# Image of poset with Hausdorff interval topology

Given a poset $(P,\leq)$ the interval topology $\tau_{\text{int}}(P)$ 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 $\uparrow x = \{y\in P: y\geq x\}$.

Let $P, Q$ be posets and $e:P\to Q$ be order-preserving and surjective. Assume that $(P,\tau_{\text{int}}(P))$ is Hausdorff. Does $(Q,\tau_{\text{int}}(Q))$ have to be Hausdorff?

• You probably want to exclude the case of finite discrete $P$ - or are you only interested in infinite posets? Jan 23, 2015 at 13:09
• Any finite poset has discrete (therefore Hausdorff) interval topology. And the image of any finite poset is finite, so the question has a positive answer for finite posets. I suspect the answer to the question is positive for all posets, but I don't know. Jan 23, 2015 at 13:28

The answer is no, not necessarily.

For a counterexample, let $Q$ be any atomless complete Boolean algebra, and let us view it via Stone's theorem as a field of sets, so that $Q$ is a subalgebra for some set $X$ of the power set algebra $P=P(X)$, which is atomic.

By Proposition 5 in this paper of Northam (due originally to Katetov 1951), a Boolean algebra is Hausdorff in the interval topology just in case every non-zero element sits over an atom. So $P$ is Hausdorff, but $Q$ is not.

But meanwhile, we have a surjective order-preserving map $f:P\to Q$, defined by $f(x)=$ the join in $Q$ of the elements of $Q$ that are below $x$ in $P$. We use the completeness of $Q$ in order to know that this join exists in $Q$.

• In that paper of Northam the topology is defined slightly differently: closed sets are generated by those ${\downarrow}x$ and ${\uparrow}y$ which are infinite, together with those ${\downarrow}x\cap{\uparrow}y$ which are finite. Is this the same topology?? Jan 24, 2015 at 6:37
• @მამუკა ჯიბლაძე Yes, it is the same topology. The author just want to say that $[a, b]$ is a finite interval in the sense that $a$ and $b$ are not infinite, probably because it is the way Frink presented "closed intervals". Of course, this is unnecessary since "finite" intervals are obviously the intersection of two infinite intervals. Jan 24, 2015 at 17:40

Observe that $e$ sends an element $\downarrow \alpha$ to an element of the form $\downarrow \alpha'$, and an element $\downarrow \beta$ to an element of the form $\downarrow \beta'$. Now, if $x'$ and $y'$ are distinct in $Q$, and if $x,y$ are such that $e(x)=x'$ and $e(y)=y'$, then there exists a finite intersection $U$ of the elements of the sub base you gave above, and a finite intersection $V$ of these elements, such that $U\cap V = \emptyset$. Let us denote $U=u_1\cap u_2\cap\ldots\cap u_k$, with $u_i = P-s_i$ and $s_i=\downarrow \alpha$ or $s_i=\uparrow \alpha$, and $V = v_1\cap \ldots v_r$, with $v_j = P - t_j$ and $t_j=\downarrow \beta$ or $t_i = \uparrow \beta$. Then by booleanity, $P=\cup_i s_i \cup \cup_j t_j$. It follows that $e(P)=Q=\cup_i e(s_i) \cup \cup_j e(t_j)$, and conclude by booleanity that $Q$ is Hausdorff.

• It is not necessarily the case that $e(\downarrow \alpha)=\downarrow e(\alpha)$; only $\subseteq$ holds in general. If instead of $e(s_i)$ and $e(t_j)$ you mean things like $\downarrow e(\alpha)$, then they will still cover $Q$ and you get two disjoint open sets in $Q$, but you can't be sure that these sets contain $x'$ and $y'$. Jan 23, 2015 at 14:33
• Yes, actually, my proof is wrong. But maybe it is now not too difficult to find a counter example. Jan 23, 2015 at 14:41