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First of all, I define accumulation point as: Let $a$ be an accumulation point of $A$. $\forall \ \epsilon > 0$, $B_\epsilon(a) \setminus \{a\} $ contains an element of $A$.

T/F question: If $a$ is an accumulation point of $A \subset \mathbb{R}$, then there exists a sequence $(a_n)$ in $A$ converging to $a$.

I'm tempted to say true by using this proof:

For each $n \in \mathbb{N}$, construct $B_\epsilon(a)$ such that $\epsilon = 1/n$. Then pick a point out of the ball such that $a_n \neq a$. Repeating this procedure yields a sequence and this sequence converges to $a$.

Two questions:

(1) Am I right

(2) Is it necessary that $A$ is a subset of $\mathbb{R}$? Why or why not?

user1691278
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  • yes, that is the classic proof. 2) A needs the be a closed metric space but it needn't be R. You need a space where $B_\epsilon(x)$ will always have points other than$x $. And you need a space where cauchy sequences must converge.
    • – fleablood Dec 05 '17 at 06:38
  • For (2), does that mean I need a complete metric space? What does it have to be closed? – user1691278 Dec 05 '17 at 06:40
  • I think one only needs a metric space for 2) to work. – Gribouillis Dec 05 '17 at 06:40
  • Actually, I was wrong about the second and third proof would work in Q. As a is in the set. But you do need the space to be "dense" so that all B_e have more than one point. – fleablood Dec 05 '17 at 06:42
  • @fleablood You don't need this. If $a$ is isolated, it is not an accumulation point according to the above definition. – Gribouillis Dec 05 '17 at 06:43
  • It will not work on the discrete metric. I meant complete, not closed. That was a typo. But it doesn't have to be complete. Just dense. As the accumulation point is assumed to be in A, we don't have to worry about a sequence not converging. It will converge to a. – fleablood Dec 05 '17 at 06:46
  • Doh. You are right. Never mind. Any metric space only because B_e needs to make sense. But there is a caveat. There are metric space where B_e/{a} can be empty. But then a can't be an accumulation point then. And there are metrics where cauchy sequences don't converge to known points. But then there will not be a point a, to take a series of neighborhoods around. – fleablood Dec 05 '17 at 06:53