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1. (Hi, I’m new)

I added some examples relating too simple to be simple to the idea of unbiased definitions. The point is that we often define things to be simple whenever they are not a non-trivial (co)product of two objects, and we can extend this definition to cover the “to simple to be simple case” by removing the word “two”. The trivial object is often the empty (co)product. If we had been using an unbiased definition we would have automatically covered this case from the beginning.

I also noticed that the page about the empty space referred to the naive definition of connectedness as being

“a space is connected if it cannot be partitioned into disjoint nonempty open subsets”

but this misses out the word “two” and so is accidentally giving the sophisticated definition! I’ve now corrected it to make it wrong (as it were).

• CommentRowNumber2.
• CommentAuthorTodd_Trimble
• CommentTimeSep 29th 2015

I had to go to empty space and look at what you did to fully understand your comment, but now that I see it: good catch!

I have this occasional nagging feeling that unbiased definitions ought to be expanded in some ways, because it seems to me the process of removing bias comes in many guises. The general idea of using a Lawvere theory instead of a traditional signature for, say, the theory of groups, is a way of removing bias, by “saturating” the class of operations. Similarly, there is a way of defining Boolean algebra in an unbiased way, as a product-preserving functor from finite non-empty sets to $Set$, that removes the persistent bias that Boolean algebras are intrinsically about the number $2$. But I’m somewhat undecided what I want to say.

• CommentRowNumber3.
• CommentAuthorMike Shulman
• CommentTimeSep 29th 2015

I would also generalize “bias” from what is said at bias to cover cases when the generating operations have arities other than 0 and 2. For instance, the usual definition of heap is biased, although its basic operation is ternary.

• CommentRowNumber4.
• CommentAuthorDavidRoberts
• CommentTimeSep 29th 2015
• (edited Sep 29th 2015)

a product-preserving functor from finite non-empty sets to Set

doesn’t this then turn a theorem (baby Stone duality) into a definition? (Not that this is bad: the theorem then becomes that the definition follows from the biased definition)

• CommentRowNumber5.
• CommentAuthorTodd_Trimble
• CommentTimeSep 30th 2015

Let’s see: I understand baby Stone duality as the assertion that the category of finite Boolean algebras (including by the way the terminal one) is dual to the category of finite sets (including the empty one). There are relations between baby Stone duality and what I’m saying here, but I tend to think of it slightly differently, and the relation is not quite obvious. I’ll try to point to something on this relation in a minute, but first here’s how I think of things. Or, if you like you can skip to the end where I touch on the relation.

The usual Lawvere theory for Boolean algebras is, via baby Stone duality, equivalent to the category of finite sets of cardinality $2^n$ (inasmuch as f.g. free Boolean algebras are the ones of cardinality $2^{2^n}$). Let me denote it $Fin_{2^\bullet}$. The Cauchy completion of $Fin_{2^\bullet}$ is the category $Fin_+$ of nonempty finite sets.

The following result is easily proven, but I don’t think I’ve ever seen it in a book:

Proposition: For any category with products $C$, the Cauchy completion $\bar{C}$ also has products and the inclusion $i: C \to \bar{C}$ preserves them. Restriction along $i$ induces an equivalence $Prod(\bar{C}, Set) \to Prod(C, Set)$.

Applying this to $C = Fin_{2^\bullet}$ and $\bar{C} = Fin_+$, this gives the result that the category of Boolean algebras is equivalent to $Prod(Fin_+, Set)$.

We could go in reverse, too, and say $Prod(Fin_+, Set)$ is equivalent to $Prod(Fin_{3^\bullet}, Set)$, by the same proposition. This would lead to a presentation of Boolean algebras in terms of a somewhat different Lawvere theory biased towards $3$. i.e., there would be not $2^{2^n}$ definable $n$-ary operations on “underlying sets” of Boolean algebras, but $3^{3^n}$ many. What I am suggesting here is that there are many different underlying-set functors $Bool \to Set$ to choose from, depending on one’s bias. You could even see a Boolean algebra $B$ not as a one-sorted object, but as multisorted: for example for each prime $p$, the $p$-sort $U_p(B)$ is the value of $B$ = product-preserving functor $Fin_{+} \to Set$ evaluated at $\{1, 2, \ldots, p\}$.

I wrote some of this out at Boolean algebra; look for the section on the unbiased definition.

There is also an unbiased definition of ultrafilter. Let $i_+: Fin_+ \to Set$ be the inclusion, corresponding to the Boolean algebra we normally call $2$. For any set $X$, the product-preserving functor $Set(X, i_+ -): Fin_+ \to Set$ corresponds to the power set Boolean algebra $P X$; we could also denote this as $(-)^X: Fin_+ \to Set$. Then define an ultrafilter as a natural transformation $(-)^X \to (-)^1$. This is related to Tom Leinster’s articles on ultrafilters via codensity monads; it’s briefly touched upon here.

Among the observations there is that, passing to the $3$-biased case (and letting $M_3$ denote the Lawvere theory for that case), there are enough unary operations = functions $3 \to 3$ that an ultrafilter on $X$ can equivalently be described as not only as a 3-biased Boolean algebra map $3^X \to 3$ (which sounds complicated), but more simply as a function $3^X \to 3$ which respects the canonical action under the monoid $Fin_+(3, 3)$. This boils down to a characterization of ultrafilters mentioned by Tom here: an ultrafilter on $X$ is a collection $\mathcal{U}$ of subsets of $X$ such that for every partition of $X = X_1 + X_2 + X_3$ into $3$ nonempty subsets, exactly one of the $X_i$ belongs to $\mathcal{U}$.

Lawvere discusses similar themes here, mentioning (in somewhat obscure style) that existence of a nonprincipal $\mathcal{U}$ satisfying the analogous property but with $3$ replaced by $\mathbb{N}$ is equivalent to the existence of a measurable cardinal.

Back to Stone duality and the relation to what I just described: since it’s late here, let me just direct you to the middle of a Café conversation we had about this, starting with an elegant observation by Richard Garner here. Basically, yeah, (baby) Stone duality gives you that the category of finite sets (including the empty one) is dual to the category of f.p. Boolean algebras, and $Bool$ is thus equivalent to the category of finite-limit-preserving functors $Fin \to Set$. On the other hand, I talked about finite-product-preserving functors $Fin_+ \to Set$ on the category of nonempty finite sets. So seeing these are equivalent (directly) takes a little more work, and that’s where the conversation was going.

• CommentRowNumber6.
• CommentAuthorDavidRoberts
• CommentTimeSep 30th 2015

@Todd - thanks for clarifying!

• CommentRowNumber7.
• CommentAuthorKarol Szumiło
• CommentTimeSep 30th 2015

Oscar, maybe I’m being dense, but when you include arbitrary finite (co)products, you also include the unary ones. So if you say

$n$ is prime if whenever $n=\prod_{i=1}^k a_i$ we have $n=a_i$ for some $i$

this includes $k=1$ which would say that every number is prime. Did I misunderstand something?

2. It’s certainly true that whenever we write any number as the product of one thing then it’s equal to that thing. But I think my definition if okay because I’m requiring that it hold for all $k$. Perhaps my wording is slightly ambiguous, what I mean to say is that

$n$ is prime” $\Leftrightarrow$$\forall k \forall (a_1,\dots,a_k) :\quad n=\prod_{i=1}^k a_i \Rightarrow (\exists i:n=a_i)$

• CommentRowNumber9.
• CommentAuthorKarol Szumiło
• CommentTimeSep 30th 2015

Right, I am being dense. That’s actually a really nice observation, it always bothered me that I didn’t know a clean way of stating such definitions.

• CommentRowNumber10.
• CommentAuthorMike Shulman
• CommentTimeSep 30th 2015

This is a small point, but I think the definition is better phrased using divisibility instead of equality: $\forall k \forall (a_1,\dots,a_k) :\; n\mid\prod_{i=1}^k a_i \Rightarrow (\exists i:n\mid a_i)$. That way it generalizes better to prime ideals.

• CommentRowNumber11.
• CommentAuthorOscar_Cunningham
• CommentTimeSep 30th 2015
• (edited Sep 30th 2015)

I agree that divisibility is nicer to use than equality. EDIT: Or at least that it’s nicer when thinking about rings. Is there an analogue of divisibility when thinking about whether spaces are connected?

What to do about $0$? Under the above definition it’s prime in $\mathbb{Z}$. (In fact we could use “$0$ is prime” as a joke definition of “integral domain”). But if you think about the divisibility preorder of $\mathbb{Z}$, $0$ is sitting at the top whereas the primes are all at the bottom just above $\pm 1$. So it seems like $0$ is genuinely different from the primes and shouldn’t be one. Is there a nice way to exclude it?

• CommentRowNumber12.
• CommentAuthorTodd_Trimble
• CommentTimeSep 30th 2015
• (edited Sep 30th 2015)

One way might be to say “if $(m n) = (p)$, then exactly one of $(m), (n)$ equals $(p)$.

Sheesh, hopefully I finally got that right after a few edits. :-P

• CommentRowNumber13.
• CommentAuthorzskoda
• CommentTimeSep 30th 2015

Not quite right :) still the quotation mark is opened but not closed :)

• CommentRowNumber14.
• CommentAuthorKarol Szumiło
• CommentTimeSep 30th 2015

Oscar, the analogue of the divisibility condition for spaces says that a space $X$ is connected if and only if the representable functor $\mathrm{Top}(X, -)$ preserves (finite) coproducts.

• CommentRowNumber15.
• CommentAuthorMike Shulman
• CommentTimeSep 30th 2015

Well, $0$ ought also to be prime if we want every integer to be expressible as a product of primes and units.

• CommentRowNumber16.
• CommentAuthorZhen Lin
• CommentTimeSep 30th 2015

To be fair, $(0)$ is a prime ideal, even if $0$ is not irreducible.

• CommentRowNumber17.
• CommentAuthorColin Tan
• CommentTimeOct 2nd 2015
The concrete definition given at the page on linear independence seems not to work for an empty collection of vectors.
• CommentRowNumber18.
• CommentAuthorTodd_Trimble
• CommentTimeOct 2nd 2015
• (edited Oct 2nd 2015)

Colin, I don’t see a problem. At linearly independent set it says

The subset $S$ is linearly independent if, conversely, every $a_i = 0_K$ whenever the sum

$\sum_{i=1}^n a_i v_i = 0_V ;$

otherwise, $S$ is linearly dependent.

(Here the $v_i$ are tacitly distinct elements of $S$.) From the statement, it is vacuously true that the empty set is a linearly independent set. For it is vacuously true that every scalar $a_i = 0$, since there are no scalars to speak of in an empty sum of scalar multiples on the left side of the displayed equation.

• CommentRowNumber19.
• CommentAuthorColin Tan
• CommentTimeOct 3rd 2015

There is a tacit universal quantifier. To be pedantic, a subset S is linearly independent if, for each finite subset T of S and each $a \in K^{T}$, if $\sum_{v\in S'} a_v v = 0$, then every $a_v = 0$.

• CommentRowNumber20.
• CommentAuthorTodd_Trimble
• CommentTimeOct 3rd 2015

Yes, of course. Do you still think there’s a problem?

• CommentRowNumber21.
• CommentAuthorColin Tan
• CommentTimeOct 3rd 2015
Probably not. Still, would we like to make the tacit universal quantifiers explicit?
• CommentRowNumber22.
• CommentAuthorTodd_Trimble
• CommentTimeOct 3rd 2015

Does it look okay now?

• CommentRowNumber23.
• CommentAuthorRodMcGuire
• CommentTimeOct 3rd 2015
• (edited Oct 4th 2015)

Todd - don’t non 2 biased (finite) Boolean lattices, (“Oolean”), correspond to number divisor lattices?

$2^n$, a 2oolean, is the divisor lattice of any product of n different primes - $div( \prod_{i=1}^n p_i)$

$3^n$, a 3oolean, is the divisor lattice of any product of n different squared primes - $div( \prod_{i=1}^n p{_i}^2)$

$k^n$, a koolean, is the divisor lattice of any product of n different $k-1$ powers of primes - $div( \prod_{i=1}^n p{_i}^{k-1})$

$1^n = 1^m = div(1) = 1$ is not a special “too simple” case.

and a (potentially) mixed bias Oolean lattice is the divisor lattice of any arbitrary number. Does your approach also apply to the mixed bias case? Is $div(0)$ an exceptional case for the “too simple” number $0$?

EDITED: mainly to change some wrong “some”s to “any”s.

• CommentRowNumber24.
• CommentAuthorColin Tan
• CommentTimeOct 4th 2015
It's great, Todd!
• CommentRowNumber25.
• CommentAuthorTodd_Trimble
• CommentTimeOct 4th 2015

(Rod, I haven’t forgotten your question, but I’d like to think more before replying. If Lawvere were here, he’d have a great deal to say, I’m sure.)

3. I think I found another example: We should say that the zero ideal isn’t principal. This is because it’s generated by no elements of the ring rather than just one. This has the advantage of making the statement of unique factorization easier: “The monoid of principal ideals is isomorphic to a free monoid”. It might seem that it makes the definition of principal ideal ring a bit less nice, namely “the minimal number of generators of any ideal is 1 or less”, but this actually makes the definition analogous to that for a Noetherian ring “the minimal number of generators of any ideal is finite”.

• CommentRowNumber27.
• CommentAuthorTodd_Trimble
• CommentTimeFeb 4th 2016

I’m conflicted on this one. You’re right that it’s generated by no elements. But it’s also generated by $0$ (there’s no law that says the generators have to be linearly independent). It would take a little rewriting over at ideal, which covers cases besides rings, to make it come out the way you want. (Maybe not much rewriting, but I’m not convinced it would be a good idea in the end.)

Somehow the argument about free (commutative) monoids doesn’t sway me. I think I’m happy referring to the monoid of nonzero principal ideals.

Just by-the-bye: one nice characterization of principal ideal domains is that they’re commutative rings where all ideals are free modules.

• CommentRowNumber28.
• CommentAuthorMike Shulman
• CommentTimeFeb 4th 2016

My initial inclination is also that the zero ideal is principal, because we generally talk about generators for ideals rather than “bases”. Sometimes the nullary cases are clarified by looking at constructive mathematics: what’s the right notion of “principal ideal” there?

• CommentRowNumber29.
• CommentAuthorTobyBartels
• CommentTimeSep 15th 2016

On prime numbers: I don't want to say that $0$ is a prime number. Saying that $1$ is not a prime number is a valid correction of the classical definition, which did not quite get at what it was trying to get at. But saying that $0$ is a prime number is going beyond that. (One way to see this is that general ring theory has a concept that generalizes that of prime number; it's just that this concept is called ’irreducible element’ rather than ’prime’.) I don't know why the meaning of the word ’prime’ changed in the transition from elementary number theory to ring theory, but it did.

• CommentRowNumber30.
• CommentAuthorTobyBartels
• CommentTimeSep 15th 2016

I'm also inclined to take the zero ideal to be principal, for the reason that Mike gives. And I'm the most constructivist person in this conversation, I think. But constructivism isn't giving me any insights here. A principal ideal is still just an ideal that is generated by a singleton subset. (Constructivism also gives us a principal antiideal in a ring-with-apartness, as one anti-generated by a singleton subset, but that doesn't help any.)

• CommentRowNumber31.
• CommentAuthorTobyBartels
• CommentTimeSep 15th 2016

The definition at integral domain allowed for the trivial ring, but I fixed that (and expanded some other stuff).

• CommentRowNumber32.
• CommentAuthorTobyBartels
• CommentTimeAug 26th 2018

Nothing wrong with Oscar’s definition of path-connected.

4. Rephrased the field example; ’except’ may be taken to imply 0 is an exception, i.e. the full non-naïve definition is assumed already.

Anonymous

• CommentRowNumber34.
• CommentAuthorGuest
• CommentTimeNov 2nd 2019
The article says:
A space is path connected if for each (Kuratowski)-finite subset there is a path passing through every point of that subset.

Then the empty space is not path connected because it has no paths at all and hence no path through the empty subset.

This argument leads me to the conclusion that no space is path-connected, because every space has the empty set as
a finite subset. The traditional definition of a path-connected space is one that admits a path between any pair of its
elements.
• CommentRowNumber35.
• CommentAuthorGuest
• CommentTimeNov 2nd 2019
Sorry. I should have added: Rob Arthan (rda@lemma-one.com) to the comment I just posted (about empty spaces).
• CommentRowNumber36.
• CommentAuthorMike Shulman
• CommentTimeNov 3rd 2019

I don’t follow your conclusion. Yes, any space $X$ has a finite subset $\emptyset$, so the “unbiased” definition on the page says that for it to be path-connected there must exist a path in $X$ that passes through every point of $\emptyset$. But this is true as soon as $X$ is nonempty, since any path in $X$ (such as a constant path) vacuously passes through every point of $\emptyset$.

• CommentRowNumber37.
• CommentAuthorTobyBartels
• CommentTimeDec 2nd 2019

Add structure and a paragraph beginning ’This should be distinguished from barring the trivial object entirely.’.