Proving intersection and union of a particular indexed set

Question

I'm having trouble proving these, can anyone help?

The question is as follows:

For each $n \in \mathbb{N}$, let $A_n = \lbrace{ k \in \mathbb{Z} ; k^2 <= n}\rbrace$

Prove that:

1) $\bigcap A_n$ = $\lbrace 0, 1, -1 \rbrace$

2) $\bigcup A_n = \lbrace ...,-2,-1,0,1,2,...\rbrace = \mathbb{Z}$

Can anyone show me how to prove these?

Thanks in advance.

Answer 1

1) The fact that $\{-1,0,1\}\subset \bigcap_{n}A_n$ is clear. For the converse inclusion, if $k\in \bigcap_{n}A_n$, then $k^2\leq n$ for all $n\in\mathbb N^*$. In particular, $k^2\leq 1$, and thus $k\in\{0,1,-1\}$.

2) The fact that $\bigcup_{n\in\mathbb N}A_n\subset \mathbb Z$ is clear. Let $k\in\mathbb Z$. In particular, $k^2\in A_{k^2}$. The claim follow.

Answer 2

Hints:

1. Let $$x\in\bigcap_{n=1}^\infty A_n$$ Then, $x\in A_n$ for every $n\in\mathbb{N}$. What can you get from here?
2. Let $$x\in\bigcup_{n=1}^\infty A_n$$ Then, there exists some $n_0\in\mathbb{N}$ such that $x\in A_{n_0}$. What can you get from this?

Edit: After your request:

For the first: let $$x\in\bigcap A_n$$ Then, as you said, $x\in A_1$, so $x\in\{-1,0,1\}$. So $$\bigcap A_n\subseteq \{-1,0,1\}$$ Now, since $$\{-1,0,1\}\subseteq \bigcap A_n$$ it comes that $$\bigcap A_n=\{-1,0,1\}$$ Try to use somthing similar for the second. The most usual way to show that two sets are equal is to use the following:

$$A=B\Leftrightarrow A\subseteq B\text{ and }B\subseteq A$$