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• CommentRowNumber1.
• CommentAuthorTobyBartels
• CommentTimeSep 6th 2012

A base of a topological space $X$ need not be closed under intersection. But it is laxly closed in a sense. Does this concept of laxness have a name, perhaps in some categorified context?

Specifically, if $U, V$ belong to the base, then $U \cap V$ need not belong, but some subset of $U \cap V$ must belong. That is, we must have $W$ with $W \subseteq U \cap V$, that is $W \leq U \wedge V$, that is $W \to U \times V$. (We take it for granted that the power set of $X$ has intersections/meets/products. We are looking at a full subcategory of this power set and considering in what sense it’s closed under products.)

• CommentRowNumber2.
• CommentAuthorTodd_Trimble
• CommentTimeSep 6th 2012

Cofiltered?

• CommentRowNumber3.
• CommentAuthorTobyBartels
• CommentTimeSep 7th 2012

Well, a base is cofiltered, but that term says too much. I want to say that the base is [???] under intersection. There may not be a word for that. I could invent ‘lax-closed’.

• CommentRowNumber4.
• CommentAuthorMike Shulman
• CommentTimeSep 7th 2012

What does “cofiltered” say that is more than what you want?

• CommentRowNumber5.
• CommentAuthorTobyBartels
• CommentTimeSep 7th 2012

It says ‘intersection’. I want a word to put before ‘under intersection’.

OK, maybe you all are saying that ‘cofiltered’ is that word: cofiltered under intersection. Of course, that must be what you’re saying!

All right, but can I use this word if intersection is replaced with something that is not a semilattice operation? I feel like ‘cofiltered’ should only be used in that context, and that it generalises to non-semilattice posets rather than to non-semilattice binary operations.

• CommentRowNumber6.
• CommentAuthorTobyBartels
• CommentTimeSep 7th 2012

Here’s another example. Take the operation of addition on the poset of real-valued functions of one real variable (which I’m just trying to keep from being too simple). Consider a set $X$ of such functions. Suppose that, whenever $a, b \in X$, we have some $c \in X$ with $c \leq a + b$. So $X$ is not closed under addition. Should I say that $X$ is lax-closed under addition, or is there anything better to say?

• CommentRowNumber7.
• CommentAuthorMike Shulman
• CommentTimeSep 7th 2012

Ah, I see what you’re getting at. “Cofiltered” (or “codirected”) refers only to when the binary operation is intersection (and it makes sense even when intersections don’t exist).

Offhand I can’t think of a better name than “lax-closed”, except that perhaps “colax-closed” would be better. Since if $X\subseteq Y$ has this property and you choose such a $c$ for each $a,b\in X$ (or more generally for all $a_1, \dots ,a_n\in X$ — and maybe you need a separate assumption to be able to do this when $n=0$) then you’ll get a function $T X \to X$, where $T$ is the free-monoid monad, and a transformation

$\array{ T X & \hookrightarrow & T Y \\ \downarrow & \neArrow & \downarrow \\ X & \hookrightarrow & Y }$

going in the direction that would make the inclusion $X\hookrightarrow Y$ into a colax map. (Although there’s probably no reason for $T X \to X$ to be a map of posets, or to make $X$ into a $T$-algebra (even a lax or colax one).)

• CommentRowNumber8.
• CommentAuthorTobyBartels
• CommentTimeSep 8th 2012

Thanks, I’ll say ‘colax’.

1. Something seems odd about Proposition 4.1 and this discussion. When talking about the base of a topology, one takes the base sets to be open, as far as I know. This implies in particular that the intersection of base sets, being open, must be a union of base sets. It is not enough to require that the intersection of base sets contain another base set. To see this, consider a collection B = {A1, A2, A3} of subsets of X such that the union of A1 and A3 is the whole of X, while A2 is a strict subset of the intersection of A1 and A3. Then B satisfies the premise of Proposition 4.1. But B ist not a basis for any topology, since the intersection of A1 and A3 is not the union of elements of B.

Or have I made a mistake somewhere?

• CommentRowNumber10.
• CommentAuthorDavidRoberts
• CommentTimeJan 4th 2017

Are you thinking of a sub-basis?

• CommentRowNumber11.
• CommentAuthorTodd_Trimble
• CommentTimeJan 4th 2017
• (edited Jan 4th 2017)

What I think Carsten is saying is that the usual condition, to have a base $\mathbf{B}$, is that for any $U, V \in \mathbf{B}$ and any point $x \in U \cap V$, there should be $W \in \mathbf{B}$ with $W \subseteq U \cap V$ and $x \in W$.

• CommentRowNumber12.
• CommentAuthorCarsten Führmann
• CommentTimeJan 4th 2017
• (edited Jan 4th 2017)

@David No, I’m not thinking about a sub-basis. My mentioning of intersections of base sets rather springs from the proof of the following proposition, which is what I think Proposition 4.1. should be:

A collection B of subsets of X is the base of some topology on X iff every finite intersection of elements of B is an (arbitrary) union of elements of B.

Proof:

=> Let B a base of a topology - that is, every element of B is open and every open set is a union of elements of B. Then the intersection of two elements of B is also open, and therefore a union of elements of B. (There is also a trivial case for nullary intersections, which are by convention the whole space.)

<= Suppose every finite intersection of elements of B is an arbitrary union of elements of B. Then we define a topology of X by taking as open sets the arbitrary unions of B. This really is an open system: closure under aribitrary union is trivial. For closure under finite intersections, consider first the binary intersection of two unions of elements of B. We can use the boolean distributive law to obtain a union of binary intersections of elements of B. Since every such binary intersection is a union of elements of B, the whole thing is a union of elements of B. To see that X itself is open, note that X is the nullary intersection of elements of B, which by assumption is a union of elements of B.

q.e.d.

There is also a modified version of the proposition, also easily proved, which requires on the right side of the “iff” that for every element x in a finite intersection of elements of B, there must be another set of B that contains x and is contained in the intersection.

My example B = { A1, A2, A3 } from my previous comment was a counterexample to Proposition 4.1. in its current form on the nLab page.

Or have I blundered somewhere?

• CommentRowNumber13.
• CommentAuthorCarsten Führmann
• CommentTimeJan 4th 2017
• (edited Jan 4th 2017)

Todd I see your comment only now. You interpret me correctly. Plus, I think I have given a counter-example to the current version of Proposition 4.1.

• CommentRowNumber14.
• CommentAuthorDavidRoberts
• CommentTimeJan 4th 2017

My apologies, I see what you mean. Your proof looks good to me.

• CommentRowNumber15.
• CommentAuthorMike Shulman
• CommentTimeJan 5th 2017

Yes, I agree too. What’s colaxly-closed is not the base for the entire topology, but the corresponding base for the neighborhood filter at each point (which is the set of basic opens that contain that point).

2. I edited Proposition 4.1, using the formulation also suggested by Todd. (The first condition, that B must cover X, is just the case for the nullary intersection X of base sets.) I also removed the remark about “colax”, because that should probably go somewhere else.

• CommentRowNumber17.
• CommentAuthorMike Shulman
• CommentTimeJan 5th 2017

I think a colaxness remark would be appropriate here too.

3. I don’t feel competent to add that colaxness remark in the appropriate way. Does anyone else want to add one that fits?

• CommentRowNumber19.
• CommentAuthorMike Shulman
• CommentTimeJan 5th 2017