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Let $\{x_n\}_{n=1}^\infty$ be a sequence of points in $\mathbb R$. Let $X$ be a set defined as a collection of all points in the sequence $\{x_n\}_{n=1}^{\infty}$.

Is the following claim true?

$\left\{x_n\right\}_{n=1}^\infty$ converges to a limit $x^*$ if and only if the set $X$ has a limit point.

My intuition is that the claim is true but I'm not quite sure how to go about showing a rigorous proof of it.

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Think about what happens if the points in the set are isolated from each other (such as will be the case for a finite set), but the sequence is such that a tail of the sequence is just one value is repeated over and over again. –  Dave L. Renfro Sep 16 '11 at 15:45

2 Answers 2

Neither direction is true.

A counterexample for one direction $x_n = (-1)^n(1+1/n)$ (whose image has limit points but does not converge).

For the other direction consider $x_n = 0$ for all $n$ (which converges, but the image has no limit point).

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I'm afraid you are on the wrong way. For example pick $$x_n=\frac{1}{n}.$$ Then obviously the sequence converges to $0$ but $0\not \in X$.

Edit Thanks to dave to point out that $X$ has a limit point is different from $X$ contains a limit point. But even in this case $$x_n\equiv c,\quad \forall n\in\mathbb N,$$ is a sequence which clearly converges but $X=\{c\}$, which has not limit points, being finite.

Neither the converse holds. Pick for example the following: $$\begin{cases} x_1=0,\\\\ x_{2k} =\frac{1}{k},\quad k\geq 1,\\\\ x_{2k+1}=(-1)^k,\quad k\geq 1.\end{cases}$$

Then the set $X$ has $0$ as a limit point, however the sequence does not converge, since we can extract two different subsequences with limits respectively $\pm 1.$

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"For example pick $x_{n}=\frac{1}{n}$. Then obviously the sequence converges to $0$ but $0\not \in X$." Perhaps the original problem statement has changed, but right now it reads "$X$ has a limit point", not "$X$ has a limit point that belongs to $X$". –  Dave L. Renfro Sep 16 '11 at 15:55
    
A limit point of a subset is either a point which is repeated infinite times or an accumulation point, so i can't understand your objection since $X$ has nor infinitedly repeated points nor accumulations. –  uforoboa Sep 16 '11 at 15:59
    
Thinking more about this, I can see how "$X$ has a limit point" could be read as "$X$ contains a limit point". These kinds of marginally ambiguous wordings are what good textbook editing should take care of, but to do so requires an editor with a decent background in mathematics (which I imagine is quite rare). –  Dave L. Renfro Sep 16 '11 at 16:03
    
I'm not english native speaker so i can't understand the difference between the two meanings which besides seems to be quite subtle. Can you explain me more? –  uforoboa Sep 16 '11 at 16:04
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If someone says "the set of numbers has an even integer", the meaning is (almost certainly) "there is a set; each element of this set is a number; one of the elements of this set is an even integer". On the other hand, usually when one says "the set has a limit point", one means that there exists a limit point of the set (which may or may not belong to the set). That is, one means that the set of limit points of the set is not the empty set. So it seems to me that the use of "has" might be very difficult for a non-native English speaker to interpret. –  Dave L. Renfro Sep 16 '11 at 16:12

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