Prove that every sequence of real numbers has at least one limit point 
*

*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


*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 $x_n$


*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.


*Sequence cannot be finite and unbounded in $\mathbb{R}$
Please check my proof for any errors.
 A: 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 $n_1<\ldots<n_m$ 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 $n_1<n_2<\ldots$ 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 $a$.
