Show $\bigcup _{i\varepsilon \mathbb{Z}}\left[ i,i+1\right) =\mathbb{R}$ My proof idea. $(\Rightarrow)$. Let $x\in\bigcup _{i\in \mathbb{Z}}\left[ i,i+1\right)$. We need to show that $x\in\left( -\infty ,\infty \right)$. Actually, is it clear?

$\left( \Leftarrow \right)$. Let $x\in\left( -\infty ,\infty \right)$. We need to show that $x\in\bigcup _{i\in \mathbb{Z}}\left[ i,i+1\right)$. How can I show?
I think the left inclusion is trivial since $[i, i+1) \subset \Bbb R \ \forall i \in \Bbb Z$. Now, for the other inclusion, pick $x \in \Bbb R$, and assume $x &gt; 0$. Then by the Archimedean principle there exists some $n \in \Bbb N : n &gt; x$. Furthermore, since $\Bbb N$ is well-ordered, we can pick the least such $n$. Then $n &gt; x \ge n - 1 \implies x \in [n-1, n)$, and obviously $n-1 \in \Bbb Z$. For $x &lt; 0$ a very similar argument can be made taking $n &gt; - x$ (be careful with half open intervals extremes, though). The case $x=0$ is trivial.
$\left( \Leftarrow \right)$. $\forall x \in (-\infty, +\infty)$, always $\exists i \in \mathbb Z$, such that $i \le x &lt; i+1$, thus $x \in [i, i+1)$, thus $x \in \cup_{i\in \mathbb Z}[i,i+1)$. Thus $(-\infty, +\infty) \subset \cup_{i\in \mathbb Z}[i,i+1)$