= Solution
Both $\varnothing$ and $X$ satisfy the condition. For an arbitrary <intersection> of members, either every member contains $a$, in which case the <intersection> contains $a$, or one does not, in which case that member is countable and so is the <intersection>. The empty <intersection> is $X$.
For a finite <union>, either a member contains $a$, so the <union> does too, or all members omit $a$ and are countable, so their finite <union> is countable. Thus the <closed set> axioms hold and define a <topology>. This is a <Fortissimo space>: every point other than $a$ is isolated, while the <open neighbourhoods> of $a$ have countable complement.
Back to article page
