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When I am learning, one thing I am puzzled is the definition. For example, we define $0$ as $\emptyset$. But when we use set language, how could we know we are talking about $0$ or the empty set. Another example is the definition of order pair $\{\{a\},\{a,b\}\}$, how do we know we are talking about order pair or just a set with this structure.

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If you define $(a,b)$ to mean $\{\{a\},\{a,b\}\}$ then the ordered pair and the set with that structure are the same thing. – Henry Jul 9 '14 at 8:29
As always in language, context is everything. – Henno Brandsma Jul 9 '14 at 8:31
A priori, "$0$" is not a set. To use that symbol in a set theoretic context it must be given a definition in terms of sets. Which set would you define it to be? For the latter, your use of "or" is misleading -- it is most useful to remember that we are talking about both at once so that any tool useful for either is available. – Eric Towers Jul 9 '14 at 14:27
up vote 23 down vote accepted

In ordinary everyday mathematics, a number and a set are different kinds of things that follow different kinds of rules. Similarly an ordered pair and a set are different kinds of things.

It doesn't make sense to ask what the elements of a number is. And it doesn't make sense to ask what the cosine of $\varnothing$ is -- or rather, it does make some sense to apply a function to a set of numbers rather than a single number, but we get different results. So we get $$ \cos(0) = 1 \qquad\text{whereas}\qquad \cos(\varnothing)=\varnothing $$

In everyday mathematics it doesn't even make sense to ask whether $0$ and $\varnothing$ are equal, because they are different kinds of thing -- claims such as $0=\varnothing$ or $0\ne\varnothing$ are neither true or false, they are just confused.

Similarly, writing down the set $\{0,\varnothing\}$ in ordinary mathematics would usually be a sign of confused thinking. We don't want to put different kind of things into the same set, because then the meaning of set operations such as $\{0,\varnothing\}\setminus\{0\}$ would begin to depend on things like $0=\varnothing$ which are themselves ill-typed claims. Likewise, we don't even ask whether $\varnothing\in\{0,1,2\}$, because the truth of that again depends on the ill-typed $0=\varnothing$.

We know whether we're talking about one or the other because they are different things, and its up to the mathematician to be sure which kind of things he's speaking about at any given time.

The business about using 0 as an abbreviation for $\varnothing$ and $(a,b)$ as an abbreviation of $\{\{a\},\{a,b\}\}$ comes in at one very specific point, namely when we're using axiomatic set theory to model the rest of mathematics.

In that very specific situation we deliberately choose to forget the difference between numbers and sets and pairs (etc.) in order to gain the technical advantage that it's much simpler when our axioms only have to speak about one kind of things (namely sets), than if we had to use new axioms for each kind of things mathematics speaks about.

This advantage is an advantage for set theory, however. It's not at all helpful outside set theory, in the ordinary mathematics we're modeling -- so it would be counterproductive to insist in ordinary mathematics that $0$ is interchangeable with $\varnothing$.

The fact that in the set-theoretic formalization of mathematics $(0,1)$ and $\{1,2\}$ happen to be represented by the same object (namely the set $\{\{\{\}\},\{\{\},\{\{\}\}\}$) is a property of the set-theoretic formalization, not a property of the underlying mathematics that is being modeled. When doing ordinary mathematics there is no reason to care about how such things work.

It is good for a mathematician to know how the set-theoretic formalization of things works, if only because "arguments that can be modeled in standard set theory" agrees pretty well with "arguments that most mathematicians agree are valid mainstream arguments", and it is easier to verify that the argument you have in mind can be formalized in set theory than to go out and poll a lot of mathematicians.

However, as long as you're confident that what you're doing is mainstream, and you're keeping different kinds of things separate, there's absolutely no reason to get hung up on the technical details of the possible set-theoretic representations of things.

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Very good answer. – Asaf Karagila Jul 9 '14 at 12:30
I know it is frowned upon to make too many "great stuff!" comments -- but this, as Asaf has already said, is great stuff. – Peter Smith Jul 9 '14 at 13:12
@StevenStadnicki: What I'm saying is that it's not uncommon to use the notation $f(A)$ to mean $\{f(x)\mid x\in A\}$ when $A$ is a subset of the domain of $f$. If one uses that notation, $f(\varnothing)$ does get a meaning. But of course nobody says you have to use it. – Henning Makholm Jul 9 '14 at 18:14
@DougSpoonwood "A number is after all a nullary function, and functions are collections of n-tuples" Really? Really??? Well, to make just a local point: these claims are just the kind of ill-typed offences against the working mathematician's typing discipline that Henning Makholm is complaining about (and lots of mathmos round my neck of the woods would echo him). – Peter Smith Jul 10 '14 at 13:58
@DougSpoonwood "Iif we have a function f such that f(0)=1, and f(1)=0, then the elements of the function ..." That assumes functions are sets. Well, many -- following knowingly or unknowingly in the footsteps of Great-Uncle Frege -- will insist that that is already a mistake. Functions can be modelled inside set-theory by sets (well, there is nothing else to play with there); but that doesn't mean that they are sets. – Peter Smith Jul 10 '14 at 14:30

Since $0=\emptyset$, there is no difference in talking about $0$ and talking about the empty set. When you are talking about the empty set, you are talking about $0$, and when you are talking about $0$, you are talking about the empty set.

Your question is puzzling. Basically, it's as if I would ask: "I know Barack Obama is the first black president of USA, but how could I know if we are talking about Obama or about the first black president of USA."

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Sorry for the downvote, but I believe the poster is asking about the difference between how set theory is defined and how it is used. He clearly considers $0 = \emptyset$ to be unsound and is asking how logically to reconcile the way set theory is defined and how it is used. To say that it is puzzling that the poster doesn't concede that $0 = \emptyset$ comes across as dismissing the poster's valid question to me, although he may feel differently. – DanielV Jul 9 '14 at 12:47
You may be right, but since the poster is obviously not responsive, we will never know – 5xum Jul 9 '14 at 13:59
I have not seen foundations of maths text books claim that 0 = ∅, or even that using 0 is an abbreviation for ∅. Rather 0 is defined as the cardinality of the empty set ∅. – MattClarke Jul 10 '14 at 1:35
@Matt: Have you seen any books about modern set theory? – Asaf Karagila Jul 10 '14 at 10:36
Yeah, my mistake. It's been too long since I read that stuff and memory has failed me once again. – MattClarke Jul 10 '14 at 22:15

The other answerers more or less say the question is dumb or irrelevant. I dont think that is true, and Paul Halmos in Naive Set Theory (which the OP could read) has this to say:

"It is easy to locate the source of the mistrust and suspicion that many mathematicians feel toward the explicit definition of ordered pair given above. The trouble is not that there is anything wrong or anything missing; the relevant properties of the concept we have defined are all correct (that is, in accord with the demands of intuition) and all the correct properties are present. The trouble is that the concept has some irrelevant properties that are accidental and distracting. The theorem that $(a, b) = (x, y)$ if and only if $a = x$ and $b = y$ is the sort of thing we expect to learn about ordered pairs. The fact that $\{ a, 6 \} \in (a, 6)$, on the other hand, seems accidental; it is a freak property of the definition rather than an intrinsic property of the concept.

The charge of artificiality is true, but it is not too high a price to pay for conceptual economy. The concept of an ordered pair could have been introduced as an additional primitive, axiomatically endowed with just the right properties, no more and no less. In some theories this is done. The mathematician's choice is between having to remember a few more axioms and having to forget a few accidental facts; the choice is pretty clearly a matter of taste. Similar choices occur frequently in mathematics; in this book, for instance, we shall encounter them again in connection with the definitions of numbers of various kinds. "

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I don't say that. Not more and not less. I'd be happy if you wouldn't put words in my mouth, especially when these words are mild insults like dumb. – Asaf Karagila Jul 9 '14 at 12:23

When you talk about the real numbers, how do you know if you are talking about an ordered field, or about a topological space?

It's the context. Whether or not you are talking about $0$ or about $\varnothing$ is irrelevant, both are the same objects and have the same properties. One "unfortunate" result is that the following is true, $\langle 0,1\rangle\subseteq 3$, or even worse, $3-1=\langle0,1\rangle$.

But context dictates that the $-$ sign is set difference, and not ordinal subtraction. And that's why it's better to use $\setminus$ anyway for that.

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When you define a theory, you don't want it to have contradictions in it. One way to reduce the chance of having a contradiction (or at least make it more believable that it is consistent) is to reduce the number of assumptions that you make. So if you can, you want to only define axioms relating to sets, rather than axioms relating to sets and natural numbers and ordered sets and everything else all at the same time.

So suppose we define:

  • $S$ is the assumptions relating to sets
  • $O$ are the assumptions relating to ordered sets
  • $N$ are the assumptions relating to natural numbers (such as peano arithmetic)

If you assume $S$, $O$, $N$, and everything else you want to discuss all at the same time then you are asking for trouble. You might get away with it on pencil and paper, but a formal logic like on a computer would probably discover a paradox.

So we don't do that.

We use the following important property in logic about consistency:

If $A$ and $B$ are sets of propositions, and $A \vdash B$, then $B$ is no less consistent than $A$

Introduce definitions $O_\text{def}$ and $N_\text{def}$ which define ordered sets and natural numbers in terms of sets. $0 = \emptyset$ would be an example of a definition in $N_\text{def}$. These only contain definitions, and not axioms. Then, if we can establish

$S, O_\text{def}, N_\text{def} \vdash O, N$

then we can conclude that

$N$ and $O$ are no less consistent than $S$, $N_\text{def}$, and $O_\text{def}$

But since definitions are vacuous (their consistency is considered self evident as they are just assigning names), we can conclude

$N$ and $O$ are no less consistent than $S$

We can establish that natural number theory etc are just as consistent as set theory. Furthermore, we only have to examine the consistency of set theory on it's own without being concerned about the axioms of the implied theories. And we don't need to keep $0 = \emptyset$ for the sake of maintaining the consistency result. $O_\text{def}$ and $N_\text{def}$ are just temporary definitions used to prove $S, O_\text{def}, N_\text{def} \vdash O, N$, the consistency result holds even we don't keep $0 = \emptyset$ as an assumption.

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