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To Prove: Each sequence that is unbounded above / below has a subsequence that converges (improperly) to $+ \infty / −\infty$. Conclude that each sequence of real numbers has a properly or improperly convergent subsequence.

Whenever a sequence is unbounded, we have that it has an increasing/decreasing subsequence. Why is this the case? For any number $M$ you give me, there exists an $m\in \mathbb{N}$ such that $a_m$>M. We choose the subsequence $\{a_i, a_j, a_k \dots \}$ such that the index set is strictly increasing/decreasing and also the terms are strictly increasing/decreasing. Does this argument work? how can I formalise this better?

An example of my method applied would be the following sequence: $\{0,1,2,1,0,3,4,5,4,3,6 \dots \dots \}$ My method would pick the following subsequence $\{0, 1,2,3,4,5,6 \dots\}$. This is the sequence $a_l=l$ where $l$ is now a new index.

Whenever a sequence is bounded it has a convergent subsequence by Bolzano Weierstrass (it has an accumulation point, hence the subsequence converges to this accumulation point.)

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    $\begingroup$ Simply build recursively a subsequence $(a_{n_j})$ such that $a_{n_0}=a_0$, say, and, for every $j$, $$a_{n_{j+1}}\geqslant a_{n_j}+1$$ $\endgroup$ – Did Nov 25 '18 at 6:15
  • $\begingroup$ That's a nice way to put it. $\endgroup$ – Wesley Strik Nov 25 '18 at 6:18
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We build a recursive subsequence defined by: $$a_{n_0}=a_0 \land n_{j+1} > n_j \land a_{n_{j+1}} \geq a_{n_j}+1 $$ We can do so because the set is unbounded, so we can always handpick the next term that is greater than our previous one, because there should never exist a bound. Suppose we would not be able to pick a greater term, we would get a contradiction since $a_{n_j}+1$ would be a bound for the entire sequence, which was supposed to be unbounded.

We also notice that the index set is strictly increasing and that the subsequence itself is increasing, the question is, is it increasing fast enough?

Notice that we take some arbitrary term for the sequence and handpick the next one that is at least greater than the current one plus $1$. This looks very familiar: $$ a_{n+1} = a_{n} + 1$$ Is equivalent to linear growth, or also in a direct formula: $$ a_n = n + a_0$$ We have thus desired that our sequence grows at least as fast as this sequence. However, we know that this sequence diverges. Because or minorant diverges, our original subsequence must also diverge. $\square$

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