# Distinguishing equality and isomorphism as relations

Is this relational characterization of equality in Wikipedia accepted?

The identity relation is the archetype of the more general concept of an equivalence relation on a set: those binary relations which are reflexive, symmetric, and transitive. The relation of equality is also antisymmetric. These four properties uniquely determine the equality relation on any set S and render equality the only relation on S that is both an equivalence relation and a partial order. [My italics]

If one follows wikipedia, antisymmetry is defined in terms of equality, hence the definition is circular (impredicative?):

In mathematics, a binary relation $R$ on a set $X$ is antisymmetric if there is no pair of distinct elements of $X$ each of which is related by $R$ to the other. More formally, $R$ is antisymmetric precisely if for all a and b in $X$

if $R(a,b)$ and $R(b,a)$, then $a = b$, or, equivalently,

if $R(a,b)$ with $a ≠ b$, then $R(b,a)$ must not hold.

In posets antisymmetry is typically defined in terms of isomorphism, not equality, particularly when categorified. In any poset, such as $\bf 3$, itself a category (as well as an object of $\bf Poset$) distinction between objects should only be resolved up to isomorphism, not equality - the latter is considered an evil property. Regarding this, the n-lab entry on partial-order includes the statement:

the antisymmetry law, saying that $x \leq y$ and $y \leq x$ imply $x=y$, is evil in a certain sense. (On the other hand, it is not evil if taken as a definition of equality.)

So I wrote "typically" above to consider the parenthetical. Of course the objects of $\bf 3$ are not equal, but they are also not isomorphic (would have to look in pre-order categories. Here I'm focusing attention on the relation between equality and isomorphism).

Note that the n-lab entry on equality considers "different kinds of equality" complicating the issue.

By substituting $\cong$ for $=$ in both Wikipedia entries, doesn't isomorphism also yield both an equivalence and an order relation?

NOTES: it's interesting for such a basic, fundamental relation to be so convoluted, see eg Barry Mazur's definition in "When is one thing equal to another thing" ("up to unique iso") and Andreas Blass et al's "When are two algorithms the same". Historically, the issue also intrigued Leibniz who proposed the principle of identity of indiscernibles. Physically, nature at low quantum numbers seems to obey this principle: we cannot tell apart any two protons say, or water molecules. But already there are combinatorially ~10^200 possible molecules of less than 50 Daltons composed of H, C, O, N, S and maybe a couple other elements (~10^120 times more than the estimated number of atoms in the universe).

SARCASM: perhaps equality is unique up to unique isomorphism?

-
you might be interested in reading about Makkai's notion of first order logic with dependent sorts. It is a logic with no equality. It relates to both issues you raise in your question. –  Ittay Weiss Jan 24 '13 at 1:28
I think the first quote defined "equality" as a special type of relation. Its definition depends on the actual equality defined on elements of the set $S$. So, I'd say the definition is not circular in the sense that it defines a type of relation from the predicate "=". However, if the collection $\{(x, y) \in S \times S\ |\ x = y\}$ can always be defined, the definition above can be considered somewhat circular in the sense that one can obtain the predicate from the relation. Anyway, I'd say technically it is not circular. –  Tunococ Jan 24 '13 at 1:28
@IttayWeiss, thanks for the ref. Will check it out. Something tells me "first order logic" is similarly circular, but we'll see. –  alancalvitti Jan 24 '13 at 2:15
@Tunococ, what do you mean by "="? –  alancalvitti Jan 24 '13 at 2:15
Is it that rare/surprising that Wikipedia is at times inaccurate, circular and imprecise? –  Martin Jan 24 '13 at 4:18

The antisimmetry article in Wikipedia that you mentioned is written reasonably well and the definitions it gives are not circular.

In order to see this, one has to understand the implicit assumptions made by many mathematicians, including the author of the the above mentioned article.

Mathematicians, like journalists, have to write their papers/articles using a language. This means they need to follow a syntax if they wish to be understood by their readers.

A journalist writing an article for the New York Times is assumed to write according to the rules of the English language (a natural language) and will make deductions according to the rules of common sense logic.

A mathematician writing an article is assumed to write according to the rules of ...What?

Well, most mathematicians, unless they explicitly specify otherwise, write their papers using - consciously or unconsciously - the syntax rules of a first (or higher) order language with equality and reason according to classical logic.

Classical logic means roughly that they take for granted the law of excluded middle, so that they can use common sense reasoning and describe things that should or could exist, even if nobody knows - in practice - how to build them.

First (or higher) order language with equality means that the equality sign is a logical symbol, so they can write atomic formulas like $$a=b$$ and they can use simple substitution rules like $$a=b \rightarrow \phi (a)=\phi (b)$$ for any well formed formula (wff) $\phi$ containing $a$.

All this is very well explained in this Wikipedia article and also in this one.

Armed with this understanding, let's look again at the statement in the antisymmetry article:

if R(a,b) and R(b,a), then a = b

Here the author follows the (all too easily) condoned practice of mixing the standard first order language with equality with natural (English) language symbols. What is really meant is: $$R(a,b) \land R(b,a) \rightarrow a = b$$

This wff contains logical symbols $$\land , \rightarrow , =$$ Terms (in this case, just simple variables) $$a, b$$ Atomic formulas $$R(a,b), R(b,a), a=b$$ $R$ is a binary predicate in a (tacitly) assumed signature $\Sigma$ containing $R$, but not $=$. It is a wff as I said. The author also probably assumes some set theory axioms (ZFC presumably) and the deductive system of classical logic.

I want to stress that the equality sign here is just a logical symbol which happens to have the properties of the identity relation in set theory and the semantics of identity, just like the $\land$ symbol is a logical symbol which happens to have the properties of an operation in a boolean algebras theory and the semantics of the conjunction (and).

The Equality article in Wiipedia that you mention, tries to give different interpretations and uses to the concept of equality in different mathematics branches. I believe it is better to read it in the light of what I wrote above.

Category theory

Category theory is written , like set theory, in a first order language with equality. So all that has been said before, continues to hold.

"In categories like $\bf Poset$ antisymmetry is typically defined in terms of isomorphism, not equality."

does not make much sense to me, for the simple reason that in $\bf Poset$ there is no obvious definition of antisymmetry. $\bf Poset$ is a collection of posets and inside each individual poset we have an antisymmetric relation. This is exactly like saying that in $\bf Grp$ there is no obvious multiplication operation. In general one should not confuse $\bf Poset$ with a poset, like one should not confuse $\bf Grp$ with a group or $\bf Set$ with a set. Marc van Leeuwen is his Feb 19 comment to this question expresses this same point of view.

Indeed one can consider a single poset as a category, same like considering a single set (or monoid or group) as a category. These constructions are simply functors from $\bf Poset$ (or $\bf Set$ or $\bf Mon$ or $\bf Grp$) to $\bf Cat$ the category of small categories. they could be called $\bf Cat$-images. Morphisms in $\bf Poset$ can be considered as morphisms (functors) in $\bf Cat$ between the corresponding $\bf Cat$-images. Same story with preorders and the category $\bf Pre$.

In any category $\bf C$, one can consider the isomorphism relation and so partition the objects of the category into equivalence classes and choose one representative for each class. One can then consider the full subcategory of the representatives and this subcategory is called the (a) skeleton of $\bf C$.

One generally thinks that , if a concept or property is really categorical, then it should be preserved in this skeleton too.A category and its skeleton are examples of equivalent categories. The skeleton of a preorder (considered as a category) is a poset (considered as a category).In general categorical concepts should hold in equivalent categories That's how past and present categorists decide whether a new concept/definition is worth considering.

-
Magma, I edited my Q to clarify the relation between posets, antisymmetry and up-to-isomorphism (I'm aware of the distinction between objects and cats). Please consider rereading it. Thank you for your informed comments on cat. equivalence, but in my opinion it does not answer the question: is equality as a binary relation (as the Wikip. entry states) the only equivalence and order relation? Or is isomorphism as a binary relation also have these properties? –  alancalvitti Mar 24 '13 at 20:18
I knew @alancalvitti that most of what I wrote was probably known to you, but perhaps it will be useful for others. I have reread you question and I am not sure I understand what exactly is your issue here. I covered the linguistic/logic part, I think. So I just want to stress that in mathematics, also in set theory and also in cat theory, it is very clear and distinguishable when two "things" are equal, when they are not equal but they are "isomorphic" and when they are not equal and not isomorphic. Continued.... –  magma Mar 25 '13 at 1:46
Continued. Cat theory is nothing special. It is a theory with its axioms written in first order logic with equality. The equality sign appears directly in the axioms and is necessary in order to define composition and the dom and cod functions. You then define isomorphisms and it just happens that all important properties are conserved by isomorphisms, like all important topological properties are conserved by homeomorpisms (which are - not surprisingly - isomorphisms in $\bf Top$). –  magma Mar 25 '13 at 1:47
While you're correct that equality is necessary to define domain and codomain, it's also the case that equality is viewed as an "evil" relation, unlike isomorphism. In any case that's a soft distinction. The hard distinction is that since 2 objects in a given category may be isomorphic but not equal, the obvious inference is that isomorphism is not the same as equality (as relations). Yet isomorphism seems to satisfy what the Wikip. article on equality claims is a unique property of equality, namely: the only relation which is both an equivalence and an order. –  alancalvitti Mar 25 '13 at 18:15
@alancalvitti aha! Now I understand exactly what is the crux of your question. The isomorphism relation is reflexive, symmetric and transitive, but it is definitely not antisymmetric! So it is not a partial order. So it does not compete with the unique position held by equality, which is instead antisymmetric (besides being symmetric). Conclusion: the Wikipedia article is correct in saying that equality is the only reflexive, symmetric, antisymmetric and transitive relation. –  magma Mar 25 '13 at 19:29

There's just too much confusion in the question. The link you point to uses "isomorphism" as a name for the (equivalence)-relation of mutual dominance in a pre-order; this has very little to do with isomorphisms in category theory, which are morphsims that have an inverse. Equality holds between elements of a set (when doing mathematics) or between morphisms (when doing category thoery), but asking for equality between objects of a category is kind of not-done in category theory (an object is of course equal to itself and not equal to any other object, but this is rather a pointless thing to say). And there is no relation between equality and unique isomorphisms; if an object has nontrivial automorphisms then isomorhisms involving it are never unique (without further qualification).

And yes, using "anti-symmetry", whose definition refers to equality, as a means to characterise the equality relation is quite circular. It is like defining a relation $R$ to be "equimorphous" if $a \mathrel R b$ holds if and only if $a=b$, and saying equality is the unique equimorphous relation. To me trying to characterise equality is a somewhat curious goal in the first place; you cannot well consider sets and relations at all without already knowing about equality.

-
Re: "isomorphism" as a name for the (equivalence)-relation of mutual dominance in a pre-order; this has very little to do with isomorphisms in category theory, which are morphsims that have an inverse." - I disagree w/ your interpretation, since in the category of pre-orders, the arrows are between related pairs, and isomorphic pairs are those related by a pair of antiparallel arrows - so each has an inverse, hence both are iso's (categorically). So the concepts that you say have little in common actually are the same. Can you explain? –  alancalvitti Feb 18 '13 at 17:50
You are confusing a preordered set, which can be viewed as a very particular kind of category in which hom-sets never have more than one element, with the category of preorders, which is an entirely different kettle of fish. Indeed in a preorder-as-category all that is needed for an arrow to be an isomorphism is that the reverse homset is not empty, but this is not a reason to call the relation of having mutual dominance in a pre-order "isomorphism", nor to associate isomorphisms in general with anything involving pre-orders. –  Marc van Leeuwen Feb 18 '13 at 21:18
Yes there's a distinction between a pre-ordered set versus preorder, so let's talk specifically about pre-ordered sets, or even posets where anti-symmetry appears explicitly. What are the isos here? Can you give an example where isomorphism does not correpond to order relation equivalence and/or vice-versa? –  alancalvitti Feb 18 '13 at 21:30
I don't understand the question. Lots of isomorphisms have nothing to do with preorders; the fact that $\def\Z{\Bbb Z}\Z/6Z\cong\Z/2\Z\times\Z/3\Z$ (Chinese remainder theorem) for instance. And conversely the fact that $(6,10)\leq(15,25)\leq(6,10)$ in this answer to another question has nothing to do with any isomorphism. –  Marc van Leeuwen Feb 18 '13 at 22:04
What categories are related to the above examples you give? Obviously the concept of isomorphism is abstract, but I was specifically referring to isomorphism in the category $\bf Poset$. Can you give an example of isomorphism in $\bf Poset$ which is not equivalent to the order equivalence? –  alancalvitti Feb 18 '13 at 23:21