1. ## Proof with sets

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

2. ## Re: Proof with sets

Originally Posted by Borkborkmath
"Let I be the set of real numbers that are greater than 0. For each x$\displaystyle \in$I, let $\displaystyle A_x$ be the open interval (0,x). Prove that $\displaystyle \cap$$\displaystyle _{x\in I}$$\displaystyle A_x$ = $\displaystyle \emptyset$, $\displaystyle \cup$$\displaystyle _{x\in I}$$\displaystyle A_x$ = I. For each x $\displaystyle \in$ I, let $\displaystyle B_x$ be the closed interval [o, x]. Prove that $\displaystyle \cap$$\displaystyle _{x\in I}$$\displaystyle B_x$ = {0}, $\displaystyle \cup$$\displaystyle _{x\in I}$$\displaystyle B_x$ = I$\displaystyle \cup${0}."
All of these follow from two basic facts:
$\displaystyle \left( {\frac{1}{n}} \right) \to 0\;\& \,(n) \to \infty$.

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

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

3. ## Re: Proof with sets

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

4. ## Re: Proof with sets

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