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Recently I learned about the Bolzano-Weierstrass theorem. The theorem is the following:

In $\mathbb R$ every bounded sequence contains a convergent subsequence.

A sequence $a_n$ is bounded if $a_n \in [-C,C]$ for some $C$. The proof that I saw was doing a bisection of $[-C,C]$ into subintervals of decreasing length. I understand the proof but I found another proof that seems slightly less complicated but I don't know if it is correct. Please can someone read my proof and tell me if it is correct?

Proof: If $a_n$ is bounded then $A = \{a_n \mid n \in \mathbb N \}$ is bounded. By the axiom of completeness $a = \sup A$ exists. By the definition of $\sup$ for every $k$ there is $a_{n_k} \in A$ with $|a_{n_k} - a|<{1 \over k}$. The $a_{n_k}$ are a convergent subsequence of $a_n$.

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    $\begingroup$ In addition to what @Did said, the "bisection of intervals" proof is an important and useful proof technique, which I think you should take some time to get used to. $\endgroup$
    – Prahlad Vaidyanathan
    Commented Dec 27, 2013 at 14:10
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    $\begingroup$ Also, a set cannot contain repeated elements, where a sequence can, e.g. $a_{n}=1, \forall n$ vs $A=\left \{ 1 \right \}$. So I am not sure how will you extract the sub-sequence from the set you constructed from the original sequence? $\endgroup$
    – rtybase
    Commented Dec 27, 2013 at 14:41
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    $\begingroup$ $a_n=1/n$ is bounded, and $\sup\{a_n:n\in\mathbb N\}=1$, but I don't think there is a subsequence converging to $1$. $\endgroup$
    – bof
    Commented Dec 27, 2013 at 14:41
  • $\begingroup$ I don't think the definition of sup used here is correct. It should be: $a=\sup A \Leftrightarrow \forall \varepsilon >0,\exists x\in A:a-\varepsilon \leq x$ and $y\leq a, \forall y\in A$ $\endgroup$
    – rtybase
    Commented Dec 27, 2013 at 14:50
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    $\begingroup$ blue's mistake is that the definition of $\sup$ doesn't guarantee the subindices will follow any appropriate order. $\endgroup$
    – Pedro ♦
    Commented Dec 27, 2013 at 15:42

4 Answers 4

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My mistake was that I assumed there are infinitely many $a_n$ near the $\sup$ of the set. But this is not implied by the definition of $\sup$.

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    $\begingroup$ This is the only pertinent answer to the question. $\endgroup$
    – Christian Blatter
    Commented Jan 21, 2014 at 12:01
  • $\begingroup$ @ChristianBlatter The answer by LutzL is helpful to me because it shows why my approach does not work and how to rescue it. $\endgroup$
    – blue
    Commented Jan 21, 2014 at 12:12
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If you take the sequence $(-2,2,-1,1,-1/2,1/2,-1/4,1/4,...)$ your argument using the supremum completely breaks down. However, you can rescue your approach by considering the limes superior, for instance in the form of

$$s=\limsup_{n\to\infty} a_n=\lim_{n\to\infty}\sup_{k\ge n}a_k=\inf_{n\in\mathbb N}\sup_{k\ge n}a_k.$$

For $\varepsilon_j=2^{-j}>0$ and $N_j\in\mathbb N$ you will find an index $k_j\ge N_j$ with $s-ε<a_{k_j}\le s$. Then find the next index with $N_{j+1}=k_j+1$.

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  • $\begingroup$ Where you wrote $N$ should it be $N_j$? $\endgroup$
    – blue
    Commented Jan 21, 2014 at 11:38
  • $\begingroup$ Yes. Corrected it. $\endgroup$
    – Lutz Lehmann
    Commented Jan 21, 2014 at 16:18
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The completeness of the real line and Bolzano-Weierstrass theorem are actually equivalent hence invoking the former to prove the latter is questionable.

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    $\begingroup$ How is this questionable? $\endgroup$
    – Brian
    Commented Dec 27, 2013 at 14:22
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    $\begingroup$ @Did Well, we need one of those to be talking about the real numbers. I really don't see how it is questionable. $\endgroup$
    – Pedro ♦
    Commented Dec 27, 2013 at 14:31
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    $\begingroup$ @Did Yes, but we usually take the axiom of completeness as given or prove it via the construction of the reals, depending on the course. So, completeness is given or proven without mention of Bolzano-Weierstrass, then we use completeness in this proof. I see nothing questionable or circular about that. Another example: Axiom of Choice and Zorn's Lemma are equivalent, so is it questionable to prove Zorn's lemma, even though we accept Choice as true in ZFC? $\endgroup$
    – Brian
    Commented Dec 27, 2013 at 14:40
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    $\begingroup$ @Did Also, how else do we prove equivalence than by using one to prove the other? So, proving equivalence after proving the truth of one of them proves the other. $\endgroup$
    – Brian
    Commented Dec 27, 2013 at 14:41
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    $\begingroup$ I don't think it is "deeply wrong", in particular if that axiom is presented in beginners courses of real calculus for non-mathematicians (which as most courses almost everywhere), @Did . For mathematicians there's usually an added explanation about these matter and whether the sup is presented axiomatically or not students know there are other approaches. $\endgroup$
    – DonAntonio
    Commented Dec 27, 2013 at 15:25
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The problem with your argument is that you cannot state that $a_{n_k}$ is a convergent sequence.

(Hint: it may not even be a sequence in $(a_{n})$).

All you know is that for every $k$ there is an element $somewhere$ in the sequence $(a_{n})$ that is in a neighbourhood of size ${1 \over k}$ of $a$.

However your argument can be made correct if you identify the property such elements $a_{n_k}$ must have in order to create a convergent sequence, and show that such a sequence exists.

Hint: When is a bounded sequence necessarily convergent?

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