Results 1 to 5 of 5

Math Help - [SOLVED] Real Analysis Proof: Intersection and Union of an Indexed Set

  1. #1
    Member ilikedmath's Avatar
    Joined
    Sep 2008
    Posts
    98

    Exclamation [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) ).

    ---
    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.
    Last edited by ilikedmath; October 12th 2008 at 07:12 PM. Reason: Edited my 'work' so far
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member
    Joined
    Oct 2008
    Posts
    147
    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.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member ilikedmath's Avatar
    Joined
    Sep 2008
    Posts
    98
    Quote Originally Posted by terr13 View Post
    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!
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,617
    Thanks
    1581
    Awards
    1
    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?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member ilikedmath's Avatar
    Joined
    Sep 2008
    Posts
    98
    Quote Originally Posted by Plato View Post
    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.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Real Analysis: Union of Countable Infinite Sets Proof
    Posted in the Differential Geometry Forum
    Replies: 4
    Last Post: September 14th 2010, 05:38 PM
  2. intersection and union proof
    Posted in the Number Theory Forum
    Replies: 1
    Last Post: February 25th 2010, 01:59 PM
  3. Replies: 2
    Last Post: November 21st 2009, 12:51 AM
  4. Proving the intersection and union of the indexed...
    Posted in the Discrete Math Forum
    Replies: 1
    Last Post: May 5th 2009, 12:20 PM
  5. Replies: 20
    Last Post: October 15th 2008, 10:30 AM

Search Tags


/mathhelpforum @mathhelpforum