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Countable But Not Second Countable Topological Space

  1. Sep 29, 2007 #1
    I'm wondering if someone can furnish me with either an example of a topological space that is countable (cardinality) but not second countable or a proof that countable implies second countable. Thanks.
     
  2. jcsd
  3. Sep 29, 2007 #2

    morphism

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    Take the countable set [itex]\mathbb{N}\times\mathbb{N}[/itex]. Topologize it by making any set that doesn't contain (0,0) open, and if a set does contain (0,0), it's open iff it contains all but a finite number of points in all but a finite number of columns. (Draw a picture. If an open set contains (0,0), then it can only miss infinitely many points in a finite number of columns, while it misses finitely many points in all the other columns.)

    Now this topology doesn't have a countable base at (0,0), so it's not first countable let alone second countable.

    Source: Steen & Seebach, Counterexamples in Topology, page 54. They call it the Arens-Fort Space.
     
  4. Sep 29, 2007 #3
    Thanks! I own that book so I'll be having a look pretty soon.
     
  5. Sep 30, 2007 #4

    morphism

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    I've been thinking about this a little bit more, and I believe I have another example, although it's a bit more 'sophisticated' in that it requires a bit of advanced set theory.

    This time our countable set is X = [itex]\mathbb{N} \cup \{\mathbb{N}\}[/itex]. Take any free ultrafilter F on [itex]\mathbb{N}[/itex], and define a topology on X by letting each subset {n} of [itex]\mathbb{N}[/itex] be open, and defining nbhds of [itex]\left{\mathbb{N}\right}[/itex] to be those of the form [itex]\{\mathbb{N}\} \cup U[/itex], where U is in F. As in the previous example, this topology fails to have a countable base at [itex]\left{\mathbb{N}\right}[/itex] (because we cannot have a countable base for any free ultrafilter on the naturals), so again it fails to be first countable.

    Edit:
    Hmm... Now I'm wondering if there's a countable space that's first countable but not second countable!

    Edit2:
    Maybe that was silly. If X is countable and has a first countable topology, then the union of the bases at each of its points is the countable union of countable sets and is hence countable (and a basis for the topology). So, I'm lead to conclude that a countable, first countable space is necessarily second countable.
     
    Last edited: Sep 30, 2007
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