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Subspace topology of Rationals on Reals

  1. Oct 31, 2012 #1
    I am trying to visualize the subsppace topology that is generated when you take the Rationals as a subset of the Reals.

    So if we have ℝ with the standard topology, open sets in a subspace topology induced by Q would be the intersection of every open set O in ℝ with Q. Since each open set in ℝ is an open interval (a,b), and because between any two reals there are an infinite number of both rationals and irrationals, I'm picturing open sets in the subspace topology as infinite sets of rational numbers. What I mean is, you wouldn't ever have a singleton set. You'd have say, the infinite number of rationals included in any interval (a,b). Am I right in this thought?
  2. jcsd
  3. Oct 31, 2012 #2
    I would be careful with this. The open sets of [itex] \mathbb R [/itex] are more than just the open intervals. The open intervals do form a basis for [itex] \mathbb R [/itex] though.

    You can have a singleton set in [itex] \mathbb Q [/itex], but what I think you mean is that you can never have an open singleton set, which is correct. Furthemore, note that if [itex] \mathcal B [/itex] is a bases for [itex] \mathbb R [/itex] then [itex] \left\{ U \cap \mathbb Q: U \in \mathcal B \right\} [/itex] is a basis for [itex] \mathbb Q [/itex] in the subspace topology.
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