Editing Bolzano–Weierstrass theorem (section)

== Alternative proof ==
There is also an alternative proof of the Bolzano–Weierstrass theorem using [[nested intervals]]. We start with a bounded sequence <math>(x_n)</math>:

<gallery widths="300" heights="150">
File:Bolzano–Weierstrass theorem - step 1.svg|Because <math>(x_n)_{n\in\N}</math> is bounded, this sequence has a lower bound <math>s</math> and an upper bound <math>S</math>.
File:Bolzano–Weierstrass theorem - step 2.svg|We take <math>I_1=[s,S]</math> as the first interval for the sequence of nested intervals.
File:Bolzano–Weierstrass theorem - step 3.svg|Then we split <math>I_1</math> at the mid into two equally sized subintervals.
File:Bolzano–Weierstrass theorem - step 4.svg|Because each sequence has infinitely many members, there must be (at least) one of these subintervals that contains infinitely many members of <math>(x_n)_{n\in\N}</math>. We take this subinterval as the second interval <math>I_2</math> of the sequence of nested intervals. 
File:Bolzano–Weierstrass theorem - step 5.svg|Then we split <math>I_2</math> again at the mid into two equally sized subintervals.
File:Bolzano–Weierstrass theorem - step 6.svg|Again, one of these subintervals contains infinitely many members of <math>(x_n)_{n\in\N}</math>. We take this subinterval as the third subinterval <math>I_3</math> of the sequence of nested intervals.
File:Bolzano–Weierstrass theorem - step 7.svg|We continue this process infinitely many times. Thus we get a sequence of nested intervals.
</gallery>

Because we halve the length of an interval at each step, the limit of the interval's length is zero. Also, by the ''[[nested intervals]] theorem'', which states that if each <math>I_n</math> is a closed and bounded interval, say

: <math>I_n = [a_n, \, b_n]</math>

with

: <math>a_n \leq b_n</math>

then under the assumption of nesting, the intersection of the <math>I_n</math> is not empty. Thus there is a number <math>x</math> that is in each interval <math>I_n</math>. Now we show, that <math>x</math> is an [[accumulation point]] of <math>(x_n)</math>.

Take a neighbourhood <math>U</math> of <math>x</math>. Because the length of the intervals converges to zero, there is an interval <math>I_N</math> that is a subset of <math>U</math>. Because <math>I_N</math> contains by construction infinitely many members of <math>(x_n)</math> and <math>I_N \subseteq U</math>, also <math>U</math> contains infinitely many members of <math>(x_n)</math>. This proves that <math>x</math> is an accumulation point of <math>(x_n)</math>. Thus, there is a subsequence of <math>(x_n)</math> that converges to <math>x</math>.