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Connected Sets

  1. Apr 20, 2008 #1
    I have to present it in front of the class so I’m trying to make it as clean and correct as possible.

    If someone could just point me in the right direction for this proof or give me some examples that would be great.

    (Sorry if this is sloppy but I’ll try to make it as nice as possible)
    The problem I'm having trouble with is.

    1 Prove that QxQ c R² (The Cartesian product of the rationals in R²) is disconnected
    2 Show it is totally disconnected

    So far for 1 I have
    Let qЄQ
    By the Cartesian product we would get a sequence
    (q1, q1), (q2, q2), (q3, q3),…,(qn, qn)

    (This is where it gets a bit confusing and sloppy. I don’t know if I can take (q1, q1) and just represent it as q)

    The numbers in the sequence can be separated as
    {x | x > q}∩U1, {x | x < q}∩U2

    Therefore by the definition of connected sets it is disconnected

    For Q x Q to be totally disconnected it has at least two points and for all distinct points p1, p2 in S, the set S can be separated by two disjoint open sets U1 and U2 into two pieces S∩U1 S∩U2 containing p1 and p2

    (q1, q1) Є U1
    (q2, q2) Є U2
    (q1, q1), (q2, q2) Є S

    Therefore they are totally disconnected.
  2. jcsd
  3. Apr 20, 2008 #2


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    ??What are U1 and U2? and what definition of "connected sets" are you using?

    Again, WHAT are U1 and U2? If you are saying they are the "two disjoint open sets U1 and U2" you mention above, then you are assuming the set is disconnected to begin with.

    Since R2 is connected and you are trying to prove that Q2 is NOT, it seems to me your proof had better make use of the difference btween R and Q- and I don't see that anything you have done here wouldn't apply to R2 as well.
  4. Apr 20, 2008 #3
    Sorry the def I'm using is
    A set is connected if it cannot be separated by two disjoint sets U1 and U2 into two nonempty pieces S∩U1 and S∩U2
  5. Apr 20, 2008 #4
    Sorry I don't understand what you're saying. I'm not being sarcastic I'm just trying to get what you mean.
  6. Apr 20, 2008 #5
    Since we are talking about metric spaces, an equivalent definition of disconnected is that it contains a proper (and nontrivial) subset that is both open and closed (this is equivalent because then its complement is also both open and closed and these are disjoint)

    With this definition, a totally disconnected subset is one whose only connected subsets are the empty set and any singleton sets. (i.e., if you have any set with more than 1 element in it, then it is disconnected)

    Showing that Q^2 has a proper, non empty, subset that is both open and closed is pretty easy, you just need to demonstrate one. Showing that every subset of Q^2 with more than 1 element is disconnected is a little more difficult. I remember doing it, but I can't quite remember how at the moment.
  7. Apr 20, 2008 #6
    Thanks. I'm pretty sure I've been making this proof seem much harder then it actually is.
  8. Apr 21, 2008 #7


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    What I meant was that the set of real numbers (and so R2) is a connected set. Q2 is not connected essentially because Q is not. If replacing "Q" by "R" in your proof doesn't change the result, it can't be correct! So your proof had better involve some way in which the rational numbers are different from the real numbers.

    For example, the fact that every bounded set of real numbers has a least upper bound is not true for rational numbers, the "monotone convergence theorem" is not true for rational numbers (at first, since you were using sequences of points, I thought that was what you were using), Cauchy sequences of rational numbers do not converge (to a rational number), etc.

    One simple difference is that [itex]\sqrt{2}[/itex] is NOT a rational number. In other words, Q2= QxQ is the union of {(x,y)|x,y[itex]\in[/itex] Q, y>0, y^2> 2} and {(x,y)|x,y[itex]\in[/itex]Q, y^2< 2}. Can you show that those sets are both closed and so separated?

    That's incorrect. Any set with 2 or more points can be separated like that. You need two "separated" sets (U1 closure∩ U2 and U1∩ U2 closure are empty), not just "disjoint", or else "U1, U2 both closed sets" or "U1, U2 both open sets".
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