# Segregation of countable sequence into two countable subsequences

This is I think a very simple question about infinite sequences. I thought I knew the answer but the manipulation described below worries me.

Suppose I divide the interval $(0,\frac{1}{4})$ into infinitely many subintervals $S_n = (\frac{1}{(n+1)^2},\frac{1}{(n)^2})$, to wit: $S_1 = (\frac{1}{9},\frac{1}{4}),S_2 = (\frac{1}{16},\frac{1}{9})$, etc. Suppose there is a countably infinite subsequence $\{S_{n_i} \}$ of these intervals that interests me. I wish to segregate this subsequence by moving it to the left of the interval, so that the two countable sequences $\{S_n\} \setminus \{S_{n_i}\}$ and $\{S_{n_i}\}$ are segregated and remain in length-order, respectively. So we would have, $0,...S_{n_2},S_{n_1},0,...,S_2,S_1$.

Are any special assumptions needed to justify this manipulation (and is the situation clear)?

Thanks for any help.

Edit, example:

We have the line segment s: 0 _______ 1/4

I divide it into subintervals as described. Now suppose I want to take the subset of intervals indexed by odd n, and move them to the left of the segment. From the right at x = 1/4, I have a subsequence of intervals whose length approaches zero near (let us say) x = s, and then a new subsequence that begins at s, whose lengths approach 0 as they move towards x = 0. Does this help?

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Revised: Let $a_n$ be the length of $S_n$, so that $\sum_{n\ge 1}a_n=\frac14$. This is an absolutely convergent series, so we may rearrange the terms as we wish. Let $\langle n_k:k\in\Bbb Z^+\rangle$ be an increasing enumeration of the subscripts of the intervals that are to be ‘moved’ to the left, and let $\langle m_k:k\in\Bbb Z^+\rangle$ be an increasing enumeration of the subscripts of the remaining intervals.

For $k\in\Bbb Z^+$ let $s_k=\sum_{i=1}^ka_{m_i}$; $s_k$ is the total length of the first $k$ of the remaining intervals. Let $x_0=\frac14$, and for $k\in\Bbb Z^+$ let $x_k=\frac14-s_k$ and $T_k=(x_k,x_{k-1})$. Let $s=\sum_{i\ge 1}a_{m_i}$; then $0<s<\frac14$, and

$$\bigcup_{k\ge 1}T_k=\left(s,\frac14\right)\setminus\{x_k:k\in\Bbb Z^+\}\;.$$

Now repeat the process with the $S_{n_k}$ in the interval $(0,s)$. For $k\in\Bbb Z^+$ let $u_k=\sum_{i=1}^ka_{n_k}$. Let $y_0=s$, and for $k\in\Bbb Z^+$ let $y_k=s-u_k$ and $U_k=(y_k,y_{k-1})$. Then $\sum_{k\ge 1}a_{n_k}=\frac14-s$, since $\sum_{k\ge 1}a_k=\frac14$, and

$$\bigcup_{k\ge 1}U_k=(0,s)\setminus\{y_k:k\in\Bbb Z^+\}\;.$$

The intervals $U_k$ and $T_k$ are now arranged as you want them.

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@daniel: Yes, I now understand what you’re trying to do; the problem was that your original description didn’t distinguish between the intervals and their lengths. See my revised answer. – Brian M. Scott Jul 14 '12 at 7:41
Yes, this answers my question, much appreciated. – daniel Jul 14 '12 at 7:43

Introduce $x_i=\sum\limits_{j=i}^{+\infty}\left(\frac1{n_j^2}-\frac1{(n_j+1)^2}\right)$. Then, for every $i\geqslant1$, the interval $S_{n_i}=\left(\frac1{(n_i+1)^2},\frac1{n_i^2}\right)$ gets moved to the interval $X_i=(x_{i+1},x_i)$.

To get the new location of each interval $S_n$ with $n$ not in the sequence $(n_i)_{i\geqslant1}$, write down the collection of integers not in the sequence $(n_i)_{i\geqslant1}$ as an increasing sequence $(m_i)_{i\geqslant1}$ and introduce $y_i=\sum\limits_{j=i}^{+\infty}\left(\frac1{m_j^2}-\frac1{(m_j+1)^2}\right)$. Then, for every $i\geqslant1$, the interval $S_{m_i}=\left(\frac1{(m_i+1)^2},\frac1{m_i^2}\right)$ gets moved to the interval $Y_i=(x_1+y_{i+1},x_1+y_i)$.

One can check that $X_{i+1}\lt X_i\lt Y_{j+1}\lt Y_j$ for every $i\geqslant1$ and $j\geqslant1$, with the leftmost point of $X_i$ going to $0$ when $i\to\infty$, the rightmost point of $X_1$ being $x_1$, the leftmost point of $Y_j$ going to $x_1$ when $j\to\infty$, and the rightmost point of $Y_1$ being $x_1+y_1=1$.

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