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Let $X \subset \mathbb R$ be a seq. $\{x_n\}$ in $\mathbb R$ that is dense in $\mathbb R.$ What is the set of limit points of $\{x_n\}$ ?

Answer part:

So we know that $X \cap (x_1, x_2) \neq \emptyset $ such that $x_1 < x_2$ from the dense definition. I guess it should be an increasing sequence. What to do next ? Is it to define a subsequence from $x_1's$, $x_2's$ etc. then show that its partial limit is indeed a infinite set of ($x_1$, $x_2$, ... ) ?

Can you help me with this ?

Thanks...

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    @Analysis In the above definition of *dense* set $X$ $a$ and $b$ not necessary are elements of $X$. Every interval (a,\,b)\subset [0,\,1] \quad (a contains (infinitely many) elements of sequence $(x_n)$.2012-10-15

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Let $a\in\mathbb R$ be arbitrary. Since $X\cap (a,a+1)$ is nonempty, there is an $y_1:=x_n$ where $n$ is minimal with $x_n\in(a,a+1)$. If you have already found $y_k>a$, use that $X\cap (a,\min\{y_k,a+2^{-k}\})$ is nonempty, hence you can let $y_{k+1}:= x_n$ where $n$ is minimal with $x_n\in(a,\min\{y_k,a+2^{-k}\})$. The sequence $(y_k)$ is a subsequence of the given sequence and it should be obvious that $y_k\to a$.