# Can definition be existence (in mathematics)?

The set omega, as the comment in this question points out, can be defined as the smallest set that is closed under succession and includes the empty set.

This is enough to define it uniquely, but to find it, that is to prove that it it exists, one must use the first axiom of infinity in ZFC.

Thus it appears that definition is not existence.

But, this doesn't seem quite right. For surely the axiom of infinity was introduced into ZFC such that an argument as shown above can have existential import.

In this line of thought, it appears that definition is enough for existence.

Is this right?

I would say not, and point to our coherent understanding and experience of the mathematical infinite that is condensed dually into definition and axiom.

Is this right as well?

(This, I think, appears to have certain tangential relationships with the various forms of the ontological arguments of God, which I don't want to go into here).

(It also appears to have some affinity with the notion of essence preceding existence).

• I'm sorry, I don't understand the question. Is the question: `Does the fact that some entity can be defined entail that that entity exists?` If so, the answer is no. A unicorn is a horse with a horn, but there aren't any such things. Or perhaps the question is: `Are there any things which, by definition, exist?` There the answer is less clear. Anselm, for instance, famously argued that God was such a being, but his proof is controversial--as I'm sure you're aware.
– user5172
May 29, 2014 at 12:38
• I didn't say this in the question, but the scope of it is within mathematics - so this would rule your fictional unicorn out. I'll edit the question accordingly. May 29, 2014 at 13:27
• and as for you second comment, sure - thats why added a brief note on the notological proof of God, as well as essence preceding existence. May 29, 2014 at 19:18

In normal math (above the level of set theory) it's standard procedure to define an object, and then show that at least one such object exists.

For example, we define a Group as a set with a binary operation satisfying such-and-so. Then the very next thing we do is to give some examples of groups: the integers under addition, the nonzero reals under multiplication, the symmetries of a bed. The maids at the Hilbert Hotel are familiar with that one.

That's the entire purpose, for example, of the Dedekind cut construction of the real numbers. Nobody ever uses it. You see it once and you forget about it. Why is it important?

It's important because if we define the real numbers as an Archimedean complete ordered field, you can do all of real analysis with those axioms. They're all you need.

But you could build all of real analysis out of those axioms, and you wouldn't know if there is any such set. So just once, you go back to basics and show how, given the axioms of set theory, you can construct a set having those properties.

Once you do that, you can then just use the real number axioms. You never again care about the existence proof.

But it is essential to supply an existence proof along with a definition. I could define a purpleunicoren as a unicorn that's purple. I could describe the biology of a purpleunicorn, and write books on breeding purpleunicorns. But it would all be vacuous. There is no object that fits the definition.

Existence proofs are mandatory.

• I'm not sure to understand your argument. 1) You say: "start with the axioms for Archimedean complete ordered field [call it: A-axioms], and with them develop real analysis." But when you say: "and you wouldn't know if there is any such set"; are you meaning : the reals? But the reals are the elements in the domain of every model of the axioms for Archimedean complete ordered fields. 2) When you say: "you go back to basics and show how, given the axioms of set theory, you can construct a set [I assume : satisfying the A-axioms] having those properties" you are proving that ... 1/2 May 30, 2014 at 16:31
• ... in the domain of every model satisying the axioms of set theory there are objects satisfying the A-axioms. What is the "essential" difference ? Why we are licensed to "never again care about the existence proof" ? We have "reduced" the existence of reals to that of sets, and this is a enormous step ahead for math, but from an "ontological" or "epistemological" poit of view, it seems to me that we have made no real progress. Having defined reals as "special sets" are we dispensed with an "existence" proof ?2/2 May 30, 2014 at 16:32
• @Mauro ALLEGRANZA It's my purpleunicorn example. I can write down the axioms for the reals. But what if there is no such object satisfying those axioms? Then anything I prove about the reals is vacuous. How do you know there is ANY mathematical object satisfying those axioms? We need an existence proof. Of course we have not shown absolute existence ... just existence relative to the axioms of set theory. For working mathematicians, that's sufficient. If we believe in sets then we are justified in believing in the reals. May 30, 2014 at 16:58
• @MauroALLEGRANZA it's pretty simple: you can write all the definition you want, it doesn't mean that there is even one object that matches your definition. So, you take the axiomatic definition of eg a new structure you want to define, and you use it to prove you have at least one such object. Existence is not guaranteed just because you have a definition. Once you have shown existence, you may use other equivalent definitions if more convenient. Feb 20, 2023 at 16:29

Set theories need not postulate the a priori existence of any objects or structures. ZFC does, however, postulate the existence the empty set (it's zero) and a kind of successor function based on the empty set as a starting point. The resulting set could have infinitely many junk terms that need to selected out using the Separation (Subset) Axiom, leaving only the set of natural numbers, i.e. a subset that satisfies Peano's axioms.

You could also simply postulate the existence of some Dedekind-infinite set outside of set theory -- not a huge leap of faith. (If such a set does not exist, the universe would be a very dull place indeed.) Then you can extract a subset from it that satisfies PA.

Having shown the existence of a set that satisfies PA by one of these means, you would be quite justified in beginning your development of number theory and analysis by simply defining the natural numbers using PA.

• I do not agree with "Set theories need not postulate the a priori existence of any objects or structures." In modern term, we interpret a theory with a "universe of discourse" which has a non-empty domain. Thus, if ZFC is consistent it has a model with "properties" which satisfy ZFC's axioms. Thus the universe of those models must contain an "emptyset", an infinite set, and so on. The basic stuff of those models are the sets. May 30, 2014 at 9:02
• @MauroALLEGRANZA Too many unnecessary levels of abstraction? I believe you can have a workable set theory as I describe. It is the basis for the simplified version of ZFC that I use in my DC Proof program. With it, I can say with some confidence that you can probably construct the real numbers using the Peano axioms as an initial premise. (A huge project requiring 10's of thousands of lines of formal proof, of which I have completed perhaps 3/4.) Jun 2, 2014 at 14:44

In logic and mathematics : NO

Consider the well-known Russell's Paradox : we (try to) define a set R satisfying a certain condition, only to prove a contardiction. Conclusion : that set do not exists.

Consider a "paradigmatic" case like the emptyset in ZF.

First we prove that a set with no elements exists, then we add to the language of the theory a new symbol (an individual constant) denoting it.

All "ontological" conclusions are apt to you ...

• The key problem here though is inconsistency. If you use paraconsistent logic, you can in fact define such a set - and also prove things about it. I think, the question I'm asking, lies somewhere in the philosophy of formalism. May 29, 2014 at 13:25
• @MoziburUllah - I think taht you are "eluding" the problem. As far as I know, paraconsistent set theory can prove the existence of Russell's set; but it is still interested into the existence of models (see e.g.Greg Restall, A Note on Naive Set Theory in LP (1992) or Ross Brady, The simple consistency of a set theory based on the logic CSQ (1983), both in NDJouFormLog). We can dissociate consistency from existence of a model, but still the problem remain. When in ZF we state the axiom of infinity, we are not creating by fiat an infinite set. 1/2 May 29, 2014 at 13:53
• We are only restricting the class of possible models to those with an infinite domain. But still we ask : are there such models ? The definition of infinite set is not enough to conclude with existence of an infinite set. 2/2 May 29, 2014 at 13:55