Find $\bigcap \mathcal A^c$ and $\bigcup \mathcal A^c$ with the proof.

Question

Let $A_n =\{x \in \Bbb R :-\frac {1}{n} \lt x \lt \frac{1}{n}\}$,$n \in \Bbb N$ and define the indexed family $\mathcal A^c = \{ A_{n}^{c} :n \in \Bbb N \}$. Find $\bigcap \mathcal A^c$ and $\bigcup \mathcal A^c$ with the proof.

I have no idea how to do.

If $A_{n} = \left(-\frac{1}{n},\frac{1}{n}\right)$, then $A_{n}^{c} = \left(-\infty,-\frac{1}{n}\right]\cup\left[\frac{1}{n},\infty\right)$. Now, draw the sets on the real line. — Lucas, Jan 06 '19 at 03:53

score 2 · Answer 1 · answered Jan 06 '19 at 04:59

2

$\cap_n \mathcal A_n =\cap_n (-\frac 1 n,\frac 1 n)^{c}=(\cup_n (-1,1))^{c}=(-1,1)^{c}=\mathbb R \setminus (-1,1)$. I leave it to you to show that $\cup_n \mathcal A_n =\mathbb R \setminus \{0\}$.

answered Jan 06 '19 at 04:59

Kavi Rama Murthy

311,013

Find $\bigcap \mathcal A^c$ and $\bigcup \mathcal A^c$ with the proof.

1 Answers1