I am reading a proof of this theorem:
The continuous function $f:\mathbb{R}^n \rightarrow \mathbb{R}$ is coercive if and only if for every $\alpha \in \mathbb{R}$ the set $\{x|f(x) \le \alpha\}$ is compact.
The beginning of the proof says:
Since $f$ is continuous, then the set $\{x|f(x) \le \alpha \}$ is closed. Thus, by Bolzano Weierstrass Theorem, it remains only to show that any set of the form $\{x|f(x) \le \alpha \}$ is bounded….
My understanding of Bolzano Wierstrass was that it states that a set is compact if and only every sequence in the set contains a subsequence which converges to a point in the set.
I do not understand how we are applying the Bolzano Weierstrass(https://en.wikipedia.org/wiki/Bolzano%E2%80%93Weierstrass_theorem) theorem here. What are we calling a “subsequence” in this context?
Also, as a reminder, coercive means $$\lim_{||x||\rightarrow\infty}f(x)=\infty$$
https://en.wikipedia.org/wiki/Coercive_function
How is this an application of the Bolzano Weierstrass theorem?
P.S. As a follow up question, does this proof imply that a continuous function is always has an domain which is closed? If not, then why are we allowed to say that the "contiuity of $f$ implies the closedness of the sets $\{x|f(x)\le \alpha\}$
The Bolzano–Weierstrass theorem is essentially saying that
Heine–Borel, which is probably the appropriate one here, says that
Of course, for metric spaces compact $\iff$ sequentially compact, so the difference is not very important in this case.
C.f. Wikipedia's article.
Regarding the P.S., you can go directly using the sequential version of closed: suppose that $x_n$ is a sequence of points, all of which have have $f(x_n) \leq \alpha$, and that $x_n \to x$. Then $\alpha - f(x) = (\alpha - f(x_n)) + (f(x_n)-f(x))$, and the first term is nonnegative, the second tends to zero as $n \to \infty$ by continuity, so we also have $f(x) \leq \alpha$, so the set is closed.
A continuous function need not have a closed domain ($1/x$ is continuous on $(0,1]$, which is neither open nor closed), but here the domain is the closed set $\mathbb{R}^n$, so we can do this.