$\forall n\in\mathbb N$, let $A_n=\{x\in\mathbb R\mid n-1\lt x\lt n\}$. Prove that $\displaystyle\bigcup_{n\in\mathbb N}A_n=(\mathbb R^+-\mathbb N)$.

Question

$\forall n\in\mathbb N$, let $A_n=\{x\in\mathbb R\mid n-1\lt x\lt n\}$. Prove that $\{A_n\mid n\in\mathbb N\}$ is a pairwise disjoint family of sets and that $\displaystyle\bigcup_{n\in\mathbb N}A_n=\left(\mathbb R^+-\mathbb N\right)$.

Could someone please check my proof for the first part and provide any feedback on its correctness? I'm pretty sure I'm missing something (at least in the 2nd part).

Proof (part 1)

Let $\alpha,\beta\in\mathbb N$ such that $A_\alpha\neq A_\beta$ and assume, without loss of generality, that $\alpha\leq\beta$. This means that $\alpha\leq\beta-1$. Now, $$A_\alpha\cap A_\beta=\{x\in\mathbb R\mid(\alpha-1\lt x\lt\alpha)\land(\beta-1\lt x\lt\beta)\}$$ But since $\alpha\leq\beta-1$, $$A_\alpha\cap A_\beta=\{\alpha-1\lt x\lt\alpha\leq\beta-1\lt x\lt\beta\}$$ But this is a set of numbers $x\in\mathbb R$ that satisfy $x\lt x$ which is impossible. Hence $A_\alpha\cap A_\beta=\emptyset$ and so $\{A_n\mid n\in\mathbb N\}$ is pairwise disjoint.

Proof (part 2)

Let $a,b\in\mathbb N$ such that $a\neq b$ and assume, without loss of generality, that $a\lt b$. Since the family is pairwise disjoint, $$A_a\cap A_b=(a-1,a)\cap(b-1,b)=\emptyset$$ This means that $a\notin\displaystyle\bigcap_{n\in\mathbb N}A_n$. Hence, $\forall a\in\mathbb N$, $a\notin\displaystyle\bigcap_{n\in\mathbb N}A_n$. Now, $\forall a\in\mathbb N$, $(a-1,a)\in\bigcap_{n\in\mathbb N}A_n$. Hence $\bigcap_{n\in\mathbb N}A_n=\left(\mathbb R^+-\mathbb N\right)$.

Answer 1

Your first part is generally correct, I have one small issue with it. You start with

Let $\alpha,\beta\in\mathbb N$ such that $A_\alpha\neq A_\beta$ and assume, without loss of generality, that $\alpha\leq\beta$.

The fact that $\alpha \neq \beta$ is implicit with $A_\alpha\neq A_\beta$, but you should state it explicitly. Also, the $\alpha\leq\beta$, although technically correct, would be better expressed as $\alpha \lt \beta$ to help emphasize this point.

For the proof of the second part, you have tried proving for the intersection of the sets instead of the union. Thus, it doesn't work. Instead, you could state something like let $a \in \mathbb{N}$. If $a \in \bigcup_{n\in\mathbb N}A_n$, then there exists a $b \in \mathbb{N}$ such that $a \in A_{b}$, which means $b - 1 \lt a \lt b$, but this is not possible, so $a \not\in \bigcup_{n\in\mathbb N}A_n$. Next, consider any $r \in \mathbb R^+-\mathbb N$. Then $r = m + x$ where $m \ge 0$ is an integer, so $m + 1 \in \mathbb{N}$, and $0 \lt x \lt 1$. You have that $m \lt r \lt m + 1$, so $r \in A_{m+1}$ and, thus, $r \in \bigcup_{n\in\mathbb N}A_n$. This shows that all natural numbers are not in this union, but all other positive real numbers are, concluding that $\displaystyle\bigcup_{n\in\mathbb N}A_n=\left(\mathbb R^+-\mathbb N\right)$ is true.