What is the smallest possible basis for the finite complement topology on $\mathbb{R}$? Let $\mathbb{R}$ denote the set of all real numbers, and let $\mathscr{T}$ be a collection of subsets of $\mathbb{R}$ such that $U \in \mathscr{T}$ if and only if $\mathbb{R} \setminus U$ is either finite or all of $\mathbb{R}$. Then $\mathscr{T}$ is a topology on $\mathbb{R}$, called the finite complement, or co-finite, topology. 
Now $\mathscr{T}$ itself is a basis for this topology, but can we determine a basis for $\mathscr{T}$ that is contained in every other basis for $\mathscr{T}$? 
 A: No, a non-discrete T$_1$ topology cannot have a minimal base.
For suppose $\mathscr{B}$ is a base for such a topology. Then we can choose $B \in \mathscr{B}$ with $|B|\gt 1$, since a topology with a base consisting only of singleton sets is discrete. 
We show that $\mathscr{B}$ is not minimal by showing that 
$$ \mathscr{B}^\prime \colon= \mathscr{B} \setminus \{B\} $$ 
is a base for the topology, and we show that by showing that $B$ is a union of sets in $\mathscr{B}^\prime$.
Consider any point $x \in B$; we have to find a set $B^\prime \in \mathscr{B}^\prime$ such that 
$$ x\in B^\prime \subseteq B.$$ 
Since $|B|\gt1$, we can choose a point $y\in B$ such that $y \neq x$. As the topology is T$_1$, so there is an open set $U$ such that $x \in U$ and $y \not\in U$. 
Since $\mathscr{B}$ is a base, there is a set $B^\prime \in \mathscr{B}$ such that $$ x \in B^\prime \subseteq B\cap U.$$ 
Since $y \not\in U$, therefore $y \not\in B \cap U$ either and thus $y \notin B^\prime$, but as $y \in B$, so we have $B^\prime \neq B$. Thus $B^\prime \in \mathscr{B}$ and $B^\prime \neq B$; that is, 
$$ B^\prime \in \mathscr{B}^\prime.$$
In particular, the cofinite topology on an infinite set is a non-discrete T$_1$ topology, so it has no minimal base.
A: Let $\mathscr{B}_0=\{U\in\mathscr{T}:|\Bbb R\setminus U|\text{ is even}\}$ and $\mathscr{B}_1=\{U\in\mathscr{T}:|\Bbb R\setminus U|\text{ is odd}\}$. $\mathscr{B}_0$ and $\mathscr{B}_1$ are both bases for $\mathscr{T}$, but $\mathscr{B}_0\cap\mathscr{B}_1=\varnothing$, so $\mathscr{T}$ has no base that is a subset of both $\mathscr{B}_0$ and $\mathscr{B}_1$ and therefore no smallest base.
