaystyle\inf_{y \in X} ||f(y)|| = ||f(x_0)||$. Note that $||f(x_0)|| \neq 0$, so that $||f(x_0)||>0$.

This means that $\displaystyle\inf_{y \in X} ||f(y)|| = \delta> 0$. Hence, if $\epsilon = \frac \delta 2$, it follows that $||f(y)|| > \epsilon$ for all $y \in X$.

Answer 3

$f(X)$ is compact and $0 \notin f(X)$. Since $f(X)^c$ is open, there is some $\epsilon>0$ such that $B(0,\epsilon) \cap f(X) = \emptyset$.

In particular, $\|f(x)\| \ge \epsilon$ for all $x\in X$.