To show that $\bigcup_{n\in \mathbb{N}}[-n,n]=\mathbb{R}$

Question

We are asked to show that $\bigcup_{n\in \mathbb{N}}[-n,n]=\mathbb{R}$. This is in the context of families of sets. We have previously covered proofs, sets and boolean operators. I've managed to use one of the definitions of families of sets provided in the book to obtain that:

$$ \bigcup_{n\in N}[-n,n]=\{x:x \textrm{ belongs to at least one set} \{x\in \mathbb{R}:-n\leq x \leq n\}\}. $$

My attempts of a proof suffers from a lack of tools i think. But i think it is obvious why the union equals $\mathbb{R}$ because if one consider that one of the sets is $[-\infty,\infty]$, then x must be in all of the smaller sets. Any ideas for making this a rigorous proof?

Answer 1

$\mathbb{R}$ is a whole space, hence one direction of inclusion is obvious, namely $$\bigcup\limits_{n \in \mathbb{N}} [-n, n] \subseteq \mathbb{R}$$

The other direction. Pick $x \in \mathbb{R}$ and show it belongs to some interval, for example to $[-\lceil|x|\rceil, \lceil|x|\rceil]$.

Answer 2

If $x \in \mathbb R$, then, by the Archimedean property of the reals, there is $n \in \mathbb N$ such that $|x| \le n$, hence $x \in [-n,n]$.

Can you proceed ?