# A nice proof of Bolzano-Weierstrass

Following is a very nice proof for Bolzano-Weierstrass that I discovered. I think that any proof of B-W would require an infinite construction, and this one is no exception--However I feel that the infinite construction in this proof is easier to the intuition than the constructions in other proofs.
The theorem:
If is a sequence of numbers in the closed segment , then it has a subsequence which converges to a point in .
Let's take an arbitrary point , which is somewhere between the points and . Observe the segment . It may contain a finite number of members from the sequence and it may contain an infinite number of them. If we were to take the point to be , the segment would obviously contain an infinite number of members from the sequence. If we take to be , the segment would contain at most only one point from the sequence. Introducing the set : Sidenote: A segment containing zero elements from will be considered to contain a finite number of them.

We know that belongs to . There is one more thing we need to note about the set . If a point belongs to , that would mean that has a finite number of members from in it, and that would therefore mean that any subset of would also have only a finite number of members from . Hence for any that belongs to , all the points between that and would also belong to . We can see that is actually a segment, starting at and ending in some unknown location in . Now we make our next move: We are now going to show that is an accumulation point of .
Let's ignore the special case , and assume that . Now we take an arbitrarily small . Observe the segment . cannot belong to since it is higher than the supremum. Hence contains an infinite number of members. Now observe the segment . must belong to , since it is smaller than the supremum of the segment . Thus contains a finite number of members from . But is a subset of . If the bigger set contains an infinite number of members and its subset contains only a finite amount, the complement of the subset must contain an infinite number of members from . We have proved that for every , the segment contains an infinite number of members from the sequence. To rephrase that, we have proven that there exists a point in the segment , such that every neighborhood of includes an infinite number of members from the sequence. We have proven that is an accumulation point of , from which it is easy to prove that there exists a subsequence of which converges to , and that spells Q.E.D.

In the special case that it can still be shown that is an accumulation point. I will not expand upon that since it is too boring.

For the benefit of the novice I will explain in detail how to prove that an accumulation point of a sequence is a limit of one of its subsequences. We already noted that for any , contains an infinite number of members from . Now we will construct a subsequence of that converges to . Take to be . Take any member in to be our first member. Take to be . Take an member in with an index number higher than the previous member to be our next member--You are guaranteed to find one since there are an infinite number of members in that segment. Take to be , and now find an member in with an index number higher than that of our previous member. Again you are guaranteed to find one. This process can continue indefinitely, thus constructing a subsequence of that converges to . (I told you an infinite construction has to come at some point!)
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