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Quick question about dense sets

  1. Mar 20, 2012 #1

    I was wondering whether a non-empty dense set has any isolation points. From my understanding, when a set is dense you can always find a third point between two points that is arbitrarily close to them so any ball you "create" around a point will contain another point hence a non-empty dense set has no isolation points.

  2. jcsd
  3. Mar 20, 2012 #2


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    Give [itex]\mathbb{Z}[/itex] the subspace topology. Notice that [itex]\mathbb{Z}[/itex] is dense in [itex]\mathbb{Z}[/itex] and that every point of [itex]\mathbb{Z}[/itex] is an isolated point.
  4. Mar 20, 2012 #3
    What if the dense subset of [itex]\mathbb{R}[/itex] in question is [itex]\mathbb{R} \setminus \mathbb{Q}[/itex]
  5. Mar 20, 2012 #4


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    In the subspace topology, the set [itex]\mathbb{R} \setminus \mathbb{Q}[/itex] has no isolated points. In the discrete topology, every point in [itex]\mathbb{R} \setminus \mathbb{Q}[/itex] is an isolated point.

    Edit: Although in the discrete topology you lose density in the reals. The point is that you certainly can give sets topologies which make them dense AND which give them isolated points. In the usual topologies though, a dense set in the reals will have no isolated points.

    For example, let [itex]\mathscr{T}[/itex] be the usual topology on [itex]\mathbb{R}[/itex]. Then the collection [itex]\mathscr{B}=\mathscr{T} \cup \{\sqrt{2}\}[/itex] is a basis for a topology on [itex]\mathbb{R}[/itex]. In this topology [itex]\mathbb{R} \setminus \mathbb{Q}[/itex] is dense and [itex]\sqrt{2}[/itex] is an isolated point.
    Last edited: Mar 20, 2012
  6. Mar 20, 2012 #5
    OK. Thank you.
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