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In Peddechio & Tholens Categorical Foundations they quote PT Johnstone in their chapter on Frames & Locales:

...the single most important fact which distinguishes locales from spaces: the fact that every locale has a smallest dense sublocale. If you want to 'sell' locale theory to a classical topologist, it's a good idea to begin asking him to imagine a world in which any intersection of dense subspaces would always be dense. Once he has contemplated some of the wonderful consequences that would follow from this result you can tell him this world is exactly the category of locales.

Q. My classical topology in somewhat rusty and neither Johnstone nor the book in which this quote is embedded in expand upon 'the wonderful consequences'. What might they be?

It seems to me that one obvious result that would be a triviality is Baires Category theorem. Also, given that position and momentum observables are represented by densely defined unbounded operators, is this result useful there, directly or indirectly?

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  • $\begingroup$ It seems that you have already answered your own question... Every space is a Baire space, isn't that great? $\endgroup$ – Najib Idrissi Nov 4 '18 at 11:03
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    $\begingroup$ @NajibIdrissi I think the OP would like you to elaborate on this: not everyone is going to immediately going to see why this would be great, even if it's obvious to you :) $\endgroup$ – Yemon Choi Nov 4 '18 at 14:51
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    $\begingroup$ Of course, the key is the generalization/reinterpretation of what is meant by "subspace" here: see ncatlab.org/nlab/show/sublocale $\endgroup$ – Yemon Choi Nov 4 '18 at 14:53
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An example of an application of this appeared in Alex Simpson's paper called "Measure, Randomness and Sublocales" where Alex used this fact to resolve the Banach-Tarski paradox.

Recall that $\mathbb Q$ and $\mathbb R\setminus \mathbb Q$ are disjoint as subspaces of $\mathbb R$ but they are not disjoint as sublocales (since $\mathbb R$ has the smallest dense sublocale). A similar situation happens when decomposing a solid ball with a Banach-Tarski decomposition. The individual parts have disjoint sets of points but they intersect as sublocales.

A consequence of this is rather philosophical. Measure theory resolves the paradox by allowing to attribute size/measure only to certain subsets of a space, that is, to the measurable subsets (as opposed to all subsets). Alex Simpson suggests another approach. Instead of restricting the subsets which we can measure, we can enlarge the measure function to all "abstract subspaces", that is, sublocales. (To be more precise, Alex Simpson based his paper around $\sigma$-sublocales, which are even more general, but he claimed that the arguments could be also adapted for sublocales.)

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    $\begingroup$ I just showed a bunch of friends that paper and they were absolutely blown away. $\endgroup$ – Harry Gindi Dec 5 '18 at 10:19
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    $\begingroup$ This work is indeed mind-blowing. Independently of Simpson, Olivier Leroy did the same in the 90's. See: hal.archives-ouvertes.fr/hal-00741126/document (in French) $\endgroup$ – user56097 May 21 at 0:05
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I like to call this result the localic Baire category theorem, and it plays essentially the same role as Baire category theorem: it lets you "construct" object by showing that some spaces are non-empty because they are dense. Of course because we are working with locales one cannot necessarily "construct" an object at the end, because a non-empty locale does not neccessary have a points. But often having a non-empty space of "Something" is as good as having a non-explicit construction of a "Something".

Note that the usual Bair category theorem can be seen as a mixture of this localic Baire category theorem and some classical results saying that locales satisfying some properties have enough points. And if you stare at the usual proof of Baire category theorem you will see that one can clearly recognize the proof of these statement of existence of points)

So in addition to all the usual application of Baire category theorem, Here is the simplest and most famous example of this:

Let $I$ be an infinite set (preferably uncountable).

Consider the space $I^{\mathbb{N}}$ to the space of sequences with values in $I$ (with the product topology). For each $i \in I$, consider $D_i \subset I^{\mathbb{N}}$ the subset of sequence which take the value $i \in I$ at least once. One easily see that each $D_i$ is a dense open subset of $I^{\mathbb{N}}$.

Consequence: $$ S:=\left( \bigcap_{i\in I} D_i \right) \neq \emptyset $$

Of course the intersection means "intersection of sublocales" and by non-empty I mean that the resulting sublocale is not the empty sublocales. And it is non-empty because $\overline{S} = I^{\mathbb{N}}$.

This $S$ is essentially by definition "The space of surjections $\mathbb{N} \twoheadrightarrow I$ (indeed $f \in D_i$ means that $i$ is in the image of $f$, so $f \in S$, means that all $i \in I$ are in the image.)

So for any uncountable set, one has a non empty space of surjection from $\mathbb{N}$ to $I$. Of course this space cannot have any point if $I$ is not countable.

I like to sum up this by the fact that in locales theory "every discrete space is (geometrically) countable", but this more of a slogan.

More concretely this is directly related to the fact that any set can be made countable in a forcing extension of set theory, which is the key idea that lead to the proof By cohen of the independance of the continuum hypothesis. (though of course forcing and Cohen's proof existed before one knew about the connection with this geometric point of view, but it is still true that these proofs are essentially the same as these observations, a discussion of Cohen's proof in this language can be found in Moerdijk&MacLane sheaves in geometry and logic).

Another example is the fact that the space (read locale) of "Lebesgue Generic real number" is dense in the space of real number. Indeed Lebesgue generic real number are the things in the intersection of all sub-locales of density $1$ in $\mathbb{R}$, and these are all individually dense.


Edit: I'm trying to address Gerhard Paseman comment:

Let me start with this space of "generic real number" $G \subset \mathbb{R}$ defined as the intersection of all subspaces of density $1$ that I mentioned at the end. (Note to the experts:I am always confused on whether this space is already boolean or not, I'm assuming it is, but if it is not please tell me. If it is not then some of the things that I say below might require to move to a certain Boolean cover of $G$)

It has some quite nice properties that are easy to explain: For example bounded continuous functions from ${G}$ to $\mathbb{C}$ are exactly the same as element of $L^{\infty}(\mathbb{R},\mathbb{C})$.

The fact that $G$ has no points simply corresponds to the fact that you can't evaluate a function in $L^{\infty}$ (in localic terms: these are functions defined only on generic reals)

more generally things defined over it are things defined "almost everywhere" and that for example one can define what are bundle of Hilbert spaces over $G$ and under countability condition they are exactly the same as measurable fields of Hilbert spaces on $\mathbb{R}$ with morphisms between them being already defined up to equality almost everywhere.

But the notion of bundle of Hilbert spaces over $G$ makes sense without countability conditions and is considerably more robust than the notion of measurable fields of spaces when we drop the countability assumption (it is equivalent to the notion of $W^*$-modules over $L^{\infty}(\mathbb{R},\mathbb{C})$ ).

But of course, all those require first to show that $G$ is non-empty.

An important applications of $S$ is for forcing in set theory. Though forcing existed before this kind of consideration this gives a very nice geometric picture of what forcing is (and this is completely equivalent to the usual formulation of forcing, its just a different language).

Essentially, by replacing "sets" with "sheaves over the Space $S$" (or more precisely, over some boolean cover of $S$, or looking at double negation sheaves) one obtains a new model of set theory in which $I$ is actually countable. This is what Set theorists call a forcing extension that collapse $I$ and $\omega$.

informally (though it can be made formal with things like internal logic) its because working with sheaves over $S$, means working with structure which depends on a parameter $s \in S$, and informally such a parameter is the data of a surjection $\mathbb{N} \rightarrow I$

Other kind of forcing corresponds to other space similar to $S$, and one generally proves they are non-empty by the same argument.

The fact that space $S$ is non-empty is really equivalent to the fact that one can collapse cardinals by forcing, so it is a bit hard to find application outside of logic. But I'll see if I can think of something nice enough to be explained here.

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  • $\begingroup$ I read this and I think "So those people who point out Cantor's proof is flawed are right? The reals are actually countable?" More to the technical point, I would appreciate this answer more if I were told a use for S or for an 'empty' sublocale like S. Do you have any such use to tell those of us unfamiliar with locales? (Maybe I don't understand intersection.) Gerhard "Or, Which Axiom Is Different?" Paseman, 2018.12.04. $\endgroup$ – Gerhard Paseman Dec 4 '18 at 16:44
  • $\begingroup$ Thank you. I think you should keep trying, but (at this writing) I get the feeling S is also potentially useful and not just convenient. I especially like the link to "almost everywhere". If you could tie this directly into logics that used quantifiers like (there exist infinitely many), that would appeal to my not quite mature understanding of mathematical logic, and may reach many others as well. Gerhard "As Opposed To Mature Understanding" Paseman, 2018.12.04. $\endgroup$ – Gerhard Paseman Dec 4 '18 at 17:47
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The reason why the minimal dense sublocale of a frame is important is because it is a complete Boolean algebra and complete Boolean algebras are very special kinds of frames. Furthermore, the nucleus corresponding to a minimal dense sublocale gives an example of a frame homomorphism $f:L\rightarrow B$ and we shall see why such homomorphisms are essential to point-free topology. In this post, I will mainly talk about the importance of frame homomorphisms $f:L\rightarrow B$ for the sake of generality.

Complete Boolean algebras make point-free topology much more elegant than it would otherwise be, and the minimal dense sublocale of a frame is an important part of the relation between complete Boolean algebras and point-free topology.

Complete Boolean algebras satisfy some of the highest separation axioms, and they even satisfy some other notable peculiar point-free topological properties. Complete Boolean algebras are always regular, completely regular, normal, zero-dimensional, paracompact, ultraparacompact, extremally disconnected, $P$-frames, etc. Complete Boolean algebras are also notable because all atomless complete Boolean algebras are completely point-free (the points in a complete Boolean algebra are precisely the generic ultrafilters on that complete Boolean algebra).

If $\kappa$ is an uncountable regular cardinal, then we say a regular space $(X,\mathcal{T})$ is a $P_{\kappa}$-space if the intersection of less than $\kappa$ many open sets is still open. We say that a frame $L$ is weakly $\kappa$-distributive if it satisfies the property $x\vee\bigwedge_{i\in I}y_{i}=\bigwedge_{i\in I}(x\vee y_{i})$ whenever $|I|<\kappa$. Therefore, the notion of a weakly $\kappa$-distributive frame is a generalization of the notion of a $P_{\kappa}$-space (Caution: We need to be careful since there are multiple inequivalent but natural ways of generalizing the notion of a $P_{\kappa}$-space to point-free topology).

$\textbf{Theorem:}$ A regular space $(X,\mathcal{T})$ is a $P_{\kappa}$-space if and only if the frame $\mathcal{T}$ is weakly $\kappa$-distributive.

$\textbf{Theorem:}$ A regular frame $L$ is a complete Boolean algebra if and only if it is weakly $\kappa$-distributive for all uncountable regular cardinals $\kappa$.

One should think of a complete Boolean algebra as therefore like a space where the arbitrary intersection of open sets is open, and complete Boolean algebras in a sense behave like discrete spaces. In fact, every sublocale of a complete Boolean algebra is both an open and a closed sublocale. A regular frame is a complete Boolean algebra precisely when it has no proper dense sublocale.

If $B_{L}$ is the minimal dense sublocale of a frame $L$, then there is a surjective frame homomorphism $L\rightarrow B_{L}$ defined by $x\mapsto x^{**}$.

Let $\mathfrak{b}(a)$ denote the smallest dense sublocale of $L$ containing $a$. Then there is a surjective frame homomorphism $\phi_{a}:L\rightarrow\mathfrak{b}(a)$ defined by $\phi_{a}(x)=(x\rightarrow a)\rightarrow a.$ Then $\mathfrak{b}(a)$ is a complete Boolean algebra and if $S\subseteq L$ is a sublocale which is a complete Boolean algebra, then $S=\mathfrak{b}(a)$ for some $a$.

Fact: Every frame $L$ embeds as a subframe of a product of complete Boolean algebras. In particular, the mapping $\phi:L\rightarrow\prod_{a\in L}\mathfrak{b}(a)$ defined by $\phi(x)=(\phi_{a}(x))_{a\in L}$ is an embedding.

The above fact in a sense of the extension of the fact that every topology $(X,\mathcal{T})$ embeds as a subframe of $\{0,1\}^{X}$. One could also show that every regular frame $L$ is a subframe of a product of complete Boolean algebras using forcing, but such an argument is more difficult and less efficient in terms of the cardinality of the complete Boolean algebras required.

The congruence tower

We shall now describe a construction that allows us to extend a frame $L$ to a much larger frame $M$ but where the frame homomorphisms $f:L\rightarrow B$ are in a one-to-one correspondence with the frame homomorphisms $g:M\rightarrow B$. This construction is only interesting because for each frame $L$ there are many interesting frame homomorphisms $f:L\rightarrow B$ including the frame homomorphism from $L$ onto the minimal dense sublocale of $L.$

Suppose that $L$ is a frame. Then a congruence on $L$ is an equivalence relation $\simeq$ such that if $v\simeq w,x\simeq y$, then $v\wedge x\simeq w\simeq y$ and if $x_{i}\simeq y_{i}$ for $i\in I$, then $\bigvee_{i\in I}x_{i}\simeq\bigvee_{i\in I}y_{i}$. Let $\mathfrak{C}(L)$ denote the lattice of all congruences on the frame $L$. Then $\mathfrak{C}(L)$ is itself a zero-dimensional frame. Define a mapping $\nabla_{L}:L\rightarrow\mathfrak{C}(L)$ by letting $(x,y)\in\nabla_{L}(a)$ if and only if $x\vee a=y\vee a$. Then $\nabla_{L}$ is an injective frame homomorphism such that each $x\in L$ is complemented in $\mathfrak{C}(L)$.

$\textbf{Theorem:}$ Suppose that $L$ is a frame, $B$ is a complete Boolean algebra, and $f:L\rightarrow B$ is a frame homomorphism. Then there is a unique frame homomorphism $\overline{f}:\mathfrak{C}(L)\rightarrow B$ such that $f=\overline{f}\nabla_{L}$.

The above theorem would not be as interesting if frame homomorphisms $f:L\rightarrow B$ with $B$ Boolean were rare or difficult to construct.

Suppose that $L$ is a frame. Then define $\mathfrak{C}^{\alpha}(L)$ for each ordinal $\alpha$ by letting $\mathfrak{C}^{0}(L)=L$, $\mathfrak{C}^{\alpha+1}(L)=\mathfrak{C}(\mathfrak{C}^{\alpha}(L))$ and where if $\gamma$ is a limit ordinal, then $\mathfrak{C}^{\gamma}(L)$ is the direct limit of $(\mathfrak{C}^{\alpha}(L))_{\alpha<\gamma}$ where the transitional mappings are produced by the mappings of the form $\nabla_{\mathfrak{C}^{\alpha}(L)}$ and direct limits of such mappings. Let $\nabla_{L}^{\alpha}:L\rightarrow\mathfrak{C}^{\alpha}(L)$ be the canonical embedding.

$\textbf{Theorem:}$ Suppose that $L$ is a frame, $B$ is a complete Boolean algebra, $\alpha$ is an ordinal, and $f:L\rightarrow B$ is a frame homomorphism. Then there is a unique frame homomorphism $\overline{f}:\mathfrak{C}^{\alpha}(L)\rightarrow B$ such that $f=\overline{f}\nabla_{L}^{\alpha}$.

We now have examples of towers of structures that, unlike the automorphism group tower, do not stop growing.

$\textbf{Theorem:}$ There is a frame $L$ such that the congruence tower $(\mathfrak{C}^{\alpha}(L))_{\alpha}$ never stops growing.

$\textbf{Forcing and the congruence tower}$

The existence of many frame homomorphisms $\phi:L\rightarrow B$ play a very important role in how frames behave when you put them into forcing extensions and they are the basis of a field which I have worked on called Boolean-valued point free topology which is essentially about considering frames in forcing extensions (for the set theorists, since frames are complete Boolean algebras, the Boolean-valued model approach to forcing works quite well.). The frame homomorphisms $\phi:L\rightarrow B$ should be thought of as the points that you add to the frame $L$ when you pass $L$ to turn it into a frame a forcing extension using the Boolean-valued model approach to forcing.

Let $\textrm{Sp}_{B}(L)$ be the set of all frame homomorphisms $\phi:L\rightarrow B$. Then $\textrm{Sp}_{B}(L)$ is a $B$-valued structure where we set $\|\phi=\theta\|=b$ when $b$ is the largest element in $B$ with $\phi(x)\wedge b=\theta(x)\wedge b$ for each $b\in B$. Since $\textrm{Sp}_{B}(L)$ is a $B$-valued structure, one should consider $\textrm{Sp}_{B}(L)$ as an object in the Boolean-valued universe $V^{B}$.

$\textbf{Theorem}:$ Suppose that $L$ is a frame. Then the mapping $(\nabla_{L}^{\alpha})^{*}:\textrm{Sp}_{B}(\mathfrak{C}^{\alpha}(L))\rightarrow\textrm{Sp}_{B}(L)$ is a bijection. In particular, $$V^{B}\models\text{There is a bijective continuous function from $\textrm{Sp}_{B}(\mathfrak{C}^{\alpha}(L))$ to $\textrm{Sp}_{B}(L)$.}$$

This correspondence is startling since the frame $\mathfrak{C}^{\alpha}(L)$ could have arbitrarily large cardinality.

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  • $\begingroup$ -1: I was looking for an answer in the style of Johnstones essay; that's why I quoted it... $\endgroup$ – Mozibur Ullah Mar 13 '19 at 12:21
  • $\begingroup$ @MoziburUllah. So is there anything factually incorrect about my post? $\endgroup$ – Joseph Van Name Mar 13 '19 at 13:55
  • $\begingroup$ A complete disregard for the notion of a complete Boolean algebra is a bad mathematical practice. $\endgroup$ – Joseph Van Name Mar 13 '19 at 16:58

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