# A proof of Bolzano-Weierstrass theorem

I was trying to prove Bolzano-Weierstrass theorem and fount an argument, but I am not completely sure if it is not wrong.

Consider a bounded sequence of real numbers $$(x_n)$$, now we exclude the following two "trivial cases":

First, the case where the sequence is monotone because in this case the sequence is convergent and any of it's subsequences will be convergent.

The second case, where the sequence "becomes" monotone, this is $$\exists n_0 \in \mathbb{N}$$ such that the subsequence ($$x_{k_i}$$) where $$k_0 = n_0, k_1 = n_0 +1, ...$$ is monotone, in this case, this subsequence is monotone and bounded, thus, it is convergent.

So, we'll consider the case where the sequence is neither monotone nor becomes monotone. Thus, we can say that if we take an arbitrary element of the sequence $$x_{j_1}$$ then we can find some other element of the sequence, $$x_{j_2}$$ such that $$x_{j_1} > x_{j_2}$$ and $$j_2 > j_1$$. We can continue this procedure indefinitely by means of the hypothesis that the sequence does not become monotone. So, we can construct a decreasing subsequence $$(x_{j_k})$$ that is convergent because it is bounded and montone. Also, notice that we can construct also an increasing subsequence starting from $$x_j$$ with the same argument, thus, a bounded sequence that does not become monotone has more than one different convergent subsequences that converges to different values.

• The sequence not being eventually monotone doesn't mean it has a decreasing subsequence. Consider $1,3,2,4,3,5,4,6,5,7,6,8,7,9,8,10...$ It never becomes monotone but also doesn't have a decreasing subsequence. – Mark Apr 20 at 13:28

Instead try to prove that every sequence in $$\mathbb{R}$$ contains a monotone subsequence. Give it a shot yourself, but if you're stuck, a proof can be found here.