# Consider the sequence $\frac{1}{2},\frac{1}{3},\frac{2}{3},\frac{1}{4},\frac{2}{4},\frac{3}{4},\frac{1}{5},\frac{2}{5},\frac{3}{5},\frac{4}{5},\ldots$

Consider the sequence $$\frac{1}{2},\frac{1}{3},\frac{2}{3},\frac{1}{4},\frac{2}{4},\frac{3}{4},\frac{1}{5},\frac{2}{5},\frac{3}{5},\frac{4}{5},\ldots$$ For which numbers $b$ is there a subsequence converging to $b$?

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Every number in $[0,1]$? – anon Feb 16 '12 at 7:06
In fact that sequence are the rational numbers in $]0,1[$ ordered in a particular form. Do you know the accumulation points of the rational numbers? – Dirk Feb 16 '12 at 7:09

Your sequence includes every rational number in $(0,1)$ infinitely many times. For any $b\in[0,1]$, there exists a sequence of rational numbers in $(0,1)$ converging to $b$. That sequence necessarily occurs as a subsequence of your sequence. Conversely, any $b\notin[0,1]$ cannot have a subsequence of your sequence converge to it, as there exists a positive number $\epsilon$ such that every element of your sequence is more than $\epsilon$ away from $b$.

Thus, the numbers $b$ for which there is a subsequence converging to $b$ are precisely those $b\in[0,1]$.

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Even if it included every rational in $(0,1]$ exactly once, the answer would be the same. – Gerry Myerson Feb 16 '12 at 10:54

Anon pointed out $[0,1]$ as the answer : your sequence goes like this : $$\dots, \frac {n-2}{n-1}, \frac 1n, \frac 2n, \dots, \frac {n-1}n, \frac 1{n+1}, \dots$$ Now every element in $[0,1]$ is in one of the intervals $[0,1/n), [1/n,2/n), \dots, [n-1/n,1]$. Let $\alpha \in [0,1]$ and define $x_n$ to be the lower bound of the interval in which $\alpha$ lies in. Since the length of those intervals is $1/n$, $|x_n - \alpha| \le 1/n \to 0$ as $n \to \infty$ and $x_n$ is a well-defined subsequence of your sequence (it preserves the ordering and everything).

You can't get any number outside $[0,1]$, since they are not limit points of the set of points in your sequence. Therefore the answer you're looking for is $[0,1]$.

Hope that helps,

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