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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?

Related: How does continuity implies closeness

Andrés E. Caicedo
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makansij
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    No, he's saying there must be a subsequence that goes to infinity. Think about $0,1,0,2,0,3,0,4,\dots$ – saulspatz Apr 21 '18 at 21:57
  • Who's "he" in this context? – makansij Apr 21 '18 at 21:58
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    Take any unbounded sequence in $S$ and consider the sequence of the norms. You obtain a sequence of positive reals that is unbounded, so it has a subsequence diverging to $+\infty$. Then you can just consider this subsequence only. – Giuseppe Negro Apr 21 '18 at 21:58
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    @Candic3 "this guy" you are quoting. – saulspatz Apr 21 '18 at 21:59
  • That's not Logan. if you follow the link, it's a proof given by a class at a university. – makansij Apr 21 '18 at 22:01
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    $$f(n)=\begin{cases}0&n\text{ even}\n&n\text{ odd}\end{cases}$$ is unbounded but doesn't go to infinity. – Thomas Andrews Apr 21 '18 at 22:04
  • Thanks guys. I guess there are two things contributing to my confusion here. 1) When the proof says "Then there must exist a sequence" I guess it should be called "subsequence" instead, right? 2) Also, ${x^v} \subset S $ with $ ||x^v|| \rightarrow \infty$ must be the notation for "subsequence", which I did not know. I thought it was the notation for "sequence" not subsequence. Is that correct? – makansij Apr 21 '18 at 22:13

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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.

Chappers
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  • okay, i think i've finally understand. Since "every unbounded sequence contains a subsequence which diverges to infinity", then we are allowed to make this claim: "If $S={x|f(x) \le \alpha }$, is unbounded, then $\exists {x^v} \in S$" such that ${x^v} \rightarrow \infty$ and hence since the norm is always positive then we can say ${||x^v||} \rightarrow \infty$, for this paricular subsequence, too. right? Source:https://math.stackexchange.com/questions/523985/prove-that-every-unbounded-sequence-contains-a-monotone-subsequence-that-diverge – makansij Apr 21 '18 at 22:37
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    $x^{\nu} \to \infty$ mean $\lVert x^{\nu} \rVert \to \infty$ (how else could you define it?). The point I'm trying to make in the answer is that there is no need to start with a sequence that you only know is unbounded: you can construct one that you know diverges in the first place. – Chappers Apr 21 '18 at 22:45