Subsequences

**dwsmith** · June 28th 2011, 07:23 AM

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.

**girdav** · June 28th 2011, 07:46 AM

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 $|z_k-x_0|\leq \varepsilon$ .

**Plato** · June 28th 2011, 07:48 AM

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 $n\text{ is even}$ .

**dwsmith** · June 28th 2011, 03:11 PM

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 $N=2(N_1+N_2)$

**Plato** · June 28th 2011, 03:19 PM

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.

**dwsmith** · June 28th 2011, 03:47 PM

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?

**Plato** · June 28th 2011, 04:41 PM

Originally Posted by dwsmith

Can you also explain this?

You do the mathematics.
Just take many cases.

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