Let $(x_n)$ be a sequence in $\mathbb{R}$, reals. It is not necessarily convergent.
a) Show that if $L$ is the limit of some convergent subsequence $(x_{f(n)})$ of $(x_n)$, then $L$ is an accumulation point of the set $\{x_n|n \in \mathbb{N}\}$. (Here we are meant to presume $(x_n)$ does not converge and that the set is infinite.)
b) Suppose that $A$ is an accumulation point of $\{x_n | n\in \mathbb{N}\}$. Is $A$ the limit of some subsequence of $(x_n)$?
For a), I am having a bit of a hard time with it because I'm not completely sure how to actually show $L$ is an accumulation point of the set.
By definition: A point $a\in\mathbb{R}$ is an accumulation point of A if, for all $r > 0$, the intersection $B(x, r) ∩ A$ contains infinitely many points. (Not required that $a\in A$)
So I need to show that the point $L\in\mathbb{R}$ is an accumulation point of the set $\{x_n|n \in \mathbb{N}\}$ if, for all $r > 0$, the intersection $B(L, r) ∩ \{x_n|n \in \mathbb{N}\}$ contains infinitely many points. So if I show that intersection $B(L, r) ∩ \{x_n|n \in \mathbb{N}\}$ contains infinitely many points, then I've proven it? I'm having a hard time actually showing this though but I guess I have an idea of what I'm aiming to do.
For b), I need to prove it or disprove it by counterexample.
I think the statement is basically asking me to prove/disprove that:
$A$ is an accumulation point if there exists a subsequence $x_{f(n)}$ such that $lim_{n→\infty}x_{f(n)}=A$?
I can't really think of a counterexample in this context of the reals, so I think it might be true. I'm not completely sure how to prove it though. I was thinking that if I somehow show that for some arbitrary $A$, an accumulation point, there exists a subsequence $x_{f(n)}$ such that $lim_{n→\infty}x_{f(n)}=A$, so this is then true for all $A$, or all accumulation points.
for b), I found this: Accumulation points of sequences as limits of subsequences? which is similar, but I didn't get some of the answers since they seemed a bit more advanced and I'm not completely sure if it's the same context.
Let's consider the set $X=\{x_n: n\in \mathbb N\}$ to be infinite.
In this case, b) holds.
There is an $x_{n_1}\in X$ such that $x_{n_1}\in B(A,1)$. Choose $x_{n_2}\in X\setminus \{x_{n_1}\}$such that $n_2>n_1, x_{n_2}\in B(A,\frac 12)$.
Continue like this, and choose $x_{n_k}\in X\setminus \{x_{n_1},\cdots, x_{n_{k-1}}\}$ such that $n_k>n_{k-1}, x_{n_k}\in B(A, \frac 1k)$ for $k>2$.
It follows that for every $k\in \mathbb N$ the following holds: $|x_{n_k}-A|\lt \frac 1k$
It follows that $\lim_{k\to \infty} x_{n_k}=A$
If the set $X$ is finite, a) is not true as the definition of accumulation point stated in OP doesn't hold ($B(L, r)\cap X$ can only be finite).