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Prove that the limit points of open interval $A=(2,3)$ subsets of the real numbers, are all the points of the interval $(2,3)$ including $2,3$.

This is my solution (sketch). I consider a succession $a_n=x-1/n$ where $x\in A\cup \{2,3\}$, $a_n$ converges to $x$ for each $x$ in A. Then each $x$ in A is limit point for A.

What do you think about my solution? Thanks.

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up vote 3 down vote accepted

It works fine for elements of $A \cup \{2, 3\}$ such that for all $n$, $a_n \in A$. This is clearly not the case for example, for $x = 2$.

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Thank you. How would you solve the exercise? – Mark Oct 10 '12 at 18:26
Your sequence already works for $3$. It should be easy to come up with a similar one for $2$. For any element $x$ of $A$, one way would be to consider the interval $]2, x]$ and choose the sequence $(b_n)_n$ where $b_n = x - 1/(n + k)$ with $k$ chosen big enough, so that $2 < x - b_1$. – user25784 Oct 11 '12 at 11:06

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