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?

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- Sep 3rd 2009, 12:51 PMtttcomraderUnion 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 PMsiclar
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 PMPlato