# [SOLVED] Real Analysis Proof: Intersection and Union of an Indexed Set

• Oct 12th 2008, 06:28 PM
ilikedmath
[SOLVED] Real Analysis Proof: Intersection and Union of an Indexed Set
For each http://qaboard.cramster.com/Answer-B...2087503987.gif let http://qaboard.cramster.com/Answer-B...0833750886.gif. (Note Q is the set of all rationals).
Prove that:
(1) http://qaboard.cramster.com/Answer-B...2712509613.gif (that is the intersection over the entire index is the singleton set with 0).

and prove that

(2) http://qaboard.cramster.com/Answer-B...8025003118.gif (that is the union over the entire index is the set of all rationals in the interval [0,1) ).

---
Okay, wow. I am utterly stumped by this. Things I do know that $E_x$ is the set of all sets with
the form [0 , x) for all x in the interval (0, 1). The intersection is what all of those sets have in common, and I can see that it is 0; but I can't figure out the proof. Since it is proving two sets equal, I would assume having to "chase elements" and prove the set on the right side is a subset of the left and vice versa. We did a proof similar to this in class, and our professor proved it using contradiction. So then maybe I will also need to use contradiction to prove these two statements?

As for the union proof, I'm also thinking of a chasing elements proof. But again, where to start is what I am struggling with.
I listed out some sets like:
$E_{1/16}$ = [0, 1/16)
$E_{1/8}$ = [0, 1/8)
$E_{1/4}$ = [0, 1/4)
$E_{1/3}$ = [0, 1/3)
$E_{1/2}$ = [0, 1/2)
So now I kind of 'see' better how the union would be everything including 0 but not reaching 1 because x $\in$ (0, 1).

Any helps, hints, tips, and/or suggestions on getting started are greatly appreciated!
• Oct 12th 2008, 11:08 PM
terr13
I'm taking Real Analysis as well, so my answers may not be correct.

For the intersection, you can say take a x, and then you have the set E_x. Take any element e \in E_x as your new x, and you get a new set. You can continue this and by Archimedean property the smallest such element would be {0}, so their intersection would be {0}. You can probably do the same with the other side.
• Oct 13th 2008, 06:38 AM
ilikedmath
Quote:

Originally Posted by terr13
I'm taking Real Analysis as well, so my answers may not be correct.

For the intersection, you can say take a x, and then you have the set E_x. Take any element e \in E_x as your new x, and you get a new set. You can continue this and by Archimedean property the smallest such element would be {0}, so their intersection would be {0}. You can probably do the same with the other side.

Okay, thanks. I'm still having a hard time "seeing" things to begin with. This notation is really throwing me off. :|

In class our professor did a proof similar to this:

Proof: Note that for all x in [1, ∞), -1/x < 0 < 1/x.
Thus 0 is in Ex for all x in [1, ∞) http://qaboard.cramster.com/Answer-B...4587503126.gif.
From here we split the y's into cases: If y > 0 and if y < 0.
In both cases we came to a contradiction which ended up completing the proof.
I don't quite "see" how to do that though. I am really confused and lost here. :| Any help at all is extremely helpful. Thank you for your time!
• Oct 13th 2008, 06:57 AM
Plato
Note that $y \ne 0 \Rightarrow \left( {\exists k \in \mathbb{Z}^ + } \right)\left[ {\frac{1}{k} < \left| y \right|} \right]$.
That means that $y \notin \left( {\frac{{ - 1}}{k},\frac{1}{k}} \right)$.
Can you finish?
• Oct 13th 2008, 07:07 AM
ilikedmath
Quote:

Originally Posted by Plato
Note that $y \ne 0 \Rightarrow \left( {\exists k \in \mathbb{Z}^ + } \right)\left[ {\frac{1}{k} < \left| y \right|} \right]$.
That means that $y \notin \left( {\frac{{ - 1}}{k},\frac{1}{k}} \right)$.
Can you finish?

Thanks for looking.

We already completed that proof in class. I'd only included it here as an example of a proof that might be similar to the proof I have to do for this certain problem.

I was wondering if I should approach proving the intersection and union of this problem (with x in (0,1)) the way the other proof (with x in [1, inifinity)) was approached. In other words, do I also use contradiction and end up with 2 cases?

What is troubling me most is that I can't "see"/"read" the notation to where I can understand what it means. So that causes trouble in my getting started in the proof.