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• CommentRowNumber1.
• CommentAuthorMike Shulman
• CommentTimeNov 30th 2016

I added some discussion to Hausdorff space of how the localic and spatial versions compare in classical and constructive mathematics, including in particular the fact that I just learned (in discussion with Martin Escardo and Andrej Bauer) that a discrete locale is Hausdorff iff it has decidable equality.

• CommentRowNumber2.
• CommentAuthorMike Shulman
• CommentTimeFeb 3rd 2017

I added to Hausdorff space a theorem characterizing some localically strongly Hausdorff spaces in terms of apartness relations, which is a sort of dual or converse to the theorem I recently added to apartness relation.

• CommentRowNumber3.
• CommentAuthorTim Campion
• CommentTimeOct 1st 2018

Clarified a confusing remark about separatedness in different categories. A separated scheme is certainly not the same thing as a scheme whose underlying Zariski locale is Hausdorff. I doubt even that a separated scheme is the same thing as a scheme whose underlying Zariski locale is separated over the classifying topos for local rings.

• CommentRowNumber4.
• CommentAuthortphyahoo
• CommentTimeMay 4th 2020
• (edited May 4th 2020)
Some questions on "6. Beyond topological spaces Hausdorff locales"

first sentence: "The most obvious definition for a locale X to be Hausdorff is that its diagonal X→X×X is a closed (and hence proper) inclusion."

-- Is "inclusion" same as "embedding of topological spaces" nlab page? If so I would link explicitly.

Is this definition incorrect because of the edge case discussed subsequently? If so I would say it is incorrect (or imprecisve) straight away.

At https://ncatlab.org/nlab/show/closed+subspace#constructive_mathematics there is "This constructive variety of notions of closed subspace gives rise to a corresponding variety of notions of Hausdorff space when applied to the diagonal subspace." I understand this definition is classical, but I wonder if that link has anything relevant to the current topic.

second sentence: "However, if X is a sober space regarded as a locale, this might not coincide with the condition for X to be Hausdorff as a space, since the Cartesian product X×X in the category Loc of locales might not coincide with the product in the category Top of tpological spaces (the Tychonoff product)."

Can this be rephrased as "However, if LX is the locale of opens of a sober space X, closed diagonal of LX might not correspond with Hausdorff X, since the Cartesian product in the category Loc of locales might not coincide with the product in the category Top of topological spaces (the Tychonoff product)"

-- Is "cartesian product XxX in the category Loc" the same as "product in the category Loc" ?

A concrete example where the two products are not the same, would be helpful here. https://ncatlab.org/nlab/show/Hausdorff+implies+sober#strictness_of_implication there would be my best guess as a starting point, but it's a just a guess. Is there an example here, or elsewhere, where the two notions of product are not the same?
• CommentRowNumber5.
• CommentAuthorDmitri Pavlov
• CommentTimeMay 4th 2020

Re #4: This notion of “inclusion” is discussed at sublocale.

• CommentRowNumber6.
• CommentAuthorUrs
• CommentTimeJan 21st 2021

I have added statement and proof (here) that in the category of Hausdorff spaces, inclusions of dense subsets are epi.

Briefly mentioned the converse and added pointer to

• Jérôme Lapuyade-Lahorgue, The epimorphisms of the category Haus are exactly the image-dense morphisms (arXiv:1810.00778)
• CommentRowNumber7.
• CommentAuthorTodd_Trimble
• CommentTimeJan 25th 2021
• (edited Jan 25th 2021)

Surely a much cleaner proof is to recognize that the equalizer of two morphisms $f, g: X \rightrightarrows Y$ can be computed as a pullback

$\array{ E & \to & Y \\ \downarrow & & \downarrow \mathrlap{\delta} \\ X & \underset{(f, g)}{\to} & Y \times Y }$

where $\delta: Y \to Y \times Y$ is a closed inclusion, by Hausdorffness. By continuity, the pullback inclusion $E \to X$ is also closed. By assumption, this inclusion contains a dense subset $A \hookrightarrow X$. Being therefore dense and closed, $E \to X$ is all of $X$. Hence $f = g$ on all of $X$.

(By the way, this description of an equalizer as a pullback should probably replace what is currently at equalizer. I can get to this later.)

• CommentRowNumber8.
• CommentAuthorTodd_Trimble
• CommentTimeFeb 17th 2021

Put in the simplified proof.

• CommentRowNumber9.
• CommentAuthorUrs
• CommentTimeFeb 17th 2021

It looks like the proof I had written got deleted. I have re-instantiated it and added brief indication for the reader that one proof is abstract category-theoretic and the other is concretely point-set topological.

• CommentRowNumber10.
• CommentAuthorTodd_Trimble
• CommentTimeFeb 17th 2021