Union of a set

**tttcomrader** · September 3rd 2009, 12:51 PM

This is from a proof in my book:

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

Why is this true?

**siclar** · September 3rd 2009, 12:59 PM

Originally Posted by tttcomrader

This is from a proof in my book:

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

Why is this true?

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

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

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

**Plato** · September 3rd 2009, 01:00 PM

Originally Posted by tttcomrader

This is from a proof in my book:

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

Why is this true?

One way is completely obvious.
$a \in A\, \Rightarrow \,f(a) > 0\, \Rightarrow \,\left( {\exists n} \right)\left[ {\frac{1}{n} < f(a)} \right]\, \Rightarrow \,a \in A_n$

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