I want to show that $A=\cup_{n=1}^{\infty} A_n$

Question

Let $A=\{x\in X: \varphi(x)>0\}$ and

$A_n=\{x\in X: \varphi(x)\geq1/n\}$ (increasing)

Since $A_n$ is increasing $\lim A_n =\cup_{n=1}^{\infty} A_n$ .....$\lim 1/n=0$ — Andrew, Sep 28 '19 at 15:53

score 2 · Answer 1 · answered Sep 28 '19 at 15:32

2

It is clear that $A_n\subseteq A$ so

$\bigcup_{n}A_n\subseteq A$.

Let $x\in A$, then $\phi(x)>0$, so there exists a natural number $N$ such that

$\phi(x)\geq \frac{1}{N}$

Thus $x\in A_N$

answered Sep 28 '19 at 15:32

Federico Fallucca

1 Answers1