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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?

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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 $W\notin C_0\cup C_1$.


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.


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 $B$.

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  • $\begingroup$ Thanks for this amazing answer -- I always enjoy your answers very much! $\endgroup$ – Dominic van der Zypen Oct 7 '16 at 6:02

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