1. Limited time only! Sign up for a free 30min personal 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!

Homework Help: True/false questions about open/closed sets

  1. Oct 27, 2009 #1
    1. The problem statement, all variables and given/known data

    although i'm supposed to provide explanations for these, i just want to see if my intuition about them is correct..

    1) For any set [tex]A \subseteq R, \overline{A}^{c}[/tex] is open

    2) If a set A has an isolated point, it cannot be an open set

    3) Set A is closed if and only if [tex]\overline{A} = A[/tex]

    4) If A is a bounded set, s=sup A is a limit point of A

    5) Every finite set is closed

    6) An open set that contains every rational #s must necessarily be all of R.

    2. Relevant equations

    3. The attempt at a solution

    1) True
    2) True
    3) True
    4) True
    5) False
    6) False
  2. jcsd
  3. Oct 27, 2009 #2
    I'm assuming we're only working with subsets of the real line.

    I agree with your answers to 1,2,3,6. For 5 I disagree as all one-point sets are closed and finite unions of closed sets are closed. As for 4 I agree for some definitions of limit points (I have seen different definitions in different books). If you require that all neighborhoods of x intersect A in some place OTHER than x, then I disagree since {1} is bounded, sup{1} =1, but 1 is not a limit point of {1}. If you only require that every neighborhood of x intersect A, then I agree.
  4. Oct 27, 2009 #3
    Yeah, you're right for 4 as well, since I'm using the first definition you mentioned. The set doesn't have to be "continuous".
  5. Oct 29, 2009 #4
    Hmmm... I'm thinking that maybe 6 is true.

    All rational #s by themselves are closed and so the (countably) finite union of them would be closed... but is it possible to construct an "open" set that just contains rational #s?

    Edit: actually... regardless of whether it's possible to create an open set containing rational numbers, any set containing rational numbers should contain irrationals... right? Sorry I'm not sure how to explain it better.
    Last edited: Oct 29, 2009
  6. Oct 29, 2009 #5


    User Avatar
    Homework Helper

    i assume we're talking about open as a subset of R here?

    I'm thinking its true as well, as any real number can be approximated abritrarily close by a rational, ie. the rationals are dense in the reals. So any open interval of [itex] \mathbb{R} [/itex] must contain rationals and irrational

    what about {1/3} no irrationals there....
  7. Oct 29, 2009 #6


    User Avatar
    Homework Helper

    the set of all rational numbers [itex] \mathbb{Q} [/itex] is not open in [itex] \mathbb{R} [/itex] either, as any neighbourhood will contain an irrational which is not in [itex] \mathbb{Q} [/itex]
  8. Oct 29, 2009 #7
    Oops, I meant to say a set containing "every" rational numbers, as the question asks.

    (And yes I'm talking about subsets of R.)

    As you said, the rationals are dense in the reals, so [tex]R=\overline{Q}[/tex]. So an open set containing every rationals also contains every irrationals, since any open set contains rationals and irrationals (and hence the reals)
  9. Oct 29, 2009 #8
    [itex](\sqrt{2},\infty)[/itex] and [itex](-\infty,\sqrt{2})[/itex] are open so
    [tex]U=(-\infty,\sqrt{2})\cup(\sqrt{2},\infty) = \mathbb{R} - \{\sqrt{2}\}[/tex]
    is open and contains all rationals, but U is not [itex]\mathbb{R}[/itex].
  10. Oct 29, 2009 #9


    User Avatar
    Homework Helper

    good point, missed that
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook