Let $A_{x}= (x-1, x+1)$ and $\mathcal{A}=\{A_{x} \ | \ x \in \mathbb{R}^{+}\}$. Find $\bigcup \mathcal{A}$ and $\bigcap \mathcal{A}$.

Question

Let $A_{x}= (x-1, x+1)$ and $\mathcal{A}=\{A_{x} \ | \ x \in \mathbb{R}^{+}\}$. Find $\bigcup \mathcal{A}$ and $\bigcap \mathcal{A}$.

Attempt. $\bigcap \mathcal{A} = \emptyset$.

Proof. Suppose not, that is, there exists an $e$ in the intersection. So, $e \in (x-1,x+1)$ for all $x \in \mathbb{R}^+$. Hence, if $e=2$, then $e \in (1,3)$. Thus, if $e=4$, then $e \notin (3,5)$, which is a contradiction.

$\bigcup \mathcal{A}=(-1, \infty)$.

Proof. Let $e \in \bigcup \mathcal{A}$. Then $e \in (x-1,x+1)$ for some $x \in \mathbb{R}^+$. This implies that $e \in (-1, \infty)$ since $(x-1,x+1) \subseteq (-1, \infty)$. Now, let $e \in (-1, \infty)$. Then $e \in (x-1,x+1)$ since $(x-1,x+1) \subseteq (-1, \infty).$

Is the logic correct especially for the second proof?

Answer 1

You have the right idea, but this doesn't work:

Now, let $e \in (-1, \infty)$. Then $e \in (x - 1, x + 1)$ since $(x - 1, x + 1) \subseteq (-1, \infty)$.

You haven't shown that given arbitrary $e$ in $(-1, \infty)$ that there exists $x \in \mathbb{R}^+$ such that $e \in (x - 1, x + 1)$. For this, you can consider two cases: either $e \in (-1, 0]$ or $e \in (0, \infty)$. In the latter case, simply take $x = e \in \mathbb{R}^+$; surely $e \in (e - 1, e + 1)$. In the former case, take $x = \frac{1}{2} (e + 1) \in \mathbb{R}^+$. Note that $\frac{1}{2} (e + 1) - 1 = \frac{1}{2} (e - 1) < e$ since $e > -1$ and $\frac{1}{2} (e + 1) + 1 = \frac{1}{2} (e + 3) > e$ since $e \le 0 < 3$.

EDIT: You also seem to have a few typos. For instance, in the part about the intersection, it should be $\color{red}{e} \in (x - 1, x + 1)$ for all $\color{red}{x} \in \mathbb{R}^+$.

Answer 2

To prove that $\bigcap\mathcal A=\varnothing$, you may notice that, for example, $$\bigcap\mathcal A \subseteq A_1 \cap A_3 = \varnothing.$$

To prove that $\bigcup\mathcal A=(-1,+\infty)$, pick $e>-1$, and notice that if $e\leq0$, then $e\in A_{(e+1)/2}$; if $e>0$, then $e\in A_e$. In any case, $e \in \bigcup\mathcal A$. Conversely, if $e \in \bigcup\mathcal A$, then certainly $e>-1$.