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I've been trying to solve a kinda long problem for a few days and can't quite finish it.

Prove the equivalence between

(1) Every bounded sequence has a convergent subsequence.
(2) All cauchy sequences are convergent.
(3) If $ (I_n)_{n\geq 1} $ is a family of closed nested intervals so that $|I_n| \rightarrow 0 $ as $n \rightarrow \infty$ there exist a unique $x$ so that $x \in \bigcap_{n=1}^\infty I_n$.
(4) Every upper bounded non-empty set has a supremum.
(5) Every monotone, upper bounded sequence has a supremum.

I managed to get $(1)\rightarrow(2)\rightarrow(3)$ and $(4)\rightarrow(5)\rightarrow(1)$ so I obviously would love to prove $(3)\rightarrow(4)$.

I manage to get a decreasing sequence since if there's no supremum then for every upper bound $s$ there exist at least one $\varepsilon > 0$ so that $s - \varepsilon$ is also an upper bound. So I can choose one upper bound $s$ and consider the sequence $(s - \varepsilon_n)_{n \in \mathbb{N}}$ that gets closer to the set as $n$ grows, where $s - \varepsilon_n$ is an upper bound for all $n$.

Now I'd like to construct a monotonically increasing sequence of elements of the set so that $|A_n - (s - \varepsilon_n)| \rightarrow 0$ but I'm not sure how to do this, or if my decreasing sequence is convenient for this, it does seem quite ugly to work with, but couldn't really come up with an improvement.

I've consider taking another direction and proving that (3) implies something else and that in turn that implies (4), but I couldn't find a good way of doing it. Maybe I missed something?

Any help would be greatly appreciated.

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  • $\begingroup$ Equivalence in what context? It seems likely you're talking about some sort of ordered group, but you haven't actually said. $\endgroup$ – dfeuer Aug 21 '13 at 3:28
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    $\begingroup$ @dfeuer: What in the world are you thinking?! $\endgroup$ – Ted Shifrin Aug 21 '13 at 3:34
  • $\begingroup$ @TedShifrin: Every bounded (in what sense?) sequence of whats? $\endgroup$ – dfeuer Aug 21 '13 at 3:37
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    $\begingroup$ You may be interested in Propositions equivalent to the completeness of the real numbers and in Propp's paper mentioned there. $\endgroup$ – lhf Aug 21 '13 at 3:47
  • $\begingroup$ @lhf Thanks, I'll look into that! $\endgroup$ – Bananas Aug 21 '13 at 3:51
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You're on the right track. Assuming there's no sup, you deduce that the set likewise has no maximum element. Therefore, for each $n$, you can choose an element $x_n$ of the set and an upper bound $y_n$ with $y_n-x_n<1/n$. You can also stipulate that $\{x_n\}$ is increasing and $\{y_n\}$ is decreasing.

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  • $\begingroup$ Ah, right. It seems so obvious now but for some reason choosing an element of the set so that $|I_n| \rightarrow 0$ directly never occurred to me. Thanks! $\endgroup$ – Bananas Aug 21 '13 at 3:46
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Take a closed interval that intersects the set and includes the upper bound. Divide this closed interval in half (i.e., into two closed intervals) and choose the half that has a point from the set in it; if both halves have nonempty intersection with the set, then pick the one on the right. Now do this repeatedly. Keep doing this and either you end up with only one point left from the set, i.e., the supremum, or you end up with the sort of nested intervals that you want and to which you can apply condition (3).

Can you turn this sketch into a rigorous proof?

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  • $\begingroup$ I love this solution, thanks! I don't think writing it down'll be a problem. $\endgroup$ – Bananas Aug 21 '13 at 3:42
  • $\begingroup$ This sort of method is commonly seen in proofs of (1) as well (which is sometimes referred to as the Bolzano-Weierstrass Theorem). Bound the sequence and then repeatedly divide in half; you generate a convergent subsequence by, basically, finding a binary representation for the limit. $\endgroup$ – Benjamin Dickman Aug 21 '13 at 3:53

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