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The topology of rational numbers: connected sets

  1. Nov 4, 2008 #1
    Consider the set of rational numbers, under the usual metric d(x,y)=|x-y|

    I am pretty sure that this space is totally disconnected, but I can't convince myself that the set {x} U {y} is a disconnected set.

    It seems obvious, but I can't find two non-empty disjoint open sets U,V such that U U V = {x} U {y}.

    I am sure {x},{y} are not open sets, so I need something bigger.

    Is the total disconnection only true if the the set Q is a relative topology of R?

    Thanks
     
  2. jcsd
  3. Nov 4, 2008 #2

    Office_Shredder

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    You can always map Q isometrically into R by the trivial embedding (q is mapped to q, with the standard metric in R). This is useful for the end result. Total disconnection means the only connected subsets of Q are the empty set and individual points. Suppose A is a connected subset of Q. Then there exists x,y in A s.t. x=/= y. We know there exists an irrational in between x and y. Now here's the thing... that irrational (call it z) isn't in Q, but we can still use it to our advantage as we can split A by A = {a in A | a<z in R} U {b in A | a>z in R}

    and then you can show that both of those are open.

    To clarify in your post though

    When a subset of a space is disconnected, it means you can find two open sets in the subspace. The set {x} is open in {x}U{y} (the induced topology is B subset of A subset of X, B is open in A iff there exists V open in X s.t. B = A intersect V)
     
  4. Nov 4, 2008 #3
    Thanks for the help. It was the definition of a disconnected set that was messing me up.
     
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