Proof with sets

**Borkborkmath** · September 30th 2011, 12:19 PM

"Let I be the set of real numbers that are greater than 0. For each x $\in$ I, let $A_x$ be the open interval (0,x). Prove that $\cap$ $_{x\in I}$ $A_x$ = $\emptyset$ , $\cup$ $_{x\in I}$ $A_x$ = I. For each x $\in$ I, let $B_x$ be the closed interval [o, x]. Prove that $\cap$ $_{x\in I}$ $B_x$ = {0}, $\cup$ $_{x\in I}$ $B_x$ = I $\cup$ {0}."

**Plato** · September 30th 2011, 12:45 PM

Originally Posted by Borkborkmath

"Let I be the set of real numbers that are greater than 0. For each x $\in$ I, let $A_x$ be the open interval (0,x). Prove that $\cap$ $_{x\in I}$ $A_x$ = $\emptyset$ , $\cup$ $_{x\in I}$ $A_x$ = I. For each x $\in$ I, let $B_x$ be the closed interval [o, x]. Prove that $\cap$ $_{x\in I}$ $B_x$ = {0}, $\cup$ $_{x\in I}$ $B_x$ = I $\cup$ {0}."

All of these follow from two basic facts:
$\left( {\frac{1}{n}} \right) \to 0\;\& \,(n) \to \infty$ .

That means that if $x\in\mathcal{I}$ then $\left( {\exists k \in \mathbb{Z}^ + } \right)\left[ {\frac{1}{k} < x} \right]$ .
So $x \notin \left( {0,\frac{1}{k}} \right)$ . That is the first.

Moreover, for the second: $\left( {\exists K \in \mathbb{Z}^ + } \right)\left[ {K > x} \right]$ so $x\in (0,K)$ .

**Borkborkmath** · October 2nd 2011, 09:21 AM

I might be wording this wrong, but the idea is similar to finite subcovers?

**Plato** · October 2nd 2011, 09:31 AM

Originally Posted by Borkborkmath

I might be wording this wrong, but the idea is similar to finite subcovers?

What? This question is about generalized intersections and unions.
What do covers have to do with that?
You are asked to show $\bigcap\limits_{x \in I} {A_x } = \emptyset \;\& \;\bigcup\limits_{x \in I} {A_x }=\mathcal{I}$

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