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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.

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  • $\begingroup$ If one considers $(x_n)$ a constant sequence say, a,a,a,a,..., then how is the statement in a) true? $\endgroup$
    – Koro
    Commented Mar 6, 2022 at 18:37
  • $\begingroup$ @Koro if it's a constant sequence, then the subsequence would also just be $a,a,a,a...$, so the limit of the subsequence would be $a$, which is the limit of the sequence $(x_n)$. So I need to show that the point $a$ is an accumulation point of the set $\{x_n|n \in \mathbb{N}\}$ if, for all $r > 0$, the intersection $B(a, r) ∩ \{x_n|n\in \mathbb{N}\}$ contains infinitely many points. But would it contain infinitely many points? because the set will just be $a,a,a,a...$, so the intersection would just be $a$? $\endgroup$
    – eddie
    Commented Mar 6, 2022 at 19:20
  • $\begingroup$ But in case of constant sequence the set {$a_n:n\in \mathbb N$} is just a singleton set and equals {a} so the intersection with B(a,r) can't be infinite. I think you want to assume the set {$a_n:n\in \mathbb N$} infinite in a). $\endgroup$
    – Koro
    Commented Mar 6, 2022 at 19:34
  • $\begingroup$ @Koro I suppose it's meant to be infinite then and not some constant sequence then. I didn't really consider this much because the question says to prove it, so I took it as true by assumption. $\endgroup$
    – eddie
    Commented Mar 6, 2022 at 19:41

1 Answer 1

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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).

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    $\begingroup$ @eddie: the way the definition of accumulation point in given in OP, I'd say not necessarily. Consider the sequence $(x_n)$ given as 1,2,1,3,1,4,1,5,.... for example which has $(x_{2n})$ as subsequence converging to 1 but 1 is not an accumulation point of the sequence. $\endgroup$
    – Koro
    Commented Mar 7, 2022 at 16:35
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    $\begingroup$ @eddie: Glad that you found it helpful. It may happen that you wrote the question here incorrectly. So I think you should again look up the material to see how they defined accumulation point for sequences and not for sets. $\endgroup$
    – Koro
    Commented Mar 7, 2022 at 17:51
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    $\begingroup$ @eddie: still the example in my comment is a counterexample to a). $\endgroup$
    – Koro
    Commented Mar 8, 2022 at 4:32
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    $\begingroup$ @eddie:Instead of considering the convergent subsequence mentioned in a), you could consider the convergent subsequence that is not eventually constant (that is, if $(y_n)$ is the convergent subsequence then for every $N\in \mathbb N$, there is some n>N such that $y_N\ne y_n$). $\endgroup$
    – Koro
    Commented Mar 8, 2022 at 17:22
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    $\begingroup$ @eddie: in that case, you just have to use the definition of limit. Since the subsequence converges to some L. By definition of limit, for every $\epsilon\gt 0,$ there is an N such that $x_n\in (L-\epsilon, L+\epsilon)$ for all $n\ge N$. $\endgroup$
    – Koro
    Commented Mar 14, 2022 at 4:23

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