# Union of a set

• Sep 3rd 2009, 12:51 PM
Union of a set
This is from a proof in my book:

If $\displaystyle A = \{ x : f(x) > 0 \}$, define $\displaystyle A_n = \{ x : f(x) > \frac {1}{n} \}$, then $\displaystyle A = \bigcup ^ \infty _ 1 A_n$

Why is this true?
• Sep 3rd 2009, 12:59 PM
siclar
Quote:

This is from a proof in my book:

If $\displaystyle A = \{ x : f(x) > 0 \}$, define $\displaystyle A_n = \{ x : f(x) > \frac {1}{n} \}$, then $\displaystyle A = \bigcup^\infty_1 A_n$

Why is this true?

To show set equality we just need to show inclusion, right? One inclusion is obvious:

Clearly $\displaystyle A_n\subseteq A$ for each $\displaystyle n$ so
$\displaystyle A \supseteq \bigcup^\infty_1 A_n$.

The other inclusion is not much harder. Let $\displaystyle x\in A$. Then $\displaystyle f(x)>0$ so by the Archimedean principle there exists an integer $\displaystyle M$ such that $\displaystyle Mf(x)>1$, and so $\displaystyle f(x)>\dfrac{1}{M}$ thus $\displaystyle x\in A_M\subseteq \bigcup^\infty_1 A_n$. Therefore $\displaystyle A \subseteq \bigcup^\infty_1 A_n$ and so $\displaystyle A = \bigcup^\infty_1 A_n$!
• Sep 3rd 2009, 01:00 PM
Plato
Quote:

If $\displaystyle A = \{ x : f(x) > 0 \}$, define $\displaystyle A_n = \{ x : f(x) > \frac {1}{n} \}$, then $\displaystyle A = \bigcup ^ \infty _ 1 A_n$
$\displaystyle a \in A\, \Rightarrow \,f(a) > 0\, \Rightarrow \,\left( {\exists n} \right)\left[ {\frac{1}{n} < f(a)} \right]\, \Rightarrow \,a \in A_n$