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My lecturer left this as an exercise and didn't go through it, I couldn't find it anywhere online so how is it done? Any help appreciated.

Original question: "Let $(x_n)$ be a real sequence satisfying that every subsequence of $(x_n)$ does not converge in $\mathbb{R}$. Prove that $|x_n|\rightarrow \infty$ as $n \rightarrow \infty$

Proof:

By contradiction, if $|x_n|$ does not tend to $\infty$ (i.e. negation of $|x_n| \rightarrow \infty$), then $\exists M > 0$, such that for all $N$ natural numbers, $\exists m > N$, such that $|x_m| \le M$ => $\exists$ subsequence $(x_{n_k})$ such that $|x_{n_k}| \le M$.

My main question is how to prove the last implication, it's been left as an exercise.

Falcon
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Sam Connell
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  • Welcome to NSE. In future please use MathJax to format your posts. – saulspatz Nov 07 '20 at 19:51
  • Do you know Bolzano-Weiertraß theorem ? – TheSilverDoe Nov 07 '20 at 19:55
  • Look at the proof of the lemma here – saulspatz Nov 07 '20 at 19:58
  • Yes, this is actually an example for the introduction to Bolzano-Weierstrass Theorem, my lecturer just proved it and this is the following example. Is it proven by BW? If so, why did he say it's a proof by induction? – Sam Connell Nov 07 '20 at 19:58
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    I presume this is a lemma used in the proof of Bolzano-Weierstrass – saulspatz Nov 07 '20 at 20:00
  • Maybe, the rest of the proof is a follows: "So (xnk) is a bounded sequence. By B-W theorem there exists a convergent subsequence (xnkj) of (xnk) . Contradiction, as (xnkj) is a subsequence of (xn) which converges in R. – Sam Connell Nov 07 '20 at 20:02
  • @saulspatz He managed to prove B-W without using this theorem by doing the following method: "Suppose (xn) is bounded, so there exists M> 0 such that |xn| <= M for all n>= p. By Lemma 3.2 (that every real sequence has a monotone subsequence), (xn) has a monotone subsequence. So we have -M <= xnk <= M for all k is a natural number. By the Monotone Sequence theorem, (xnk) is convergent, i.e. there exists x0 is a real number such that xnk -> x0 as k -> infinity – Sam Connell Nov 07 '20 at 20:05
  • @saulspatz the proof you linked isn't an induction proof I don't think, is there a way of doing it by induction – Sam Connell Nov 07 '20 at 21:35
  • Why not? Instead of saying, "and so on", frame it as an induction proof. – saulspatz Nov 07 '20 at 22:04

1 Answers1

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We know that because $x_{n} \not \to \infty$, when $n\to\infty$, that there exists a $M$ such that for any $N\in\mathbb{N}$ we can find a natural number $m>N$ such that $|x_{m}|\leq M$.

Lemma. $(x_{n}: n\in \mathbb{N})$ has a bounded sub-sequence $(x_{n_{k}} : k\in \mathbb{N})$.

I leave it to you to fill in the details of the following:

Proof sketch: Set $n_{1}$ to be $\min\{n\in \mathbb{N}: |x_{n}|\leq M\}$. This is made possible by the non-emptiness of the set $\{n\in \mathbb{N}:|x_{n}|\leq M\}$ and applying the Well Ordering Principle. Argue inductivly that $n_{k}:=\min\{n\in \mathbb{N}: |x_{n}|\leq M \ \text{and} \ n>n_{k-1}\}$ defines the index of the sought after sub-sequence $(x_{n_{k}}: k\in \mathbb{N})$. That is, prove that $n_{k}$ exists for all $k$ and conclude that each term in $(x_{n_{k}}: k\in \mathbb{N})$ bounded by $M$.