$\lim_{n\to\infty}\sup x_n= \max(x,y)$ and $\lim_{n\to\infty}\inf x_n= \min(x,y)$

Question

I am trying to prove this:

Let $x_n$ a real sequence. Suppose that the subsequence $x_{2n}$ converges to $x$ and $x_{2n+1}$ converge to $y$. Show that:

$\lim\sup_{n\to\infty} x_n= \max(x,y)$ and $\lim\inf_{n\to\infty} x_n= \min(x,y)$

After this, I have to show that if both subsequences converge to $x$, then $x_n$ converges to $x$, but I could prove this part.

I know I am supposed to show my work, but I am stucked! Any hint is very welcomed.

Answer 1

I will prove this using my favourite definition of the Limit Superior and Limit Inferior. Drop a comment if you need more help.

Let $ \mathscr L_{(x_n)} = \{l \in \Bbb R \ | \ \text{there is a subsequence of $ (x_n) $ which converges to $l$}\}$

We will prove that $ \mathscr L_{(x_n)} = \{ x, y \} $.

Well one direction is easy since we know that there are two subsequences of $(x_n)$ - namely, $(x_{2n})$ and $x_{(2n + 1)}$, respectively - which converge to $ x $ and $y$. Hence, $ \{ x, y \} \subseteq \mathscr L_{(x_n)} $.

Now prove that there is nothing in $ \mathscr L_{(x_n)} $ other than $x$ and $y$. To this end, suppose there exists $ p \in \mathscr L_{(x_n)} $ such that $ p \not \in \{x, y\} $. Then there is, by definition, a subsequence, $(x_{n_i}) $ of $(x_n)$ such that $ \lim \limits_{i \to \infty } x_{n_i} = p $. Now let $\epsilon = \min \{ |p - x|, |p - y| \}$. Then there is $I \in \Bbb N$ such that $ i \gt I \implies | x_{n_i} - p | \lt \frac{ \epsilon}{2} $. But this would mean that for $i \gt I$,

$$ \frac{\epsilon}{2} = \epsilon - \frac{\epsilon}{2} \le | x - p |- \frac{\epsilon}{2} \lt | x - p | - | x_{n_i} - p | \le | x - p - x_{n_i} + p | = |x_{n_i} - x| $$

and this contradicts the fact that $(x_{2n}) \to x$.

Hence, $$ \mathscr L_{(x_n)} = \{ x, y \} $$

Now if you will grant me that a finite set has a maximum and a minimum then,

$$ \lim \sup x_n = \sup \mathscr L_{(x_n)} = \max \mathscr L_{(x_n)} = \max \{x, y\} \;\; \text{and} $$

$$ \lim \inf x_n = \inf \mathscr L_{(x_n)} = \min \mathscr L_{(x_n)} = \min \{x, y\} $$

Answer 2

Let $(x_n)$ be a real sequence. Suppose that the subsequence $(x_{2n})$ converges to $x$ and $(x_{2n+1})$ converge to $y$. Suppose for the moment that $x\neq y$ and let $(x_{n_k})$ be any convergent subsequence of $(x_n)$. Then it can only converge to $x$ or $y$. This follows from the fact that, by the definition of a subsequence, at least one of the sets $\{k: n_k \text{ is even}\}$ or $\{k: n_k \text{ is odd}\}$ will be infinite.

1. If $\{k: n_k \text{ is even}\}$ is infinite, then there is a subsequence of $(x_{n_k})$ converging to $x$, and since $(x_{n_k})$ is convergent, $x$ is also its limit.
2. If $\{k: n_k \text{ is odd}\}$ is infinite, then there is a subsequence of $(x_{n_k})$ converging to $y$, and since $(x_{n_k})$ is convergent, $y$ is also its limit.
3. If both $\{k: n_k \text{ is even}\}$ and $\{k: n_k \text{ is odd}\}$ are infinite, there are subsequences of $(x_{n_k})$ converging to $x$ and $y$. So $(x_{n_k})$ and $(x_n)$ can not be convergent.

The $\limsup$ and $\liminf$ are the $\sup$ and $\inf$ of the subsequential limits respectively. Since there are only $x$ and $y$, we have $$\limsup_{n\to\infty} x_n= \max(x,y), \\ \liminf_{n\to\infty} x_n=\min(x,y).$$

If $x=y$, then $(x_n)$ converges to $x=y$, see here for example.