- #1

- 32

- 0

- Thread starter Nolen Ryba
- Start date

- #1

- 32

- 0

- #2

morphism

Science Advisor

Homework Helper

- 2,015

- 4

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,

- #3

- 32

- 0

Thanks! I own that book so I'll be having a look pretty soon.

- #4

morphism

Science Advisor

Homework Helper

- 2,015

- 4

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.

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:

- Last Post

- Replies
- 12

- Views
- 3K

- Last Post

- Replies
- 20

- Views
- 3K

- Last Post

- Replies
- 4

- Views
- 6K

- Last Post

- Replies
- 10

- Views
- 1K

- Last Post

- Replies
- 1

- Views
- 1K

- Replies
- 12

- Views
- 4K

- Last Post

- Replies
- 4

- Views
- 2K

- Replies
- 2

- Views
- 1K

- Replies
- 3

- Views
- 2K

- Replies
- 2

- Views
- 1K