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Does the set A= {$1+\frac{(-1)^n}{n}:0\leq n\leq10,n\in\mathbb{N}$} have any accumulation points in $\mathbb{R}$?

My guess is no since this is a finite set.
So there exist $\epsilon$ : $\forall a\in A, (a-\epsilon,a+\epsilon)\cap A=\varnothing$.

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Any finite set has no accumulation point, you're right. –  the symplectic camel Jan 25 '12 at 19:59
    
What is $A$? Is it the set you defined? –  the symplectic camel Jan 25 '12 at 20:03
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Why do you guess that? There is a great chance that the evidence that lead you to conjecture that will lead to an actual proof. It is impossible to know, until you tell us. –  Mariano Suárez-Alvarez Jan 25 '12 at 21:15
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Your guess is correct, but the statement that follows it is clearly false: $(a-\epsilon,a+\epsilon)\cap A$ cannot be empty, as it certainly includes $a$. Do you mean $(a-\epsilon,a+\epsilon)\cap(A\setminus\{a\})$? –  Brian M. Scott Jan 25 '12 at 21:29
    
$1+\frac{(-1)^0}{0}$ is not usually regarded as a real number. –  Henry Jan 26 '12 at 0:25

1 Answer 1

It depends on what you mean by "accumulation point". The terminology is not quite standardized. The definition you suggest is (for the case of the real line):

$x$ is an accumulation point of $A$, whenever for all $\epsilon > 0$, $(x - \epsilon, x + \epsilon) \cap A \neq \emptyset$.

I (personally) would call this an adherence point, and the set of all such $x$ is the closure of $A$ (by theorem or definition, depending on your text). In that case your $A$, being a finite subset of $\mathbb{R}$, is closed, which means it equals its closure, or equivalently, the set of "accumulation points" is exactly $A$. Note that in the aforementioned definition, every $a \in A$ is an accumulation point, for any $A$, because $a$ itself is in all intervals $(a - \epsilon, a + \epsilon)$ and in $A$, of course.

Another definition for accumulation point, and more common I believe, is that (adapted for the reals again):

$x$ is an accumulation point of $A$ whenever for all $\epsilon > 0$, $(x - \epsilon, x + \epsilon) \cap ( A \setminus \{x\} ) \neq \emptyset$,

which says that every open interval around $x$ intersects $A$ in a point other than $x$ itself (which makes no difference if $x$ was not in $A$ anyway, but does make a difference for points of $A$). In that case, for any $x$, we can compute all the finitely many distances from $x$ to all points of $A \setminus \{x\}$ and use an $\epsilon$ smaller than all those to show $x$ is not an accumlation point of $A$ in the latter definition, and in that case you are right that there are no accumulation points.

Note that (for the reals and all metric spaces) $x$ being an accumulation point in the second sense implies that every open interval around $x$ contains infinitely many points of $A$, because otherwise we could use the minimal distance idea from before again to contradict the definition. This explains the name "accumulation point" a bit better, maybe, as accumulate suggests (lots) more than one... The term limit point is also used for this second version of the definition.

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