I am reading a proof for Weierstrass' Theorem as follows:
For the proof in (2), it says "since $f$ is coercive, ${z_k}$ must be bounded." I am not quite sure how to understand this. Does it mean that $||z_k|| \rightarrow \infty \implies \lim_{k\rightarrow \infty} f(z_k) = \infty$, contradicting $f(x) \leq \lim_{k\rightarrow \infty} f(z_k)$?. But I am wondering if $z_k \rightarrow \infty$, doesn't that mean that $f(x)$ does not exist? My understanding of lower semicontinuity is that it only deals with sequences that converge. Is my understanding wrong? Does it also mean that all sequences in lower semicontinuous function must necessarily converge from above?
A side question, for (3), for $k$ sufficiently large, why can we always find $z_k$ belonging to the desired set?
If $f$ is coercive, then it indeed satisfies
$$||z_k|| \rightarrow \infty \implies \lim_{k\rightarrow \infty} f(z_k) = \infty \,, $$ see [1]. Thus for $\gamma>\inf_A f$, the level set $L_\gamma:=\{x \in A: f(x) \le \gamma \}$ is closed and bounded, i.e., compact so we can apply part (1) with $L_\gamma$ in place of $A$. Why is it bounded? Argue by contradiction ,if not you can find $z_k$ tending to infinity inside $L_\gamma$ and then obtain a contradiction from coercivity.
For parts (2) and (3) the level set $L_\gamma$ is nonempty for $\gamma>\inf_A f$ by the definition of infimum. he rest follows from the definition of the set as a lower level set and the definition of $z_k$.
[1] https://en.wikipedia.org/wiki/Coercive_function#(Extended_valued)_coercive_functions