Some axiomatic set theories formulates the system by taking an axiom 'There exists a non empty set'. I am not clear how the statement become an axiom since after all the definition of a 'set', we have lot of examples like 'the set of all students in the class having more than 50% marks in a particular exam', etc. What exactly mean by the 'existence' of such a set (example)? Why it can't make the statement 'There exists a non empty set' as a proposition?

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    $\begingroup$ In the axiomatic approach we pretend that we only know what is given by the axioms. $\endgroup$
    – littleO
    Jul 1 at 5:51
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    $\begingroup$ @MessiLio The axioms are agnostic to what "$x$ is a set" and "$x$ is an element of $y$" even mean. You can apply the theory to any definitions of those two concepts which follow the axioms. You definitely don't need the definitions to be proposed before stating the axioms. $\endgroup$ Jul 1 at 6:01
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    $\begingroup$ Which axiomatic theory are you referencing in particular? $\endgroup$ Jul 1 at 6:04
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    $\begingroup$ en.wikipedia.org/wiki/… See the axioms listed here, the article on ZFC. ZFC is the standard formal approach to sets. No axiom as you mention. It's actually not necessary. There is the axiom of infinity, but this is not necessary to derive to existence of a non empty set. $\endgroup$ Jul 1 at 6:06
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    $\begingroup$ @IsAdisplayName I'm unsure I understand your intent. If you mean you could replace the axiom of infinity with, say, the axiom of the empty set to derive the existence of a non-empty set, then I agree. If you mean you could derive the existence of a non-empty set in ZFC minus the axiom of infinity (which is nearly what it sounds like you're saying) then that's wrong. $\endgroup$ Jul 1 at 6:19

2 Answers 2


In Mathematical Induction, we Prove some $P(0)$ ; then Prove $P(n) \implies P(n+1)$ , then we claim $P(n)$ is always true.
This whole thing will collapse if we do not Prove $P(0)$.

Similarly in Set Theory, we should start by claiming that there is a set with 1 element, else we maybe talking about nothing.

Here is one such "unreal" theory :

Let M be the set of humans born on Mars.
Let N be the set of humans born on Neptune.
<< More Axioms here >>

Theorem: Every human in M is also in N. [[ Proof left as Exercise ! ]]
Theorem: Humans can be born in 2 Planets. [[ Proof : Every human in M (was born on Mars) is also in N (was born on Neptune) thus was born on 2 Planets ! ]]

Obviously, these are meaningless.

The Core Issue is when we made the Axioms, we did not claim that M & N have atleast 1 element.

Likewise, we can make Axioms about Natural Numbers, but to ensure that we are talking about something (rather than nothing), we have to make the Axiom that 1 is a Natural Number.

In Axiomatic Set Theory, we have to make the Axiom about non-empty set, otherwise, we make be talking about nothing (rather than something) and make all sorts of meaningless theorems.

Some Demonstrative Material :

[A] Empty domains

The definition above requires that the domain of discourse of any interpretation must be nonempty. There are settings, such as inclusive logic, where empty domains are permitted.

[B] Empty domain

In first-order logic the empty domain is the empty set having no members. In traditional and classical logic domains are restrictedly non-empty in order that certain theorems be valid.

[C] Free logic

A free logic is a logic with fewer existential presuppositions than classical logic. Free logics may allow for terms that do not denote any object. Free logics may also allow models that have an empty domain. A free logic with the latter property is an inclusive logic.

[[ The last reference has one Example similar to my "Mars+Neptune" Example, where "Something is Pegasus" is Proved ! ]]

  • $\begingroup$ Sir, I didn't get the proof of 1st theorem, while we are dealing with the sets $M$ and $N$ in which the number of elements has not been mentioned. $\endgroup$
    – Messi Lio
    Jul 1 at 10:17
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    $\begingroup$ Nobody was born on Mars : $M=\{ \}$ ; Nobody was born on Neptune : $N=\{ \}$ ; It is true that every element of M is an element of N. If you say that is not true, can you show me 1 element of M which is not an element of N ? Now, if the elements of M are elements of N, those elements must be born on both Planets !! This meaningless conclusion occurs because here Domain is EMPTY SET !! $\endgroup$
    – Prem
    Jul 1 at 10:25
  • $\begingroup$ You say "we should start by claiming that there is a set with 1 element, else we maybe talking about nothing". But this is not necessary. Even if we allow empty domains of discourse, we only need to claim that there exists some set in order for any model of the theory to be non-empty. In fact, asserting the existence of the empty set will suffice. You seem to be confusing the existence of a non-empty set with the model of the theory being non-empty. And even so, we can derive the existence of a non-empty set from the existence of the empty set. So I'm not sure I understand your answer. $\endgroup$ Jul 1 at 23:41
  • $\begingroup$ Additionally, I think your Mars-Neptune example is misleading. What causes the unintuitive result is not an empty domain of discourse, but rather that the sets M and N themselves are empty. This again leads me to believe you are confusing a set being empty with the domain of discourse being empty. Sure, we get the problem you illustrate with an empty domain of discourse, but we get the exact same problem in a non-empty domain of discourse for any two predicates with null extension. E.g. $\{x \in A : x \in x\} = \{x \in A : x \subset x\}$, where $A$ is some non-empty set in ZF. $\endgroup$ Jul 1 at 23:44
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    $\begingroup$ +1. If we do not have an Existence Axiom $\exists x(x=x)$ then all the other axioms of ZFC -Inf begin with $\forall$ and it is impossible to logically deduce any sentence beginning $\exists$ from them. $\endgroup$ Jul 2 at 12:40

Based on discussion in the comments, I think its worth writing this out as an answer. I should clarify, I interpret OP's question as about standard ZF(C). Since the only ZF axiom of the form OP describes is the axiom of infinity, I will focus on that. I won't worry about ZF formulated in a non classical logic.

In short, You do not need an axiom as you describe to derive the existence of some non-empty set (in standard ZF). We add the axiom of infinity so that we ensure the existence of an infinite set.

First, to clarify your confusion. Axiomatic set theories do not have sets like "the set of all students...". In ZF, the only objects are sets. So every set, except the empty one, is a set of sets. Some axiomatic theories have "urelements", which are not sets but are atom-like. Even so, you will not get a "set of all students..." as you described.

Most approaches to model theory of FOL (first order logic) stipulate that the domain of discourse is non-empty, by definition. See definition 1.1.2 of Marker's "Model Theory: An Introduction" (a standard text on the subject). Notice that FOL is independent of ZF(C). So this happens before we worry about axiomatizing ZF(C). This is not adding an axiom to ZF(C). This has to do with the definition of a model of FOL

Since ZFC lives on top of FOL, the same is true of models of ZF. Thus, empty models of ZFC are ruled out, by definition. For more, see this answer, or this Wikipedia entry at this section or this section. So it is a theorem in ZF-inf (ZF without axiom of infinity) that $\exists x(x = x)$. Applying axiom of specification with the formula $\phi :\equiv x \not= x$, we can derive $\exists x \forall z (z\not\in x)$. Applying the axiom of pairing with this newly derived set lends a set with exactly one element.

All this aside, we could tweak the definition of a model of FOL to allow for an empty domain. In this case, we would need to assert the existence of a set by hand, but it need not be a non-empty set. Asserting the existence of the empty set would allow us to derive to the existence of a non-empty one using pairing. So an axiom as OP describes simply is not necessary to establish the existence of a non-empty set.

So then why does ZF have the axiom of infinity? We add the axiom of infinity so that we ensure the existence of an infinite set. An infinite set cannot be derived with a finite application of axioms in ZF-inf. Since FOL only allows finite deductions, and thus only finite applications of the axioms of ZF-inf, we cannot derive an infinite set in ZF-inf. Since we want ZF to have infinite sets, we need to add at least one in manually. Cue axiom of infinity.

As for the other set theories you allude to, lack of a reference, I cannot speak to why they include an axiom as you describe. But, ZF(C) is the standard approach to set theory.

  • $\begingroup$ Hmmm, "Most approaches to model theory of FOL assume that the domain of discourse is non-empty" : this is "codified" as Axiom about Existence of non-empty set !! $\endgroup$
    – Prem
    Jul 1 at 6:39
  • $\begingroup$ @Prem no, this has to do with model theory and the semantics of FOL. I'll edit that part for clarity. $\endgroup$ Jul 1 at 6:40

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