1. ## Subsequences

Suppose $\{x_n\}\to x_0$ and $\{y_n\}\to x_0$. Define a sequence $\{z_n\}$ as follows: $z_{2n}=x_n$ and $z_{2n-1}=y_n$. Prove that $\{z_n\}$ converges to $x_0$.

Let $\epsilon >0$. Then $\exists N_1, \ N_2\in\mathbb{N}$ such that for $n\geq N_1, \ N_2$ we have $|x_n-x_0|<\epsilon$ and $|y_n-x_0|<\epsilon$.

I don't know what to do now.

2. ## Re: Subsequences

Let $N:=\max(N_1,N_2)$. For $n\geq N$ we have $|z_{2n}-x_0|\leq \varepsilon$ and $|z_{2n-1}-x_0|\leq \varepsilon$ hence if $k\geq 2N-1$ we have $|z_k-x_0|\leq \varepsilon$.

3. ## Re: Subsequences

Originally Posted by dwsmith
Suppose $\{x_n\}\to x_0$ and $\{y_n\}\to x_0$. Define a sequence $\{z_n\}$ as follows: $z_{2n}=x_n$ and $z_{2n-1}=y_n$. Prove that $\{z_n\}$ converges to $x_0$.

Let $\epsilon >0$. Then $\exists N_1, \ N_2\in\mathbb{N}$ such that for $n\geq N_1, \ N_2$ we have $|x_n-x_0|<\epsilon$ and $|y_n-x_0|<\epsilon$.
Let $N=2(N_1+N_2)$. If $n\ge N$ then if $n\text{ is odd}$ we have $k = \left\lfloor {\frac{n}{2}} \right\rfloor > N_2$ and $z_n=y_k$.

Use a similar idea if $n\text{ is even}$.

4. ## Re: Subsequences

Originally Posted by Plato
Let $N=2(N_1+N_2)$. If $n\ge N$ then if $n\text{ is odd}$ we have $k = \left\lfloor {\frac{n}{2}} \right\rfloor > N_2$ and $z_n=y_k$.

Use a similar idea if $n\text{ is even}$.
Why is $N=2(N_1+N_2)$

5. ## Re: Subsequences

Originally Posted by dwsmith
Why is $N=2(N_1+N_2)$
First of all, it insures absolutely that $N>N_1~\&~N>N_2$.
Therefore, we can use anyone of the statements already restricted.

6. ## Re: Subsequences

Originally Posted by Plato
then if $n\text{ is odd}$ we have $k = \left\lfloor {\frac{n}{2}} \right\rfloor > N_2$ and $z_n=y_k$.
Can you also explain this?

7. ## Re: Subsequences

Originally Posted by dwsmith
Can you also explain this?
You do the mathematics.
Just take many cases.