# Proof with sets

• Sep 30th 2011, 01:19 PM
Borkborkmath
Proof with sets
"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}."
• Sep 30th 2011, 01:45 PM
Plato
Re: Proof with sets
Quote:

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 $x\in (0,K)$.
• Oct 2nd 2011, 10:21 AM
Borkborkmath
Re: Proof with sets
I might be wording this wrong, but the idea is similar to finite subcovers?
• Oct 2nd 2011, 10:31 AM
Plato
Re: Proof with sets
Quote:

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 $\bigcap\limits_{x \in I} {A_x } = \emptyset \;\& \;\bigcup\limits_{x \in I} {A_x }=\mathcal{I}$