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• CommentRowNumber1.
• CommentAuthorMike Shulman
• CommentTimeDec 8th 2016

I added the definition of uniformly/almost located to located subspace, as well as attempted proofs that covert and compact subspaces of uniformly regular uniform spaces are uniformly located. Unfortunately covert sets and compact spaces — as opposed to compact locales — seem to be hard to come by constructively. It would be nice to have some more concrete examples. I also don’t know what to write for the “idea” section, nor what Toby had in mind for “topologically located”.

• CommentRowNumber2.
• CommentAuthorTobyBartels
• CommentTimeDec 9th 2016

I found the definition of topological locatedness thanks to Google Books. It has a similar flavour to the uniform case. Also an Idea section.

• CommentRowNumber3.
• CommentAuthorMike Shulman
• CommentTimeDec 9th 2016

Thanks! Topological locatedness seems very weak; in particular, singletons being topologically located is weaker than quasiregularity.

I’d like to see more examples of located and non-located subsets, as well as examples of what located subspaces are used for. Where should I be looking?

• CommentRowNumber4.
• CommentAuthorTobyBartels
• CommentTimeDec 9th 2016

Well, Bishop uses metrically located subsets. A lot of hereditary properties fail in general but work for located subspaces: total boundedness comes to mind. Besides that, Section 7.3 of Troelstra & van Dalen seems to be readable on Google Books.

• CommentRowNumber5.
• CommentAuthorMike Shulman
• CommentTimeDec 9th 2016
• (edited Dec 9th 2016)

I’m no longer convinced that Bridges’ “almost located” is the most natural generalization of metric locatedness to uniform spaces. If we translate almost-locatedness into a metric space, it says that for any $b\gt 0$ there exists an $a\gt 0$ such that for any $x$, either $\forall y\in S, d(x,y)\gt a$ or $\exists y\in S, d(x,y)\lt b$. But here $a$ could be much smaller than $b$, whereas metric locatedness says that for any $b\gt 0$ and any $a\lt b$, the same disjunction holds for all $x$.

It seems to me that a more natural notion of uniform locatedness would be that for any entourages $U,V,W$ such that $W\circ V \subseteq U$ we have $\neg V[S] \cup U[S] = X$. Whereas Bridges’ condition is implied by metric locatedness, I believe this condition implies metric locatedness, taking $U=U_b$ and $V = U_{a+\varepsilon}$ and $W = U_{b - a - \varepsilon}$ for sufficiently small $\varepsilon$. Bridges’ counterexample showing that metric locatedness is not a uniform invariant means that this notion of uniform locatedness is strictly stronger than metric locatedness, since it implies metric locatedness with respect to all metrics inducing the same uniformity. Moreover, I think I can prove that if $S$ is uniformly located in this sense then so is $N[S]$ for any entourage $N$, since $U[N[S]] = (U\circ N)[S]$ and if $W\circ V\subseteq U$ then $W\circ (V\circ N) \subseteq (U\circ N)$.

Unfortunately, this condition is much too strong. If the points of the discrete uniform space $2$ are located in this sense, then LEM follows: let $W$ be the diagonal $\Delta_2$, and $U = V = \Delta_2 \cup \{ (x,y) \mid P \}$ for any $P:Prop$. This suggests that we should only require it when the entourages $U,V,W$ are themselves located in some sense, but that’s getting circular. Hmm. Maybe $W\circ V \subseteq U$ should be replaced by $\neg V \cup U = X\times X$, so that “point are located” becomes trivial? But then the proof of “$S$ located $\Rightarrow$ $N[S]$ located” fails, which is what I really wanted.

Actually, I guess “$S$ located $\Rightarrow$ $N[S]$ located” should also only be expected when $N$ is located (if then), since if $S$ is a point and $N = U \cup \{ (x,y) \mid P \}$ then $N[S]$ is probably not located.

Have to go, so I’m posting this kind of half-baked, but maybe someone will have helpful thoughts…

• CommentRowNumber6.
• CommentAuthorTobyBartels
• CommentTimeDec 11th 2016
• (edited Dec 11th 2016)

Maybe we just want a base of entourages that has this property, recognizing that general entourages (being completely arbitrary supersets of basic entourages) can be unlocatably weird.

• CommentRowNumber7.
• CommentAuthorMike Shulman
• CommentTimeDec 12th 2016

Yes, I thought of that too. By itself, it’s not as strong as what I wanted, e.g. if we use the base of entourages $\{(x,y) \mid d(x,y) \lt 1/2^n \}$ then we can’t conclude, for instance, that $d(x,S)\lt 1 \vee d(x,S)\gt \frac{3}{4}$. Even requiring the base to be closed under $\circ$ doesn’t quite give this, but it might still give an interesting notion (and I think closure under $\circ$ is necessary for the $S\rightsquigarrow N[S]$ step anyway).

This is again weaker than metric locatedness, which at the moment I’m inclined to regard as a good thing: in Bridges’ example one might object that the set is only “accidentally” located with respect to one metric, but in fact I think I would argue that this set should be located (the metric space in question is itself weird, but the subset of it doesn’t involve any additional weirdness), and so the freedom to “choose an appropriate metric” is a good thing.

It does still seem that “points are located” is an additional condition for this notion of locatedness, but at least it’s not an unreasonably strong condition. I hope it would be satisfied by all metrizable or even gaugeable uniform spaces, although closing the base up under $\circ$ makes that not totally obvious.

• CommentRowNumber8.
• CommentAuthorMike Shulman
• CommentTimeDec 12th 2016

Hmm… maybe it is unreasonably strong. If $q$ ranges over a set of positive real numbers with infimum $0$, then the closure under $\circ$ of the entourage base $U_q = \{ (x,y) \mid d(x,y)\lt q\}$ for a metric space consists of the sets

$U_{q_0,\dots,q_n} \{ (x,y) \mid \exists z_1,\dots,z_n, d(x,z_1)\lt q_0 \wedge d(z_1,z_2)\lt q_1 \wedge \cdots \wedge d(z_n,y) \lt q_n \}$

where $U_{q_{n+1},\dots,q_m} \circ U_{q_0,\dots,q_n} = U_{q_0,\dots,q_m}$. There are canonical inclusions $U_{q_0,\dots,q_m} \subseteq U_{r_0,\dots,r_k}$ whenever we have an order-preserving map $\sigma:\{0,\dots,m\} \to \{0,\dots,k\}$ such that $\sum_{\sigma(j)=i} q_j \le r_i$ for all $i$. But even if an inclusion $U_{q_{n+1},\dots,q_m} \circ U_{q_0,\dots,q_n} \subseteq U_{r_0,\dots,r_k}$ is of this form, I don’t see how to show $(x,y)\in U_{r_0,\dots,r_k}$ or $(x,y)\notin U_{q_0,\dots,q_n}$. Even in the simple case $U_r \circ U_q \subseteq U_{q,r}$, we would be claiming that for any $x,y$ either there is a $z$ with $d(x,z)\lt q$ and $d(z,y)\lt r$, or $d(x,y)\ge q$. We can say that either $d(x,y)\lt q+r$ or $d(x,y)\ge q$, but in a general metric space there’s no reason the former should be “subdividable” into the existence of a $z$.