1
$\begingroup$

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.
$\endgroup$
5
  • 2
    $\begingroup$ Would you mind and expand your question? Some references, links, definitions would be really helpful. $\endgroup$ Jun 26, 2013 at 15:59
  • $\begingroup$ I added a reference that defines the notion of an Isbell Hausdorff frame. $\endgroup$ Jun 26, 2013 at 16:57
  • 1
    $\begingroup$ 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? $\endgroup$ Jun 27, 2013 at 0:55
  • $\begingroup$ 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. $\endgroup$ Jun 27, 2013 at 3:34
  • $\begingroup$ It was mostly as a test of this notification system. meta.mathoverflow.net/questions/189/when-to-add-post-notices $\endgroup$ Jun 27, 2013 at 7:26

1 Answer 1

1
$\begingroup$

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.

$\endgroup$
3
  • $\begingroup$ But P(X) has a subframe {0,a,b,1} where 'a' and 'b' are complements which is regular and hence Isbell Hausdorff. $\endgroup$ Jun 27, 2013 at 5:52
  • $\begingroup$ 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. $\endgroup$ Jun 27, 2013 at 18:47
  • $\begingroup$ I may have to contact one of the authors about this error. $\endgroup$ Jun 27, 2013 at 22:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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