Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Can I get a finite Isbell Hausdorff frame which has no Isbell Hausdorff subframe? If not possible is there one such in infinite case?

See the book [1] for definitions.

  1. Picado, Jorge, and Ales Pultr. Frames and Locales: Topology without Points. Basel: Birkhauser, 2012.
share|improve this question
Would you mind and expand your question? Some references, links, definitions would be really helpful. –  András Bátkai Jun 26 '13 at 15:59
I added a reference that defines the notion of an Isbell Hausdorff frame. –  Joseph Van Name Jun 26 '13 at 16:57
Why is there a box beneath these comments saying "This post does not cite any references or sources", when plainly it does? Did someone cause it to be there before the reference was added? How do we get rid of it? –  Tom Leinster Jun 27 '13 at 0:55
If I remember correctly, it was there before I added the reference. I don't know how to get rid of it. Perhaps, I or someone should ask this on meta. –  Joseph Van Name Jun 27 '13 at 3:34
It was mostly as a test of this notification system. meta.mathoverflow.net/questions/189/when-to-add-post-notices –  François G. Dorais Jun 27 '13 at 7:26

1 Answer 1

I claim that a subframe of even a finite Isbell Hausdorff frame is not necessarily Isbell Hausdorff. Furthermore, I claim that a finite frame is Isbell Hausdorff if and only if it is a Boolean algebra. In [1][Thm 2.3], Isbell states that every regular frame is strongly Hausdorff. In particular, since every complete Boolean algebra is a regular frame, every complete Boolean algebra is strongly Hausdorff. On the other hand, again in [1][Thm 2.3], Isbell states that the space of points in every strongly Hausdorff frame is a Hausdorff space. In particular, if $D$ is a strongly Hausdorff finite distributive lattice, then the space of points $Sp(D)$ of $D$ is a finite Hausdorff space, so $Sp(D)$ is a discrete space. Therefore the frame $\Sigma(Sp(D))$ of open subsets of $Sp(D)$ is a Boolean algebra. However, every finite distributive lattice is clearly a spatial frame, so $\Sigma(Sp(D))\simeq D$, so $D$ is a Boolean algebra.

It should note that for infinite frames, the action of taking a subframe of a frame preserves no separation axioms whatsoever. In [2. p.53], it is shown that every frame is isomorphic to a subframe of a complete Boolean algebra. Furthermore, complete Boolean algebras satisfy the strongest separation axioms since they are regular, normal, and even paracompact, and they satisfy strong zero-dimensionality axioms. In fact, complete Boolean algebras are ultraparacompact. And of course, the topologies are precisely the subframes of the power sets $P(X)$. On the other hand, a sublocale of a strongly Hausdorff locale is always strongly Hausdorff, so sublocales behave radically differently than subframes.

[1] Isbell, John R. Atomless parts of spaces. Math. Scand. 31 (1972), 5–32.

[2] Johnstone, P. T. Stone Spaces. Cambridge Cambridgeshire: Cambridge UP, 1982.

share|improve this answer
But P(X) has a subframe {0,a,b,1} where 'a' and 'b' are complements which is regular and hence Isbell Hausdorff. –  Jayaprasad P N Jun 27 '13 at 5:52
Jayaprasad. I edited my answer and I cited the original source of the notion of a Isbell Hausdorff frame. The books by Johnstone, and Picado and Pultr seem to have been a little sloppy when they said that if the frame of open sets of a space $X$ is strongly Hausdorff, then $X$ is Hausdorff. This is obviously not true. However, this fact holds in Sober spaces by the duality between Sober spaces and spatial frames. –  Joseph Van Name Jun 27 '13 at 18:47
I may have to contact one of the authors about this error. –  Joseph Van Name Jun 27 '13 at 22:10

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.