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quasar987

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## Homework Statement

I tried finding a proof for the "Bolzano-Weirstrass property" of R before looking at the actual proof and I came up with something different.

The actual thing to prove is: "Every bounded sequence in R has a subsequence that converges to some points in R."

## The Attempt at a Solution

Consider a sequence [itex]f:\mathbb{N}\rightarrow \mathbb{R}[/itex] bounded by M. Let's peek at [itex]f(\mathbb{N})[/itex]. Is it finite?

If yes, then look at the preimage of every element of [itex]f(\mathbb{N})[/itex]. There will be one that is infinite and by the well-ordering property of N, we can thus extract a constant subsequence from f.

If no, then it is denumerable and thus can be put in bijection with the set of rational numbers in (0,1) (or [0,1) or (0,1] or [0,1] depending on whether the sup and inf of [itex]f(\mathbb{N})[/itex] actually are in [itex]f(\mathbb{N})[/itex]). Also, let the bijection respect order. We can now talk of the elements of [itex]f(\mathbb{N})[/itex] as rationals in (0,1).

I will now refer to the set of rationals in (0,1) as Q'. So chose some [itex]x_1[/itex] in Q' and look at its preimage [itex]f^{-1}(x_1)[/itex]. By the well-ordering of N, we can choose the smallest integer of that set, call it [itex]n_1[/itex], and let [itex]a_1=f(n_1)[/itex] be the first element of our subsequence.

Because Q' is dense in itself, we can also select an [itex]x_2[/itex] in Q' with [itex]x_1<x_2<1[/itex]. Look at the preimage of [itex]x_2[/itex]. If all the elements of [itex]f^{-1}(x_2)[/itex] are lesser than [itex]n_1[/itex], then dump [itex]x_2[/itex] and choose some [itex]x_2'[/itex] in Q' with [itex]x_2<x_2'<1[/itex] and look at its preimage, etc. This "rechoosing" of [itex]x_2[/itex] cannot last forever. Actually, it can last at most [itex]n_1[/itex] times! So when we get an [itex]x_2[/itex] whose preimage contains an integer larger than [itex]n_1[/itex], call that smallest such integer [itex]n_2[/itex] and let [itex]a_2=f(n_2)[/itex] be the second element of our subsequence.

Generate all elements in this way. The result is a monotone increasing subsequence of f that is bounded by M and thus, by the completeness of R, converges to some point in R.

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