Definitions:
Let $a$ be an accumulation point of $A$. Then $\forall \ \epsilon >0$, $B_{\epsilon}(a) \setminus \{a\}$ contains an element of $A $.
Question:
I have two questions: if $(a_n)_{n\in N}$ is a convergent sequence in $\mathbb{R}$ then,
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Does the set $\{a_n\}$ have exactly one accumulation point? Or, could it have more than one?
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If so, does $(a_n)_{n\in N}$ necessarily converge to the said accumulation point?
I'm tempted to say no to (1), but I'm afraid that I'm missing something. My counter-example to (1) is $\{a_n\} = \{ 4, 3, 2, 1, 0,0,0,…\}$ (i.e. inserting $0$s after the 4th element). Then the set has no accumulation point and it converges to 0. Is that correct?
Best Answer
If $\{a_{n}\}$ has an accumulation point, say, $a$, and $(a_{n})$ is convergent. Then choose some $n_{1}$ such that $a_{n_{1}}\in B_{1}(a)-\{a\}$. Then choose some $n_{2}$ such that $B_{1/2}(a)-\{a,a_{1},...,a_{n_{1}}\}$, proceed in this way we have $a_{n_{k}}\rightarrow a$. Since $(a_{n})$ is convergent, one has $a_{n}\rightarrow a$.
Here I use the following definition:
$a$ is an accumulation point for $A$ if for every $\delta>0$, $(B_{\delta}(a)-\{a\})\cap A\ne\emptyset$.
And note that in the topology of ${\bf{R}}$, being such an accumulation point also implies that $(B_{\delta}(a)-\{a\})\cap A$ contains infinitely many points.