# Countable topological basis

Let $(E,\mathcal{T})$ be a topological space with a countable basis (a second countable space). If $\mathcal{U}$ is a topology on $E$ with $\mathcal{U} \subset \mathcal{T}\,$ ($\mathcal{U}$ is coarser than $\mathcal{T}$), does $(E,\mathcal{U})$ have a countable basis?

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Your definition of a basis for a topology $\mathscr{T}$ is not exactly correct. You need to say a set $O$ is open if and only if there exists $I \subset \mathbb{N}$ such that ... As stated, the union of basic element may not be open in your topology. – William Sep 22 '11 at 18:21
It's true if the topology $U$ is comes from a metric, since a metric space is separable if and only if it has a countable topological basis, and a topological space which has a countable basis is separable. But it's not necessarily the case when $U$ doesn't come from a metric. – Davide Giraudo Sep 22 '11 at 18:38
I deleted my answer. It doesn't make any sense. – William Sep 22 '11 at 18:54

The answer is not necessarily. A counterexample is the cofinite topology on $\mathbb{R}$. It does not have a countable basis but it is a subset of the usual topology on $\mathbb{R}$ which has a countable basis.

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Thanks, good counterexample ! – francis-jamet Sep 22 '11 at 18:41

No. Let $\mathcal{T}$ be the usual topology on $\mathbb{R}$, and let $\mathcal{U} = \{\mathbb{R} \setminus F:F\text{ is finite}\}$. Every finite subset of $\mathbb{R}$ is closed in the usual topology, so every member of $\mathcal{U}$ is open in the usual topology, i.e., $\mathcal{U}\subseteq \mathcal{T}$. $\mathcal{T}$ has a countable base; one such is the set of open intervals with rational endpoints. $\mathcal{U}$, however, does not. To see this, let $\mathcal{B} = \{U_n:n \in \omega\}$ be a countable subset of $\mathcal{U}$. For $n \in \mathbb{N}$ let $F_n = \mathbb{R} \setminus U_n$; each $F_n$ is finite by the definition of $\mathcal{U}$. Let $C = \bigcup\limits_{n\in\mathbb{N}}F_n$; $C$ is the union of countably many finite sets, so $C$ is countable. $\mathbb{R}$, however, is uncountable, so we can choose a point $p \in \mathbb{R}\setminus C$. Now let $U = \mathbb{R} \setminus \{p\}$. Clearly $U \in \mathcal{U}$. But for each $n \in \mathbb{R}$, $p \in U_n \setminus U$, so $U_n \nsubseteq U$. Since no member of $\mathcal{B}$ is even a subset of $U$, $\mathcal{B}$ clearly can’t be a base for $\mathcal{U}$.

Added: It is also possible to construct examples on a countable set. Let $\mathscr{D}$ be the discrete topology on $\mathbb{N}$; $\left\{\{n\}:n\in \mathbb{N}\right\}$ is a countable base for $\mathscr{D}$, and clearly any topology on $\mathbb{N}$ is a sub-topology of $\mathscr{D}$, so we need only find a topology on $\mathbb{N}$ that has no countable base.

Let $\mathscr{U}$ be a free ultrafilter on $\mathbb{N}$; then $\mathscr{U}\cup\{\varnothing\}$ is a topology on $\mathbb{N}$ with no countable base. That $\mathscr{U}\cup\{\varnothing\}$ is a topology is immediate from the definition of a filter. To see that $\mathscr{U}\cup\{\varnothing\}$ has no countable base, let $\mathscr{B} = \{U_n:n\in\mathbb{N}\}$ be any countable subset of $\mathscr{U}$. Recursively pick distinct points $x_k,y_k \in \bigcap\limits_{i\le k}U_i \setminus \left(\{x_i:i<k\} \cup \{y_i:i<k\}\right)$, let $A = \{x_k:k\in\mathbb{N}\}$, and let $B = \mathbb{N} \setminus A$; $\mathscr{U}$ is an ultrafilter, so either $A \in \mathscr{U}$, or $B \in \mathscr{U}$. But for each $k\in\mathbb{N}$, $y_k \in U_k\setminus A$, so $U_k \nsubseteq A$, and $x_k \in U_k \setminus B$, so $U_k \nsubseteq B$. Thus, neither $A$ nor $B$ even contains a member of $\mathscr{B}$, and $\mathscr{B}$ cannot be a base for $\mathscr{U} \cup \{\varnothing\}$.

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If the topologies are supposed to be Hausdorff, is there a counterexample ? – francis-jamet Sep 22 '11 at 18:59
@francis-jamet: Yes, there is: instead of using $\mathscr{U}\cup\{\varnothing\}$ as a topology, add $\mathscr{U}$ as a new point whose nbhds are the sets $\{\mathscr{U}\}\cup U$ for $U\in\mathscr{U}$. (Ah, I see now that Stefan Geschke has already added this example.) – Brian M. Scott Sep 22 '11 at 20:33
Thanks for your answer, Brian. – francis-jamet Sep 23 '11 at 1:23

I think that William Chan's example doesn't work, for the reason francis-jamet points out in his comment.
But here is something that works, along the same lines, giving a countable space.

Let $U$ be a non-principal ultrafilter on $\mathbb N$.
Consider the space $\mathbb N\cup\{U\}$ with the following basis for a topology: for all $n\in\mathbb N$, $\{n\}$ is in the basis. For all $A\in U$, $A\cup\{U\}$ is in the basis.

Let $V$ be a countable collection of sets from $U$. There is $B\subseteq$ such that for each $A\in V$, $B\setminus A$ is finite. Either $B$ or its complement is in $U$. The complement clearly includes no element of $V$. If $B\in U$, split $B$ into two infinite pieces. One of the pieces is in $U$. Both pieces contain no element of $V$. In either case, $U$ has an element that contains no member of $V$.

This information can be used to show that the space $\mathbb N\cup U$ has no countable basis for its topology. And clearly the topology is contained in the discrete topology on the space. Also note that these spaces are obtained by adding a single new point to a countable discrete space.

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This is an interesting example. It's not only Hausdorff but also hereditarily normal, if I'm not mistaken. Is it even perfectly normal? I think it depends on whether every subset of $\mathbb{N}$ is a countable intersection of elements in the ultrafilter. I wonder if this is true. – LostInMath Sep 22 '11 at 21:39
Thank you, Stefan. – francis-jamet Sep 23 '11 at 1:21
@LostInMath: Yes, it’s perfectly normal: every open set is countable, and singletons are closed (since the space is $T_1$), so every open set is an $F_\sigma$. It’s also true that every subset of $\mathbb{N}$ is a countable intersection of sets in the ultrafilter, since the ultrafilter contains every set of the form $\mathbb{N}\setminus\{n\}$ for $n\in\mathbb{N}$. – Brian M. Scott Sep 23 '11 at 3:47
@Brian: Indeed, thanks for making this clear. – LostInMath Sep 23 '11 at 9:09