Subsequences

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

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

I don't know what to do now.

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

Quote:

---

Originally Posted by dwsmith

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

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

---

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

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

Quote:

---

Originally Posted by Plato

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

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

---

Why is $\displaystyle N=2(N_1+N_2)$

Quote:

---

Originally Posted by dwsmith

Why is $\displaystyle N=2(N_1+N_2)$

---

First of all, it insures absolutely that $\displaystyle N>N_1~\&~N>N_2$.
Therefore, we can use anyone of the statements already restricted.

Quote:

---

Originally Posted by Plato

then if $\displaystyle n\text{ is odd}$ we have $\displaystyle k = \left\lfloor {\frac{n}{2}} \right\rfloor > N_2 $ and $\displaystyle z_n=y_k$.

---

Can you also explain this?

Quote:

---

Originally Posted by dwsmith

Can you also explain this?

---

You do the mathematics.
Just take many cases.