Which of these properties hold for sequences satisfying $x_n \leq y_n \leq x_{n+2}$?

Question

$\{ x_n \}$ and $\{ y_n\}$ are two sequences of real numbers such that $x_n\leq y_n \leq x_{n+2}, n=1,2,3...$ then

(a) $\{ y_n\}$ is a bounded sequence.

(b) $\{ x_n \}$ is an increasing sequence.

(c) $\{ x_n \}$ and $\{ y_n\}$ converge together.

(d) $\{ y_n\}$ is an increasing sequence.

I found this question in a book and I am not really able to reach any conclusion regarding the options.

I have concluded that since $x_n\leq x_{n+2}$, $n=1$ gives the subsequence $\{ x_{2n-1}\}$ is decreasing and similarly, $\{ x_{2n}\}$ will be decreasing. But I am not able to proceed further.

Also, I tried ruling out the options using certain examples. I think option (c) should be right.

Answer 1

Let's consider two examples:

1. Let $x_n=n$ and let $y_n=n+1$ for all $n\in \mathbb{N}$, then $x_n=n\leq y_n=n+1\leq x_{n+2}=n+2$ but $(y_n)_n$ is not bounded. In general we can't conclude (a).
2. Let $$x_n=y_n=\begin{cases}0& \mbox{ if $n$ is even,}\\ 1 & \mbox{ if $n$ is odd.}\end{cases}$$ Clearly $x_n=y_n=x_{n+2}$ for all $n\in \mathbb{N}$ and thus $x_n\leq y_n\leq x_{n+2}$ holds for all $n\in \mathbb{N}$. Clearly, both sequences are not convergent and not increasing. Hence we can't conclude (b), (c) or (d) either.

You can't conclude any of those properties.

EDIT: I misread the meaning of (c).

Let's show the following lemma.

If $a_n\to a$ and $\lim_{n\to \infty}|b_n-a_n|=0$, then $b_n\to a$.

Proof: Choose $\varepsilon>0$. By assumption, there exists an $n_0\in \mathbb{N}$ such that for all $n\geq n_0$, we have that $|a_n-a|<\frac{\varepsilon}{2}$. Similarly, there exists an $n_1\in \mathbb{N}$ such that for all $n\geq n_1$ we have that $|b_n-a_n|<\frac{\varepsilon}{2}$. Let $n\geq \max\{n_0,n_1\}$, then \begin{eqnarray} |b_n-&a|&\leq &|b_n-a_n|+|a_n-a|\\ &<& \frac{\varepsilon}{2}+\frac{\varepsilon}{2}\\ &=& \varepsilon. \end{eqnarray} This concludes the proof of the lemma.

We now show that if $(x_n)_n$ converges to some $x\in \mathbb{R}$, then so does $(y_n)_n$. Indeed, note that \begin{eqnarray} |y_n-x_n| &= & (y_n-x_n)\\ &\leq & (x_{n+2}-x_n). \end{eqnarray} Here we used that $x_n\leq y_n\leq x_{n+2}$. But as $(x_n)_n$ is convergent to $x$, the right-hand side goes to zero and thus $y_n\to x$ by the above lemma.

Conversely, you can easily show that if $y_n\to y$, then so does $x_n$. Indeed, note that \begin{eqnarray} |y_n-x_n| &= & (y_n-x_n)\\ &\leq & (y_{n+2}-y_{n-2}). \end{eqnarray} Again, as $(y_n)_n$ is convergent, the right-hand side goes to zero (as $(y_n)_n$ is a Cauchy sequence) and thus $x_n\to y$ by the above lemma.

This shows that under these conditions, $(x_n)_n$ is convergent with limit $x\in \mathbb{R}$ if and only if $(y_n)_n$ is convergent with limit $x\in \mathbb{R}$. I'll leave it to you to figure out what happens when these limits are infinite or do not exist.