1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Point set proof

  1. Sep 14, 2012 #1

    Zondrina

    User Avatar
    Homework Helper

    1. The problem statement, all variables and given/known data

    Let A and B be subsets of ℝn with A0, B0 denoting the sets of interior points for A and B respectively. Prove that A0[itex]\cup[/itex]B0 is a subset of the interior of A[itex]\cup[/itex]B. Give an example where the inclusion is strict.

    2. Relevant equations

    I know a point Q[itex]\in[/itex]S is an interior point of S if [itex]\exists N_δ(Q)[/itex] which is a subset of S.

    3. The attempt at a solution

    I've never actually attempted a problem like this, I'm wondering where to start really. Do I assume the existence of a point in A0[itex]\cup[/itex]B0 and then prove it is also contained within (A[itex]\cup[/itex]B)0 ? Any nudge in the right direction would be very helpful.
     
  2. jcsd
  3. Sep 14, 2012 #2

    jbunniii

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Yes, that's exactly what you need to do.

    So, suppose [itex]x \in A^o \cup B^o[/itex]. Then either [itex]x \in A^o[/itex] or [itex]x \in B^o[/itex]. If [itex]x \in A^o[/itex], can you prove that [itex]x \in (A \cup B)^o[/itex]?
     
  4. Sep 14, 2012 #3

    Zondrina

    User Avatar
    Homework Helper

    EDIT : I thought about it a bit, is this better?

    Suppose we pick x[itex]\in[/itex]A0. WAIT NOW... We know that A0 is the set of interior points for A so since x[itex]\in[/itex]A0 we can also say that x[itex]\in[/itex]A since x is contained within the interior of A. If x[itex]\in[/itex]A, then it follows that x[itex]\in[/itex](A[itex]\cup[/itex]B) and will be contained within the interior of A[itex]\cup[/itex]B that is x[itex]\in[/itex](A[itex]\cup[/itex]B)0?

    I know this is wrong, but maybe its a step in the right direction?
     
    Last edited: Sep 14, 2012
  5. Sep 14, 2012 #4

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Use the definition of interior point. If x is an interior point of A then there is a neighborhood of x that is contained in A. Is that neighborhood contained in AUB?
     
  6. Sep 14, 2012 #5

    Zondrina

    User Avatar
    Homework Helper

    It seems trivial in saying that. Showing it is a different story though.

    So since x[itex]\in[/itex]A0, [itex]\exists δ>0 | N_δ(x) \subseteq A [/itex]

    This is obvious to me ^, but how would I use this notion to show that x[itex]\in[/itex](A[itex]\cup[/itex]B) or does it go without saying?
     
  7. Sep 14, 2012 #6

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Goes through without much said. If x is in the interior of A then x is in A. x is in the neighborhood. If x is in A then x is in AUB. Now what about the neighborhood of x?
     
  8. Sep 14, 2012 #7

    Zondrina

    User Avatar
    Homework Helper

    Since x is contained within the interior of A, we know x is contained within A itself. The neighborhood of x is therefore contained within A.

    Since x is in A, we know that x is in A[itex]\cup[/itex]B

    So would it follow that the neighborhood of x is contained ( is a subset of ) within A[itex]\cup[/itex]B? If this were the case then since we know the neighborhood is inside A[itex]\cup[/itex]B, we know it would be contained within the interior of A[itex]\cup[/itex]B?
     
  9. Sep 14, 2012 #8

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    You are making this seem harder than it is. If the neighborhood of x is contained in A (by definition of x being an interior point), then it must be contained in AUB, right? A is contained in AUB.
     
  10. Sep 14, 2012 #9

    Zondrina

    User Avatar
    Homework Helper

    My apologies, it's been a long day so my brain isn't registering like it usually does.

    Yes i understand that the neighborhood of x being contained in A tells us that the neighborhood is also contained within A[itex]\cup[/itex]B.

    So if A [itex]\subseteq[/itex] A[itex]\cup[/itex]B that tells me that A0 [itex]\subseteq[/itex] (A[itex]\cup[/itex]B)0? Is it actually that straightforward?
     
  11. Sep 14, 2012 #10

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Yes, it is. Now you have to find the example where the inclusion is strict. You can find one by thinking about closed intervals on the real line.
     
  12. Sep 14, 2012 #11

    Zondrina

    User Avatar
    Homework Helper

    So ill just clean up everything that's been said into one post here :

    Suppose x[itex]\in[/itex](A0[itex]\cup[/itex]B0). Then x[itex]\in[/itex]A0 or x[itex]\in[/itex]B0.

    Suppose x[itex]\in[/itex]A0. Then x is an interior point of A and [itex]\exists δ>0|N_δ(Q) \subseteq A[/itex]

    Since the neighborhood of x is contained within A, it follows that it is also contained within A[itex]\cup[/itex]B.

    Now since A[itex]\subseteq[/itex]A[itex]\cup[/itex]B, it follows that A0[itex]\subseteq[/itex](A[itex]\cup[/itex]B)0. If we had chosen x[itex]\in[/itex]B0 we would have come to the conclusion that B0[itex]\subseteq[/itex](A[itex]\cup[/itex]B)0 through the exact same logic. So it follows that A0B0[itex]\subseteq[/itex](A[itex]\cup[/itex]B)0 as desired.

    Now as for the example where the inclusion is strict would I need something like :

    A = {x[itex]\in[/itex]ℝ | 0 ≤ x ≤ 1} so A0 = {x[itex]\in[/itex]ℝ | 0 < x < 1}
     
  13. Sep 14, 2012 #12

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Seems ok for the proof. To show the inclusion may be strict, you need to define B as well, yes?
     
  14. Sep 14, 2012 #13

    jbunniii

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    I would be careful with how you've worded this.

    Indeed, "since A[itex]\subseteq[/itex]A[itex]\cup[/itex]B, it follows that A0[itex]\subseteq[/itex](A[itex]\cup[/itex]B)0" is very close to the statement of what you have been asked to prove, without the proof.

    You have a neighborhood N of x such that [itex]x \subset N \subset A[/itex]. Furthermore, [itex]A \subset A \cup B[/itex], so it follows that [itex]x \subset N \subset A \cup B[/itex]. Therefore...
     
  15. Sep 14, 2012 #14

    Zondrina

    User Avatar
    Homework Helper

    Yes yes, just ensuring I needed something of that sort. Okay hmm, strict inclusion means there should be at least one element not included so would ...

    A = {x[itex]\in[/itex]ℝ | 0 ≤ x ≤ 5}
    B = {x[itex]\in[/itex]ℝ | 5 ≤ x ≤ 10}

    Work?
     
    Last edited: Sep 14, 2012
  16. Sep 14, 2012 #15

    Zondrina

    User Avatar
    Homework Helper

    Therefore the neighborhood of x is also contained within the interior of the union of A and B? Also I believe what I said is sufficiently far enough from the original question, but how would I rephrase this portion.
     
  17. Sep 14, 2012 #16

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Works. Why does it work? Spell out the reason.
     
  18. Sep 14, 2012 #17

    Zondrina

    User Avatar
    Homework Helper

    Well

    A = {x[itex]\in[/itex]ℝ | 0 ≤ x ≤ 5}
    B = {x[itex]\in[/itex]ℝ | 5 ≤ x ≤ 10}

    So

    A0 = {x[itex]\in[/itex]ℝ | 0 < x < 5}
    B0 = {x[itex]\in[/itex]ℝ | 5 < x < 10}

    So we have : A0UB0 = {x[itex]\in[/itex]ℝ | 0 < x < 10, x≠5}
    And also : (AUB)0 = {x[itex]\in[/itex]ℝ | 0 < x < 10}

    Thus : A0UB0[itex]\subset[/itex](AUB)0
     
  19. Sep 14, 2012 #18

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    That's exactly what I wanted to hear. Thanks!
     
  20. Sep 14, 2012 #19

    Zondrina

    User Avatar
    Homework Helper

    Oh man thanks so much for your patience, really though. I just wanted to understand this so badly.

    Also my final concern, is what jbuni said true? Am I too close? Or was my proof sufficient?
     
  21. Sep 14, 2012 #20

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    What you said convinces me you understand it. jbunnii's rephrasing is a better version for the proof.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Point set proof
  1. Sets and Proofs (Replies: 6)

  2. Set Proof (Replies: 3)

  3. Set Proof (Replies: 6)

  4. Set proof (Replies: 1)

  5. Set Proof (Replies: 14)

Loading...