For a compact K, prove that there exists $m > 0$ such that $f(x) ≥ m$ for all $x$ in K Pasted image for time's sake...

I'm not sure how to approach this besides using the case $f(x) = x$ on the set $(0,1)$. Would I use 1/m?
 A: By Weierstrass' theorem, $f$ attains a global minimum in $K$, say $f(x_m)=\min_{x \in K} f(x)$. By assumption, $f(x_m)>0$, and hence $f(x) \geq f(x_m)=m>0$ for every $x \in K$.
A: For each $x \in K$, let $\epsilon_x = f(x)/2 > 0$. Since $f(x) > \epsilon_x$, there is a neighbourhood $U_x$ of $x$ such that
$$
f(y) > \epsilon_x \quad\forall y\in U_x
$$
Now the collection $\{U_x : x\in X\}$ forms an open cover for $K$. Extract a finite subcover $\{U_{x_1}, U_{x_2}, \ldots , U_{x_n}\}$. Now check that
$$
m := \min\{\epsilon_{x_1}, \ldots, \epsilon_{x_n}\}
$$
works.
A: In general, if $K$ is compact and $f:K\rightarrow Y$ is continuous then $f\left(K\right)$
is a compact subspace of $Y$ (not difficult to verify just by the definition of compact). Applying this here, we find that $f\left(K\right)\subset\left(0,\infty\right)\subset\mathbb{R}$
is compact. Then it is closed too (since compact sets in a Hausdorff
space are closed) and we have $0\notin f\left(K\right)$. Then some
$m>0$ exists such that $f\left(K\right)\cap\left(-m,m\right)=\emptyset$. Combined with $f\left(K\right)\subset\left(0,\infty\right)$ this
leads to $f\left(x\right)\geq m$ for each $x\in K$.
