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

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

For each let . (Note Q is the set of all rationals).
Prove that:
(1) (that is the intersection over the entire index is the singleton set with 0).

and prove that

(2) (that is the union over the entire index is the set of all rationals in the interval [0,1) ).

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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!
Thank you for your time.

2. 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.

3. 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:

Consider for x [1 , ∞).
or .
Claim:

Proof: Note that for all x in [1, ∞), -1/x < 0 < 1/x.
Thus 0 is in Ex for all x in [1, ∞) .
Assume such that .
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!

4. 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?

5. 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.

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