How to construct sequences by induction?

Question

if $(u_{n})_{n\in\mathbb{N}}$ is an arbitrary sequence of reals. How do I start to construct, by induction, two real sequences $(a_{n})$ and $(b_{n})$ such that these criteria hold:

• $(a_{n})$ is increasing
• $(b_{n})$ is decreasing
• for all n$\in \mathbb{N}, a_{n} < b_{n}$
• the sequence $(b_{n} - a_{n})$ converges to 0.
• $u_{n} \notin [a_{n}, b_{n}]$

Answer 1

Take an arbitrary initial interval $a_0<b_0$. If this interval contains $u_0$, trim it on one side by keeping one of $\left[\dfrac{u_0+b_0}2,b_0\right]$ or $\left[a_0,\dfrac{a_0+u_0}2\right]$.

Shrink this interval, for instance with $a_1=\dfrac{2a_0+b_0}3,b_1=\dfrac{a_0+2b_0}3$. If the new interval contains $u_1$, trim it on one side by keeping one of $\left[\dfrac{u_1+b_1}2,b_1\right]$ or $\left[a_1,\dfrac{a_1+u_1}2\right]$.

And so on.

You can easily check that those intervals are non-empty, nested, do not contain the respective $u_n$ and their size converges to $0$.

Answer 2

Let $(u_{n})_{n\in\mathbb{N}}$ be a fixed sequence, wlog you can assume $u_1$ to be positive (if it is negative you can make the same argument as below, if it is 0 you can take $a_1=1$ and $b_1=2$). First step: choose $a_{1}=\frac{u_1}{4}$ and $b_1=\frac{u_1}{2}$. Now $a_{1}<b_{1}$ and $u_{1} \notin [a_{1}, b_{1}]$. Supppose you have constructed $a_n$ and $b_n$ satisfiyng the assumptions. Now $u_{n+1}$ can be in $[a_n,\frac{a_n+b_n}{2}]$ or in $[\frac{a_n+b_n}{2},b_n]$ (if it is outside $[a_n,b_n]$ you can just take $a_{n+1}=a_n+\frac{b_{n}-a_{n}}{2^{10}}$ and $b_{n+1}=b_n-\frac{b_{n}-a_{n}}{2^{10}}$). Assume $u_{n+1}\in[a_n,\frac{a_n+b_n}{2}]$ (the other case is similar), you take $a_{n+1}=\frac{a_n+b_n}{2}+\frac{b_{n}-a_{n}}{2^{10}}$ and $b_{n+1}=b_n-\frac{b_{n}-a_{n}}{2^{10}}$.You have to check that this choice of $a_{n+1}$ and $b_{n+1}$ satisfies your assumptions and use the nested interval theorem for the convergence of the sequences $(a_n)$ and $(b_n)$.