Convergent Sequence Composed of Convergent Sequences

Question

The problem reads so:

Let $\Bbb{X}: \Bbb{N} \to \Bbb{M}$ and $\Bbb{Y}: \Bbb{N} \to \Bbb{M}$ be two sequences such that $\lim\limits_{n\to\infty} x_n = L = \lim\limits_{n\to\infty} y_n$. We define the sequence $\Bbb{Z}: \Bbb{N} \to \Bbb{M}$ as $X(n/2)$ if $n$ is a pair number and $Y(2n-1)$ if $n$ is an odd number. Prove that $\lim\limits_{n\to\infty} z_n = L$

I have tried to write $n = 2k$ if $n$ is a pair number and $n = 2k-1$ if $n$ is an odd number, then the sequence $Z$ would be written as $Z= X(k)$ if $n$ was pair and $Y(k)$ if $n$ was odd.

Answer 1

Let $\epsilon > 0$. From the given limits, we know that there exists $N_1 \in \Bbb{N}$ such that $d(x_n,L) \le \epsilon$ whenever $n \ge N_1$. Idem for $N_2 \in \Bbb{N}$ and $d(y_n,L) \le \epsilon$ whenever $n \ge N_2$.

$$z_n = \begin{cases} x_{n/2} & \text{if } n \text{ is even} \\ y_{2n-1} & \text{if } n \text{ is odd.} \end{cases}$$

To prove that $\lim\limits_{n\to\infty} z_n = L$, we need $N \in \Bbb{N}$ which satisfies

\begin{cases} \dfrac{N}{2} \ge N_1 & \text{if } n \text{ is even} \\ 2N-1 \ge N_2 & \text{if } n \text{ is odd.} \end{cases}

This suggest the choice of $N = \max\left\{ 2N_1, \left\lceil \dfrac{N_2 + 1}{2} \right\rceil \right\}$ so that $d(z_n,L) \le \epsilon$ whenever $n \ge N$.

Answer 2

A sequence $(a_n)_{n\in \Bbb N}$ in $\Bbb R$ converges to $a\in \Bbb R$ iff for every $r>0$ the set $\{n\in \Bbb N: |a_n-a|\geq r\}$ is finite.

Suppose $A, B$ are infinite disjoint subsets of $\Bbb N$ and $A\cup B=\Bbb N.$ Suppose $\lim_{n\to \infty}x_n=L=\lim_{n\to \infty}y_n.$ Let $z_n=x_m$ if $n$ is the $m$th member of $A$ and $z_n=y_m$ if $n$ is the $m$th member of B. Then $\lim_{n\to \infty} z_n=L.$

Proof: For $m\in \Bbb N$ let $f(m)$ be the $m$th member of $A$ and let $g(m)$ be the $m$th member of $B.$ For any $r>0$ we have:

(i). The set $A^*=\{n\in A: |z_n-L|\geq r\}$ is the image, under $f,$ of the finite set $\{m\in \Bbb N: |x_m-L|\geq r\},$ so $A^*$ is finite.

(ii). The set $B^*=\{n\in B: |z_n-L|\geq r\}$ is the image, under $g,$ of the finite set $\{m\in \Bbb N: |y_m-L|\geq r\},$ so $B^*$ is finite.

(iii). $\{n\in \Bbb N:|z_n-L|\geq r\}=A^*\cup B^*.$