Let $A_n=(x_n-1,x_n+1)$. Does exist $\lim_n\sup A_n$?

Question

Let $\{x_1,x_2,\dots\}$ be some enumeration of the rational numbers $\Bbb Q\subset\Bbb R$ and define $A_n=(x_n-1,x_n+1)$. Does exist $\lim_\limits n\sup A_n$ and $\lim_\limits n\inf A_n$?

I've understand $\lim\limits_n\sup X_n$ as the set of elements of $X_n$ for an infinite values of $n$. In every $A_n$ exists an infinite number of rational numbers. By this intuition makes me think "If the limit exist, then we should have $\lim\sup A_n=\Bbb R$", is this correct ?

For make me understand better the problem, I've tried changing the problem looking for some equivalent sentence, but I have no ideia how to make progress on this.

Answer 1

1. To prove: $\limsup_{n}A_{n}=\mathbb{R}$.

Let $x\in\mathbb{R}$. Let $n\in\mathbb{N}$ be arbitrary. Note that $(x-1,x+1)\cap\mathbb{Q}$ is an infinite set, so there exists $k\geq n$ such that $x_{k}\in(x-1,x+1)$. Therefore $|x-x_{k}|<1$ and it implies that $x\in(x_{k}-1,x_{k}+1)$. Hence, $x\in\cup_{k\geq n}A_{k}$. Lastly, since $n$ is arbitrary, we have $x\in\cap_{n}\cup_{k\geq n}A_{k}=\limsup_{n}A_{n}$. It follows that $\limsup_{n}A_{n}=\mathbb{R}$.

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2. To prove: $\liminf A_{n}=\emptyset$.

Prove by contradiction. Suppose that there exists $x\in\liminf A_{n}$. Then, there exists $n\in\mathbb{N}$ such that $x\in\cap_{k\geq n}A_{k}$. That is, $|x_{k}-x|<1$ for all $k\geq n$. In another word, $x_{k}\in(x-1,x+1)$ for all $k\geq n$. Now, $\mathbb{Q}\setminus(x-1,x+1)\subseteq\{x_{1},x_{2},\ldots,x_{n-1}\}$, which is a contradiction because $\mathbb{Q}\setminus(x-1,x+1)$ is an infinite set.

Answer 2

The idea is valid, but it is not applied correctly.

If you want to prove that the limsup is $\mathbb R$, then you must consider any $y\in \mathbb R$ and prove that it is contained frequently in $n$ into $A_n$.

Now, since $(y-1,y+1)$ contains an infinite number of rationals, then the thesis follows.

Answer 3

Hint

$$\lim_\limits n\sup A_n = \cap_{n \geq 1} \cup_{k \geq n} A_k$$

Show that for each $n \geq 1$ you have $$ \cup_{k \geq n} A_k = \mathbb R$$

For the other

$$\lim_\limits n\inf A_n = \cup_{n \geq 1} \cap_{k \geq n} A_k$$

Show that for each $n \geq 1$ you have $$ \cap_{k \geq n} A_k = \emptyset$$