Complement of a Topology Let $(X, \tau)$ topology, I was wondering, if given $$\tau'=\{A^C \mid A \in \tau\}$$ did $\tau'$ also a topology on $X$? If so, why? Thank you.
 A: Example 1: Let $\Bbb R$ with usual topology. $\mathscr T=\{(a,b): a,b \in \Bbb R \}$ is the topology of $\Bbb R$. However $\mathscr T'$ is not the topology of $\Bbb R$. Since $\{[\frac1n,\infty]: n\in \Bbb N^+\} $ and $\{[-\infty, -\frac{1}n]: n\in \Bbb N^+\} $ are the elements of $\mathscr T'$. However their union is $\Bbb R\setminus \{0\}$, which is not the element of $\mathscr T'$. Therefore $\mathscr T'$ is not a topology on $\Bbb R$.
A: Not in general as closed sets aren't usually stable under infinite unions. If $(X,\tau)$ has this property, we say that $(X,\tau)$ is an Alexandrov space, and then this is equivalent to open sets being stable under infinite intersections.
As an elementary example of how a space can fail to be Alexandrov, take $\mathbb{R}$ with the usual topology and consider the set of open sets $$A=\left\{\left(-\infty,\frac{1}{n}\right)\cup\left(1-\frac{1}{n},\infty\right)\:|\:n\in\mathbb{N}\right\}.$$ The set of complements of elements of $A$ is $$B=\left\{\left[\frac{1}{n},1-\frac{1}{n}\right]\:|\: n\in\mathbb{N}\right\}$$ and $\bigcup_n B=(0,1)$ which is not a complement of an element of the usual topology on $\mathbb{R}$ as it is a non-trivial open set and so is not closed.
