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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.2011-09-16

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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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    I think "has a limit point" would mean "there exists a limit point" and "contains an accumulation point" would mean "there exists an accumulation point and that accumulation point belongs to the set under consideration". (This ignores whatever distinction there may be between "limit point" and "accumulation point", as these terms can vary in the literature.)2011-09-16