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?
Countable topological basis
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0I deleted my answer. It doesn't make any sense. – 2011-09-22
3 Answers
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.
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
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0Thanks for your answer, Brian. – 2011-09-23
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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0@Brian: Indeed, thanks for making this clear. – 2011-09-23