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In Tao's book Analysis 1, he writes:

Thus, from the point of view of logic, we can define equality on a [remark by myself: I think he forgot the word "type of object" here] however we please, so long as it obeys the reflexive, symmetry, and transitive axioms, and it is consistent with all other operations on the class of objects under discussion in the sense that the substitution axiom was true for all of those operations.

Does he mean that, if one wants to define define equality on a specific type of object (like functions, ordered pairs, for example), one has to check that these axioms of equality (he refers to these four axioms of equality as "symmetry", "reflexivity", "transitivity", and "substitution") hold in the sense that one has to prove them? It seems so, because of these two passages:

[In section 3.3 Functions] We observe that functions obey the axiom of substitution: if $x=x'$, then $f(x) = f(x')$ (why?).

(My answer would be "because that's an axiom", but Tao apparently wouldn't accept that.)

And after defining equality of sets ($A=B:\iff \forall x(x\in A\iff x\in B)$), Tao writes (on page 39):

One can easily verify that this notion of equality is reflexive, symmetric, and transitive (Exercise 3.1.1). Observe that if $x\in A$ and $A = B$, then $x\in B$, by Definition 3.1.4. Thus the "is an element of" relation $\in $ obeys the axiom of substitution

So he gives the exercise to prove the axioms of equality for sets. Why does one has to prove axioms? Or, put differently: if one can prove these things, why does he state them as axioms?

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  • $\begingroup$ @MauroALLEGRANZA: Your comment just states the view that one has to check axioms rather than answers my question as to why one has to check/prove axioms. So it doesn't help at all. $\endgroup$ – user7280899 Feb 1 '17 at 20:45
  • $\begingroup$ As I said, Mauro, your comments don't answer my question. I already told you why. Now you are explaining to me that there are to possiblities: first-order logic with equality and without equality. This stuff hasn't something to do with my question. $\endgroup$ – user7280899 Feb 1 '17 at 21:09
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    $\begingroup$ "People think of axioms as laws you have to follow, or true things you have to assume, and I think neither of these perspectives is correct. It's more accurate to think of axioms as a way to agree that we're talking about the same thing." - Qiaochu Yuan $\endgroup$ – Will R Feb 1 '17 at 23:03
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    $\begingroup$ They are axioms of equality; if you have an equality relation then you can assume those axioms about it. However, when you're defining a new relation, you need to prove them in order to know that your relation is an equality relation. $\endgroup$ – immibis Feb 2 '17 at 1:02
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    $\begingroup$ On a sidenote, I don't like to view $A=B\iff \forall x(x\in A\iff x\in B)$ as a definition of $=$ for sets, but instead an axiom about a peculiar property of the $\in$ relation (for example, the $\in$ of set theory differs from "is a parent of" for people, for which the corresponding statement is not true) $\endgroup$ – Hagen von Eitzen Feb 2 '17 at 16:15
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I believe Tao means that the axioms of reflexivity, symmetry, and transitivity are adequate to capture our pre-existing (this is the key) intuition about what "equality" between two objects ought to be. Let me try two contrasting examples to help unpack what I mean.

Version 1

You: Sets $A$ and $B$ are shmequal provided $x \in A \Leftrightarrow x \in B$ for all $x$.

Me: That sounds like a fine relation to investigate. Creative name, by the way.

Version 2

You: Sets $A$ and $B$ are equal provided $x \in A \Leftrightarrow x \in B$ for all $x$.

Me: Now, hold on just a second. By "equal", you mean "identical" or "exactly the same"? I'm not sure I'm ready to accept that this abstract definition captures all that. You would need to show me that the relation $x \in A \Leftrightarrow x \in B$ for all $x$ is reflexive, symmetric, and transitive before I'm willing to concede that this deserves a name like "equal".


Comment from OP

An example came to mind: we define equality for ordered pairs: $(x,y)=(a,b)⟺x=a∧y=b$. To show that equality for ordered pairs is reflexive we need to show $(x,y)=(x,y)$, which by definition means $x=x∧y=y$. But $x$ and $y$ could itself be ordered pairs. So now we are in the situation where we have to prove that every ordered pair is equal to itself but where we also have to accept this as given.

A weak response from me

I struggle to find good words to address your question, but it might help to remember that we are not checking whether $(x,y) = (x,y)$. Rather this is one of the things we insist should be the case if "equality" is to mean anything; it must apply to identical objects. Instead, we are asking "For ordered pairs, does the $x = a \wedge y = b$ property capture this self-evident truth about equality?". We find that it does: $x = x$ and $y = y$ are both true statements because we are comparing two identical objects in each case.

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    $\begingroup$ You forgot the axiom of substitution. And I don't think that one can always show that all of the axioms of equality hold (then one wouldn't need them as axioms!). For me it makes much more sense to say "everything is okay as long as a definition does not contradict the axioms of equality". $\endgroup$ – user7280899 Feb 1 '17 at 20:58
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    $\begingroup$ What do you mean by "contradict the axioms of equality"? $\endgroup$ – Austin Mohr Feb 1 '17 at 20:59
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    $\begingroup$ @user7280899 So one must check whether the definition as stated satisfies the axioms, which is what is meant by "prove the axioms of equality for sets". $\endgroup$ – Austin Mohr Feb 1 '17 at 21:01
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    $\begingroup$ @user7280899 Yes; you've defined equality of pairs (of something) in terms of equality of these somethings. For equality of pairs to be well-defined, the equality of somethings had better be well-defined to begin with. $\endgroup$ – Servaes Feb 2 '17 at 7:03
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    $\begingroup$ If the elements of the pairs are pairs themselves, they necessarily have a lower "nesting level", so by recursively applying the definition again,you eventually arrive at comparing non-pairs. $\endgroup$ – celtschk Feb 2 '17 at 7:39
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You are probably confused because you think that axioms are (by definition) statements that we take as true without any proofs. However, this word has a slightly different meaning.

Axioms are a starting point of a mathematical theory. When you build a theory, for example, Arithmetic, from scratch, you need some preliminary facts, otherwise you cannot prove anything. In Arithmetic and a bunch of other mathematical theories the described properties of equality are indeed axioms, that one does not prove. Equality is a primitive notion, and the only sensible way to actually define it is to postulate that these natural (as it seems to us humans) properties hold.

However, in set theory, these "axioms" are not the definition of equality. Rather, equality is defined via the formula above: two sets are equal when they consist of identical elements. But when we define equality in this way, there is a natural question: why we are naming this as "equality" at all? This is why we prove "axioms of equality", which we are already used to, to show that the naming "equality" is adequate. And when we prove them, they become theorems of set theory and properties of equality rather than axioms. This is because set theory is more fundamental and more powerful than most mathematical theories in a sense that you can build (almost) all mathematics based on it.

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  • $\begingroup$ In the book, Tao isn't only talking about the equality relation on sets, but also on functions, ordered pairs, and so on. $\endgroup$ – user7280899 Feb 1 '17 at 21:02
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    $\begingroup$ @user7280899 In those other cases, "equal" would be defined in a different way, and the burden would be on the person proposing the definition to show that it satisfies reflexivity, symmetry, and transitivity (and thereby deserves to be called "equal"). $\endgroup$ – Austin Mohr Feb 1 '17 at 21:04
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    $\begingroup$ @user7280899 Yes, and the functions, pairs and so on are also some objects that you expect to behave in a certain way if you studied some mathematics before. So when we define them in terms of sets, we need to prove that they behave exactly how we expect. When such a situation appears, it is common to call expected properties "axioms", despite the fact that they are not actually the axioms of the set theory. $\endgroup$ – Wolfram Feb 1 '17 at 21:07
  • $\begingroup$ @Wolfram: Your question assumes that a pure set theory where one encodes every mathematical object as a set is the only right way to go. But that hasn't anything to do with my question. $\endgroup$ – user7280899 Feb 1 '17 at 21:13
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    $\begingroup$ That's not the only right way to go. F.e., the other possible way that is in active development today (and it's considered even more powerful than set theory) is Homotopy Type Theory. But when we define equality, functions, ordered pairs etc. in any fundamental theory, we then usually prove that they satisfy these axioms. Because we could actually define them in a way they didn't satisfy them. And if it is the case, then the naming is inappropriate (even when the definitions are sensible). This is more a philosophical question than mathematical… So I do not pretend I explained it well. $\endgroup$ – Wolfram Feb 1 '17 at 21:22
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There are axioms and then there axioms. Most of the time mathematicians use the word "axiom" they mean it in a definitional sense. Instead of "An equality is a relation satisfying the reflexivity, symmetry, transitivity, and substitutivity axioms" think instead "By definition, equality is any relation for which reflexivity, symmetry, transitivity, and substitutivity hold". In other words, you can call some relation an equality if you can show that it meets the definition, i.e. that reflexivity, symmetry, transitivity, and substitutivity are true of it. So the answer to your first question is "yes". To put it another way, these "axioms" are true of equality relations, but you have to show that your relation is in fact an equality relation. This is exactly the same situation as the definition of a mathematical group for instance.

To put it an even better way, these "axioms" are axioms in the "theory of equality" and you want to show that a particular relation is a model/interpretation/semantics for that theory. Very briefly and roughly sketching, in formal logic, a theory is a collection of symbols and a collection of rules. A theory will define certain arrangements of symbols to be formulas (or sentences or terms). There will also be a notion of "theoremhood" defined by the rules. Typically the rules will have the form "if these arrangements of symbols are theorems, then this arrangement of symbols is a theorem". You could call these rules "axioms", though in this context usually only rules with no premises, i.e. that simply baldly state "this arrangement of symbols is a theorem" with no conditions, are called "axioms". However, the "axiom of symmetry", say, corresponds more closely to a rule than an axiom in this stricter sense. A theory (and the logic in which it is formulated) thus gives rise to a language.

Of course, we usually want to talk about circles and torii and other mathematical objects that we don't (usually...) think of a "arrangements of symbols". To connect a theory to some mathematical objects we use a semantics which is an assignation of mathematical objects to the arrangements of symbols in a consistent manner (usually satisfying some conditions dependent on the logic in which the theory is formulated) such that the rules are satisfied. If the rules aren't satisfied, then the assignment is not a semantics for the theory.

So Tao is (implicitly) specifying a theory of equality and these exercises are asking you to show that particular interpretations (i.e. assignments) are semantics for that theory.

So how does this relate to set theory as the "foundation" of mathematics or axioms as "self evident truths"? As far as "foundations" are concerned the situation is that mathematicians for the most part have agreed to (pretend to) work within the language of Zermelo-Fraenkel (ZF) set theory which is a theory in first-order logic. There is no question of "true" or "false" in this scenario. We simply have some formulas that are called theorems and rules for making more theorems. The rules/axioms simply define what a "theorem" is. The axioms in the stricter sense are then the starting points for deriving theorems as Wolfram said. However, we can talk about semantics for ZF set theory and to show that an assignment is a semantics we would indeed be obliged to prove the "axiom of pairing" and the "axiom of infinity" and all the other axioms of ZF set theory hold for our interpretation. This is something that is done in set theory and logic.

Pragmatically, as I stated in the first paragraph, you should just interpret "axiom" in this and most cases as "condition that needs to hold to meet the definition". The rest of this answer was more explaining how this usage of the word "axiom" is, in fact, more or less consistent with the usage in e.g. "axioms of set theory" or "axioms of geometry".

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  • $\begingroup$ Thanks. Seems to be similar to the answer of Austin Mohr. Can you answer my last question in the comments of his answer? $\endgroup$ – user7280899 Feb 2 '17 at 1:04
  • $\begingroup$ I assume you mean the comment about ordered pairs and the following one about recursion. Equality is not the best example for all this since usually equality is part of the logic rather than the theory within the logic. Furthermore, to the extent that we're implicitly working in the language of set theory, there's a built-in notion of equality. (Equality is also trickier than it seems.) Let's just talk about an equivalence relation. It's easier to see what's happening in that context. If $\sim_X$ is an equivalence relation on $X$ and similarly for $\sim_Y$, ... $\endgroup$ – Derek Elkins Feb 2 '17 at 2:08
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    $\begingroup$ ... then we can make an equivalence relation on $X\times Y$ via $$(x_1,y_1)\sim(x_2,y_2)\iff x_1\sim_X x_2\land y_1\sim_Y y_2$$ So indeed to define this equivalence relation on ordered pairs we need to be handed equivalence relations on the components. The fact that $X$ or $Y$ might themselves be cartesian products does not make this definition recursive. In fact, maybe the framework in which we're working doesn't allow cartesian products of cartesian products. It makes no difference to this definition. So any "recursion" would be coming from the framework within which we're working. $\endgroup$ – Derek Elkins Feb 2 '17 at 2:08
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I've personally come to think of axioms as little components of a big definition. For example, the axioms at the start of the Elements define what we mean by "Euclidean geometry"; the Peano axioms serve to define what we mean by "natural number", and so forth.

This gives a somewhat cleaner style than a definition that goes on for a paragraph or two with lots of conjunctions. And it serves to make the parts distinct, so that we can study the effects of maybe swapping out or changing one of them and keeping the rest the same.

So Tao is listing all the sub-parts in the definition of what we mean by "equality". What equality means is assumed in advance by these axioms. Now the question is: Does a proposed new relation really count as equality, or not? That needs to be established by proving it meets all of the criteria, that is, all of the component axioms of the definition.

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So he gives the exercise to prove the axioms of equality for sets. Why does one has to prove axioms? Or, put differently: if one can prove these things, why does he state them as axioms? $ \def\imp{\Rightarrow} \def\eq{\Leftrightarrow} $

He does not prove the axiom as stated. The axiom claims equality between two sets iff they have exactly the same members. This equality symbol "$=$" is the symbol in the foundational system itself, and there is no way you will be able to prove an axiom of the foundational system if it is independent of the other axioms. In fact, the equality symbol is part of first-order logic itself. So what exactly does Terence Tao mean?

He wrote:

One can easily verify that this notion of equality is reflexive, symmetric, and transitive (Exercise 3.1.1). Observe that if $x∈A$ and $A=B$, then $x∈B$, by Definition 3.1.4. Thus the "is an element of" relation $∈$ obeys the axiom of substitution.

This is logically not precise unless he is working in first-order logic without equality. It would be clearer to define the binary relation $\sim$ such that $A \sim B$ iff $\forall x\ ( x \in A \eq x \in B )$. Then it makes sense to ask whether $\sim$ is an equivalence relation or not, and whether it obeys substitution. It is trivial to prove that it is indeed symmetric, reflexive and transitive. In a formal system, we should think of the notion that a relation $\sim$ obeys substitution as meaning that $x \sim y$ implies that $P(x) \eq P(y)$ for any $1$-input sentence $P$. It turns out that in a first-order language we do not need to check for every $1$-input sentence, because it suffices to check for each of the non-logical symbols of the language. Since the language of set theory only has one non-logical symbol "$\in$", the precise notion of "obeys substitution" in set theory is hence:

$\forall A,B\ ( A \sim B \imp \forall C\ ( C \in A \eq C \in B ) \land \forall C\ ( A \in C \eq B \in C ) )$.

Observe that the first half of the claim is trivially true by definition of $\sim$. I don't see the second half in the quotes of Terence Tao that you have in your question. If I'm not mistaken, it is not provable, in which case Terence didn't really prove full substitutivity.

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Austin Mohr's answer is excellent; however, as it led to an extensive argument in the comments I wish to put forth a slight simplification and restatement of it which may add something.

(I posted this as a comment originally, but wanted to expand on it.)


So he gives the exercise to prove the axioms of equality for sets. Why does one has to prove axioms? Or, put differently: if one can prove these things, why does he state them as axioms?

The basic point to realize here is that English words have meaning outside of math.

You're free to invent any relation you wish, with whatever properties you wish. You can label it with whatever made-up term you want without any necessity to prove anything about it.

If you do this, you are just defining what you are talking about, and you can then proceed to say something using your stated definitions.

However, if you use an English word to name or describe your relation, then you should take into account the English meaning of the word. That is, in choosing an English word as a name, you should choose one which aligns with the properties your relation has.

And the corollary: If you want to use a particular English word as a name for your relation, you should be sure your relation has the properties that would be implied by that name.

This applies whether we are naming relations, operators, or anything else.


If I define a unary operator and call it the "inverse," but my operator has the property that repeated application of that operator will never produce the original input, I have misnamed it.

If I define a relation and call it the "equality relation," but it is not transitive nor symmetric, I have again chosen the wrong name.

(Can you misname your relations and operators? Of course you can. The only thing it will break down is your communication with other people, which is of course the only reason to have names for things in the first place.)

A relation which is not reflexive, symmetric, and transitive, will violate the English-language meaning of the word "equal," so if your relation may not have those properties then use another name for it instead.


In this case, the "equality relation" between sets has been defined in your textbook, and it is now your task to show the appropriateness of the label "equality relation" for the defined relation, by showing that it has the properties one would expect out of the English-language meaning of the word "equality."

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  • $\begingroup$ I agree very much with the sentiment in your post. But in the history of mathematics even up until today, mathematicians have always named mathematical objects rather randomly. Nobody can possibly guess the meaning of "monoid" and "group" and "ring" and "field" without knowing their definitions! Adjectives are equally opaque. For example there is nothing normal about a "normal subgroup" (it's actually special), and there are numbers that are perfect, amicable, sociable, lucky, polite, practical, sublime, fortunate, betrothed, cake... Don't forget things named after people... =) $\endgroup$ – user21820 Feb 3 '17 at 5:31
  • $\begingroup$ @user21820, okay. How about: line, point, angle (see etymology), triangle, rectangle, greater, lesser, equal (again, see etymology), set, function (etymology), empty, solid, composite, prime, divide, multiply, add, subtract, circumference, logarithm (etymology), differential, integral, tangent...I could go on. Yes, you can pick abstruse words that no one would know besides a mathematician and that are named arbitrarily. I think those are the exception, not the rule. The fact that there are more of them is beside the point; they're not actually used as often as the words I've named. :) $\endgroup$ – Wildcard Feb 3 '17 at 8:03
  • $\begingroup$ Yes agreed agreed. There are more useless terms than useful terms. It's just amusing to see that humans like one-word labels so much that they will just use all words available. =) Side-note: I don't know where the terms "logarithm" and "integral" came from; they're just labels to me haha.. $\endgroup$ – user21820 Feb 3 '17 at 8:07
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The philosophy is that axioms need to be as simple as possible, basic, and few in number. In this way we can argue about the validity of such axioms very little. In contrast, look into the history of the parallel postulate. The axioms you mention have immediate consequences; however, if we do not prove those consequences they must be accepted as part of the axiom, and hence the axiom is more complex than necessary.

For instance,

We observe that functions obey the axiom of substitution: if $x=x′$, then $f(x)=f(x′)$ (why?).

does not follow from the equality axiom alone, but also from the definition of function. Note that it does not hold for all relations, such as $r(x)=\pm \sqrt x$. Therefore, to be minimalist in axioms, we must prove that substitution holds for functions. This is where set theory and mathematics logic become quite powerful, and we undertake these simplistic problems to cut our teeth.

Likewise,

One can easily verify that this notion of equality is reflexive, symmetric, and transitive (Exercise 3.1.1). Observe that if $x\in A$ and $A=B$, then $x\in B$, by Definition 3.1.4.

is not stated in the axiom, so must either be included in the axiom or proven. It looks like, and feels like, it is part of the axiom, and the proof is almost "because the axiom says so," but the minimalist approach demands the proof.

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