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    So, confession time: It came to me when thinking over the answer I gave to a poster that I really have a somewhat unnatural love for the Bolzano-Weierstruss theorem.

    You know how they say "if all you have is a hammer, everything looks like a nail"?

    Well I say, "if all you have is B-W, every theorem looks like a sequence waiting for a convergent subsequence"...

    Want to prove a cts function f on a closed interval is bounded? Suppose it isn't, form a sequence x_n with f(x_n) going to infinity. Find a convergent subsequence. Oops, contradiction!

    Want to prove a cts function attains it's bounds? Form a sequence x_n with f(x_n) tending to the bound. Find a convergent subsequence. Done.

    Cauchy sequences converge? a_n is bounded, so a_n has a convergent subsequence. Cauchy criterion tells you a_n converges to the limit of the subsequence.

    And so on... (special mention to the Arzela-Ascoli theorem, where we take a subseqeuence of a subsequence of a ...)

    The glaring omission in the above: I'm unaware of any proof of the IVT that uses Bolzano Weierstrauss, and I'm struggling to think of one that isn't horribly unnatural.

    Any takers? Also anyone else want to share favorite B-W proofs (or indeed, other 'non-standard' analysis proofs).

    Edit: thinking about it, to prove the IVT in R, it's enough to show connectivity (of the image of f) but this isn't enough in higher dimensions. So the IVT isn't "about" compactness in the same way most of the other examples are. I don't think it precludes a BW solution but it maybe explains why it's a lot less obvious than in the other examples.
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    On a similar note: I'm quite partial to using the fact that we can form an injection between the set of finite subsets of N to N by mapping {a_1, a_2, ..., a_n} to 2^a_1 3^a_2 ... p_n^a_n (where p_k is the kth prime). This one "hammer" smashes a lot of "prove this set is countable" questions very quickly.
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    (Original post by DFranklin)
    Any takers? Also anyone else want to share favorite B-W proofs (or indeed, other 'non-standard' analysis proofs).
    Well, perhaps it's because I'm a dismal analyst, but I can't really think of a whole raft of results that depend on the B-W theorem - is it that widely used? Maybe I've just forgotten.
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    (Original post by DFranklin)

    The glaring omission in the above: I'm unaware of any proof of the IVT that uses Bolzano Weierstrauss, and I'm struggling to think of one that isn't horribly unnatural.
    A quick bit of googling around suggests that Bolzano proved the B-W theorem as a lemma on the way to his proof of the intermediate value theorem...
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    (Original post by atsruser)
    Well, perhaps it's because I'm a dismal analyst, but I can't really think of a whole raft of results that depend on the B-W theorem - is it that widely used? Maybe I've just forgotten.
    "Depend on" is a bit strong, because there are other ways you can prove them. But you can cover an awful lot of a typical real analysis syllabus using it (without jumping through hoops, either).
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    Yea maths is fun it seems like a lot of stuff is really at its fundamental the sum of a series of rational numbers that can be called something else which also can follow more rules and must have the range or domains restricted!
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    (Original post by DFranklin)
    I'm unaware of any proof of the IVT that uses Bolzano Weierstrauss, and I'm struggling to think of one that isn't horribly unnatural.

    Any takers?
    Assume continuous f:[0,1] \to \mathbb{R} with f(0) \geqslant 0 \geqslant f(1). Now consider the real numbers f(0), f(1/n), \ldots, f(1). Since f(0) \geqslant 0 and f(1) \leqslant 0 we must have some 0 \leqslant k\leqslant n-1 such that f(k/n)\geqslant 0 \geqslant f((k+1)/n) (that is, the list of numbers has to have a sign change somewhere). Set x_n = k/n \in [0,1]. Then x_{n_j} \to x by Bolzano-Weierstrass.

    By continuity f(x_{n_j}) \to f(x) \geqslant 0 since f(x_{n_j})\geqslant 0 always. But also x_{n_j} + 1/n_j \to x so f \left(x_{n_j} + 1/n_{j} \right) \to f(x) \leqslant 0 since f\left((k+1)/n_j\right) = f\left(x_{n_j} + 1/n_j \right) \leqslant 0 always.

    So f(x) = 0. Then [0,1] \cong [a,b] so the proof carries over and we may as well assume a=0,b=1 wlog.
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    bolzano was a g
 
 
 
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