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The question was posted earlier, however with low effort, so if I might start over precisely this time. I have a question from a book of theoretical questions. The definitions for the terms used are from my textbook:

Definition We say a point $a\in X$ is an accumulation point of the set $A\subset X$ ($X$ being a metric space $(X,d)$) if: $$(\forall\epsilon\in\mathbb{R^+})(\hspace{0.1cm}L(a,\epsilon)\cap(A\setminus\{a\})\neq\emptyset).$$

Definition We say a point $a\in X$ is an accumulation point of the sequence $\{a_n\}$ in metric space $(X,d)$ if: $$(\forall\epsilon\in\mathbb{R^+})(\forall m\in\mathbb{N})(\exists n\in\mathbb{N})\hspace{0.1cm}(n\geq m\hspace{0.1cm}\land a_n\in L(a,\epsilon))$$

The yes/no question says (quote):

Every accumulation point of the set of values of a sequence $\{x_n\}$ is also the accumulation point of the sequence $\{x_n\}$. Find a counterexample if false.

The reason why I'm asking is that I can't find a counterexample, because I think the statement is false, since I have read somewhere that there exist points called side points, which are not APs of the sequence, yet are APs of its set of values (though I might not have understood that clearly).

I understand that the converse (which was another question) is false, i.e. that not all APs of a sequence need to be APs of its set of values. And the counterexample was $x_n=(-1)^n$, where the APs of the sequence are $-1$ and $1$, however the set of values $\{-1,1\}$ has no APs going by the definition.

Thank you for the time.

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  • $\begingroup$ Yes they are the same. The set of values of the sequence $\{x_n\}$ is a set. Now, see if you can prove it using your $L(a,\epsilon) \cap A \setminus \{a\}$ definition. Btw, I don't know what $L(a, \epsilon)$ is, but I assume it's an open ball? $\endgroup$ – Dzoooks Jul 2 '18 at 22:07
  • $\begingroup$ Yes, it's an open ball (epsilon neighborhood). $\endgroup$ – math101 Jul 2 '18 at 22:16
  • $\begingroup$ Read this it's basically the answer: proofwiki.org/wiki/Metric_Space_Continuity_by_Open_Ball $\endgroup$ – Dzoooks Jul 2 '18 at 22:19
  • $\begingroup$ A small detail: in your definition of accumulation point, I think it should read $\forall \epsilon \in \mathbb{R}^+$ instead of $\mathbb{R}$, since $L(a,\epsilon)$ is empty for $\epsilon \leq 0$, right? $\endgroup$ – Sambo Jul 2 '18 at 23:28
  • $\begingroup$ Indeed, my mistake, thanks. $\endgroup$ – math101 Jul 2 '18 at 23:29
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The statement is true. If $ a $ is an accumulation point for the set of values of $ \{x_n\} $, and not an accumulation point for the sequence, then there would exist an $ \epsilon>0 $ and an $ m $ such that $$\{x_n:n>m\}\cap L(a,\epsilon)=\emptyset. \quad(1)$$

Equation (1) implies $$ (\{x_n\}\backslash\{a\})\cap L(a,\epsilon)\subset\{x_1,x_2,...,x_m\} $$

If we take $$\delta=\min\{|x_1-a|,|x_2-a|,...,|x_m-a|,\epsilon\},$$ then we must have $$(\{x_n\}\backslash\{a\})\cap L(a,\delta)=\emptyset.$$

However, since $ a $ is an accumulation point of the set of values of $ \{x_n\} $ and $ \delta>0 $, we see that $$(\{x_n\}\backslash\{a\})\cap L(a,\delta)\neq\emptyset,$$ a contradiction.

Thus if $ a $ is an accumulation point for the set of values of a sequence then $a$ is an accumulation point for the sequence.

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  • $\begingroup$ It's all clear now, it seemed logical, but I couldn't think of how to express it mathematically. Thank you! $\endgroup$ – math101 Jul 2 '18 at 22:55
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According to your definitions, for example, the constant sequence has no accumulation point of def1.

But if the set defined by the sequence does have an accumulation point $x$ of def1, the sequence itself must have an accumulation point $x$ of def2.

The sequence must have infinitely many values.

And for each $1/k,(k=1,2,...)$, we can "inductively" find $x_{n_k}\in L(x,1/k)$ and $n_1<n_2<...$, then we can show $x$ is an accumulation point of def2.

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  • $\begingroup$ Got it! There is no way for such an infinite set to be created without the sequence having the same accumulation point. Thank you! $\endgroup$ – math101 Jul 2 '18 at 22:57

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