I am trying to understand this question How does continuity implies closeness by this guy @Logan. In the proof linked to in the question (linked again here) Theorem 1.2, it states
Suppose to the contrary that there is an $\alpha \in \mathbb{R}^n$ such that the set $S=\{x|f(x) \le \alpha\}$ is unbounded. Then there must exist a sequence $\{x^v\} \subset S$ with $||x^v||\rightarrow\infty$.
I do not understand why this sequence must necessarilly go to $\infty$ just because it is unbounded. Do all unbounded sequences go to $\infty$?
From what I've seen (e.g. Unbounded sequence diverging to $\infty$?) that is not true. So why is it saying that in this proof?
A "subsequence" implies that you have a sequence in the first place. An unbounded sequence has a divergent subsequence. Here we have an unbounded set. What the proof is doing is choosing a sequence that is divergent. One way of doing this is as follows:
"Unbounded" means that there is no bound, i.e. there is no $R$ so that $\lVert x\rVert < R$ for every $x \in S$.
Let $R_i$ be an increasing sequence of positive real numbers that diverges to $\infty$ ( $(1,2,3,\dotsc)$ will work). Then $R_1$ is not a bound for $S$, so there is $x_1 \in S$ with $\lVert x_1 \rVert > R_1$. Similarly, $R_2$ is not a bound for $S$, so there is $x_2 \in S$ with $\lVert x_2 \rVert > R_2$. Continue doing this for each $i$, and you end up with a sequence $x_i$ with $\lVert x_i \rVert \to \infty$, as required.