$f:\mathbb R \rightarrow \mathbb R\;$ in $(\mathbb R,|.|)$ is continuous, prove that (a) and (b) are equivalent: $$(a)\;K \in R\quad is\;compact \Rightarrow f^{-1}(K)\;is\;compact$$ $$(b)\;\lim_{x\to+\infty}|f(x)|=\;\lim_{x\to-\infty}|f(x)|=+\infty$$
I tried:
$(a)\Rightarrow (b)$: by contradiction, if $f$ is bounded then $\exists [a,b]$ compact such that $f^{-1}([a,b])$ isn't compact but this contradicts (a). but what if it's not bounded but also has no limit at $x \to +\infty$?
$(b) \Rightarrow (a)$: I just know that in $(\mathbb R,|.|)$ "compact" means "closed and bounded" and I don't know what to do. Help?
(a) $\implies$ (b): Take $M\in(0,\infty)$. Then $[-M,M]$ is compact and therefore $f^{-1}\bigl([-M,M]\bigr)$ is compact too. In particular, it is bounded. So, take $N\in(0,\infty)$ such that $[-N,N]\supset f^{-1}\bigl([-M,M]\bigr)$. Then$$\lvert x\rvert>N\implies\bigl\lvert f(x)\bigr\rvert>M$$and therefore (b) holds.
(b) $\implies$ (a): Take a compact set $K\subset\mathbb R$. Then $K$ is bounded. It follows from the condition (b) that $f^{-1}(K)$ is bounded too. But $f^{-1}(K)$ is closed (since $K$ is closed and $f$ is continuous). So, $f^{-1}(K)$ is closed and bounded as it follows that it is continuous.