Given $K$ is compact and $f: K \rightarrow (0, +\infty)$ continuous then $\exists y > 0: \ f(x) > y ,~ \forall x \in K$

Question

Since $f$ is continuous and $K$ compact, we know that $f(K) \subseteq (0, +\infty)$ is compact, so it is closed and bounded. As it is bounded. Does it mean that there is some $y > 0$ such that $f(K) > y$?

Answer 1

Since $f(K)$ is closed, it contains its Infimum, let's call its infimum $a=\inf f(K)$, so we have $a\in f(K)$. We are given $0\notin f(K)$, hence $0<a$. You can choose $y=\frac{a}2>0$

$$0<y=\frac{a}2<a=\inf f(K)\le f(x),~~\forall x\in K$$

Answer 2

The claim is true, but closedness (in $\Bbb R$) is the important thing. $f(K)$ doesn’t contain zero (considered as a subset of $\Bbb R$) so its complement in $\Bbb R$ must be an open neighbourhood of zero, that is, the complement contains some $(-\delta,\delta)$ for a $\delta>0$. But then it is clear $f(x)>\delta$ for all $x\in K$.

Boundedness is not important here.

Answer 3

Suppose not. Then we can find a sequence $\{x_j\}_{j\in\mathbb{Z}^+}$ in $K$ such that

$$f(x_j)\leq\frac{1}{j}$$

for all $j\in\mathbb{Z}^+$. By sequential compactness (I'm assuming $K$ is at least a metric space, such as a subset of $\mathbb{R}^n$, since you haven't stated anything else) we can find a convergent subsequence $\{x_{j_k}\}_{k\in\mathbb{Z^+}}$ with limit $x\in K$. But then, by continuity,

$$f(x)=\lim_{k\to\infty}f(x_{j_k})\leq\lim_{k\to\infty}\frac{1}{j_k}=0,$$

contradicting that $f(x)\in(0,\infty)$.