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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 ?


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I am not familiar with the term “partial limits”, but guess that it means the same as “limit points”, or equivalently the set of limits of convergent subsequences? – Harald Hanche-Olsen Oct 15 '12 at 19:30
Here is a hint, though: If $X$ is dense then not only is $X\cap(a,b)\ne\emptyset$ for all $a<b$, but that set is infinite. For if it is finite, then you can find a smaller open subinterval of $(a,b)$ that avoids all those points … – Harald Hanche-Olsen Oct 15 '12 at 19:33
@Harald For example, for a sequence 1,0,1,0,1,... partial limits set is (1,0) and say sequence is 1,2,3,1,2,3,... its partial limit set contains 1, 2, and 3... – HarveyMudd Oct 15 '12 at 19:36
@HaraldHanche-Olsen which is the same as "the set of limits of convergent subsequences" – HarveyMudd Oct 15 '12 at 19:46
Look at the set of partial limits (limit points) of sequence $(x_n) _{n=1}^\infty =\left\lbrace {\dfrac{1}{10},\,\dfrac{2}{10},\, \ldots\, \dfrac{9}{10},\, \dfrac{1}{10^2} ,\,\dfrac{2}{10^2},\, \ldots\, \dfrac{99}{10^2},\, \ldots,\, \dfrac{1}{10^n} ,\,\dfrac{2}{10^n},\, \ldots\, \dfrac{10^n-1}{10^n},\, \ldots} \right \rbrace$. – M. Strochyk Oct 15 '12 at 20:13
up vote 1 down vote accepted

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 $y_k\to a$.

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