# Proving that every sequence of real numbers has limit point as a consequence of Bolzano-Weierstrass theorem

Our professor showed us Bolzano-Weierstrass theorem for sets only (every infinite, bounded set $$A\in\mathbb{R}$$ has at least one accumulation point in $$\mathbb{R}$$) and left us to prove that every sequence of real numbers has at least one limit point and I would be grateful if anyone would tell me if my proof is correct:

1. Sequence is infinite and bounded.

Let $$A=\{x_n|n \in\mathbb{N}\}.$$ Since $$A$$ is both bounded and infinite existence of limit point comes directly from BW theorem for sets

2. Sequence is infinite and unbounded.

Let $$G$$ be some neighbourhood of $$+\infty$$ (same applies for $$-\infty$$). For any $$M\in\mathbb{R}, \exists n\in\mathbb{N}$$ such that $$x_n\in(M,+\infty)$$ $$\forall n\geq$$ some $$n_0$$ thus there is a subsequence of $$x_n$$ that converges to infinity and so we can say that $$+\infty$$ is limit point of $$x_n$$

3. Sequence is finite and bounded

There is certain real $$a$$ such that $$x_n=a$$ for finite $$n$$.$$\implies \exists x_{n_k}=a; \forall k\in\mathbb{N}\implies lim_{k\to\infty} x_{n_k} = a$$ thus there is subsequence of $$x_n$$ that converges to some point ($$a$$) which is its limit point.

4. Sequence cannot be finite and unbounded in $$\mathbb{R}$$

I was looking also through lots of previous questions on this site that were somewhat similar but none of them have quite answered my question or I didn't understand the solution.

Please correct me if I am wrong somewhere I spent couple of hours to first find connections between BWT for sets and sequences and then prove this... Limits superior and inferior are introduced in later lectures so I am not allowed to use them.

You haven’t actually finished the first case. You know that the set $$A$$ has a limit point, say $$p$$, but you still have to show that the sequence has $$p$$ as a limit point (or as I would call it, a cluster point), i.e., that it has a subsequence converging to $$p$$. You can do this by recursively constructing the subsequence. Suppose that for $$k=1,\ldots,m$$ you’ve chosen $$n_k\in\Bbb Z^+$$ such that $$n_1<\ldots and $$|x_{n_k}-p|<\frac1k$$; there are infinitely many $$\ell\in\Bbb Z^+$$ such that $$|x_\ell-p|<\frac1{m+1}$$, so let
$$n_{m+1}=\min\left\{\ell\in\Bbb Z^+:\ell>n_m\text{ and }|x_\ell-p|<\frac1{m+1}\right\}\;.$$
This allows the recursive construction to continue, and we get a subsequence $$\langle x_{n_k}:k\in\Bbb Z^+\rangle$$ of the original sequence that converges to $$p$$. This shows that $$p$$ really is a limit point of the original sequence.
In the second case your really ought to do something similar: you need to show that you can actually get a subsequence converging to $$+\infty$$. It would suffice to show that we can find $$n_k\in\Bbb Z^+$$ for $$k\in\Bbb Z^+$$ such that $$n_1 and $$x_{n_k}>k$$ for each $$k\in\Bbb Z^+$$; this can be done by a recursive construction very similar to the one that I just did for the first case.
I think that you have a typo in your third case: I believe that you meant to say that there is an $$a\in\Bbb R$$ such that $$x_n=a$$ for infinitely many $$n\in\Bbb Z^+$$. In that case the subsequence $$\langle x_n:x_n=a\rangle$$ is a constant subsequence converging to $$a$$.