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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?

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  • $\begingroup$ You probably want to exclude the case of finite discrete $P$ - or are you only interested in infinite posets? $\endgroup$ – მამუკა ჯიბლაძე Jan 23 '15 at 13:09
  • $\begingroup$ 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. $\endgroup$ – Dominic van der Zypen Jan 23 '15 at 13:28
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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 $Q$.

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  • $\begingroup$ 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?? $\endgroup$ – მამუკა ჯიბლაძე Jan 24 '15 at 6:37
  • $\begingroup$ @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. $\endgroup$ – MikeTeX Jan 24 '15 at 17:40
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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.

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  • $\begingroup$ 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 $y'$. $\endgroup$ – Eric Wofsey Jan 23 '15 at 14:33
  • $\begingroup$ Yes, actually, my proof is wrong. But maybe it is now not too difficult to find a counter example. $\endgroup$ – MikeTeX Jan 23 '15 at 14:41

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