Let $(P,\leq)$ be a poset. The *interval topology* $\tau\_i(P)$ on $P$ is generated by
$$\{P\setminus\downarrow x : x\in P\} \cup \{P\setminus\uparrow x : x\in P\},$$
where $\downarrow x = \{y\in P: y\leq x\}$ and $\uparrow x = \{y\in P: y\geq x\}$.


If $B$ is a complete Boolean algebra, is $(B,\tau\_i(B))$ Hausdorff?



Interval topology on complete Boolean algebras
No. You can't separate $0$ from $1$ in an atomless
Boolean algebra, such as the (complete)
Boolean algebra of regular open
subsets of $\mathbb R$. 


The set of principal ideals and principal 
filters form a subbasis of closed
sets for the topology on $B$, so a typical basic closed set has the form
$I\cup F$ where $I$ is a finitely generated order ideal and $F$ is a finitely
generated order filter. Our goal is to show that there do not
exist proper basic closed subsets $C\_0=I\_0\cup F\_0$ containing $0$ and not $1$
and $C\_1=I\_1\cup F\_1$ containing $1$ and not $0$,
whose union $C\_0\cup C\_1$ equals $B$.
To obtain a contradiction, assume that
$B = C\_0\cup C\_1$ for proper basic closed sets
satisfying $0\in C\_0-C\_1$ and $1\in C\_1-C\_0$.


Since $1\notin C\_1$ and $0\notin C\_0$ we get that
$C\_0 = I\_0$ is a proper f.g. order ideal and $C\_1 = F\_1$ is a
proper f.g. order filter.


Suppose that $C\_0$ is the order ideal generated by
some finite set $X$ and that $C\_1$ is the order filter
generated by some finite set $Y$.
Let $C$ be the subalgebra of $B$ generated by $X\cup Y$. It is finite.
If the set of atoms of $C$ is $V=\{v\_1\ldots,v\_p\}$, then
for $u\_i:=\bigvee\_{j\neq i} v\_j$ we get that
$U = \{u\_1,\ldots,u\_p\}$ is the set of coatoms of $C$.
Note that $V$ consists of the cells in a finite partition
of unity, and $U$ consists of the complementary elements.


The f.g. order ideal $D\_0$ generated by
$U$ contains $C\_0$ and still does not contain $1$, while the f.g.
order filter $D\_1$ generated by $V$ contains $C\_1$ but still does
not contain $0$. This reduces the original situation to one in
which $D\_1$ is a proper order filter generated by the
cells in a finite partition of unity, while $D\_0$ is the proper f.g.
order ideal generated by the complements of single cells
in the same partition of unity.


The set $V = \{v\_1,\ldots,v\_p\}$ consists of nonempty,
pairwise disjoint, elements of $B$. If $B$ is atomless, 
then we can choose $w\_i$ such that $0<w\_i<v\_i$ for all $i$.
Let $W = \bigvee w\_i$.
For each $i$ we have $v\_i\not\leq W$,
since $v\_i\cap W = v\_i\cap w\_i = w\_i < v\_i$. Moreover,
$W\not\leq u\_j$ for any $j$, since $0<w\_j\leq W$
and $w\_j\not\leq u\_j$. Thus $W\notin D\_0\cup D\_1$.
Hence also $W\notin C\_0\cup C\_1$.




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In fact, the argument can be relativized to intervals to prove
that if $x<y$, then $x$ cannot be separated from $y$ unless
there is an atom below $y$ disjoint from $x$. This happens for every 
$x<y$ only when $B$ is an
atomic BA.
To restate this: if the topology is Hausdorff, then the BA must be atomic.




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Conversely, if $B$ is atomic, then the topology is Hausdorff. To see this,
suppose that $r\not\leq s$ in $B$. Choose an atom $t\leq r$
that is disjoint from $s$. Then the closed sets $\uparrow t$
and $\downarrow t'$ separate $r$ from $s$ and have union equal to $B$.


