# Can there be a well-defined set with no membership criteria?

Prime numbers are a well-defined set with specific membership criteria. Can the same be said about "numbers"? Aren't numbers (that is, all numbers) a well defined set but without membership criteria? Anybody can say, given a particular object, whether that belongs in the set of numbers or not. But it may not be possible to give any criteria for this inclusion.

We may want to say that the set of all numbers has a criterion and that is that only numbers shall get into the set and all non-numbers shall stay out of it. But then, this is merely a definition and not a criterion for inclusion.

Therefore my question: Can there be a well-defined set with no membership criteria?

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What exactly do you mean by all numbers? Do you mean all natural numbers? Reals? Ordinals? Cardinals? – ferson2020 Jan 31 '13 at 15:42
"All" includes natural, real ordinal etc., ad infinitum. – Ashwin Kumar Jan 31 '13 at 15:53
Such a collection of elements would not be a set; as there is no set containing all the cardinals, nor is there a set containing all the ordinals. – ferson2020 Jan 31 '13 at 15:59

This answer is the same as ferson2020's under one interpretation of the terminology in the question, but I wanted to write an answer relating it to the standard terminology and to add a proof. Let $X$ be a set. Then the following are equivalent:

(1) For some formula $\varphi$, $X$ is the unique set such that $\varphi[X]$ holds. (That is, $X$ is definable.)

(2) For some formula $\psi$, $X$ is the set of all $a$ such that $\psi[a]$ holds. (That is, $X$ has a "membership criterion.")

If (1) holds, then we can either let $\psi(a)$ say "$a \in Y$ for some set $Y$ such that $\varphi(Y)$ holds" or "$a \in Y$ for every set $Y$ such that $\varphi(Y)$ holds"—both choices of formula $\psi$ are ways of saying "$a \in X$" but without mentioning $X$ as a parameter, so they give membership criteria for $X$.

Conversely, if (2) holds, then we just let $\varphi(Y)$ say "for every set $a$ we have $a \in Y$ if and only if $\psi(a)$ holds. This formula says that $Y$ has the same elements as $X$, and by the Axiom of Extensionality this means that $Y = X$, so we have a formula that is uniquely satsisfied by $X$.

So two possible notions of the definability of a set turn out to be equivalent. (Although there is a slight wrinkle here—definability is not definable, so the equivalence of (1) and (2) is not a theorem but a meta-theorem.)

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A set being well-defined is the same thing as its membership criteria being well-defined; a set is defined by exactly which elements belong to it.

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A set must have membership criteria. Cantor defined sets by (taken from wikipedia)

A set is a gathering together into a whole of definite, distinct objects of our perception [Anschauung] and of our thought – which are called elements of the set.

As the elements in a set therefore are definite, we can describe them in a way - which is a membership critria. There can happen funny things (Take a look at Axiom of Choice and Russels Paradox, the latter showing why Cantors definition isn't enough and ZFC is needed (I sadly can't link to it now) )

In your example: The set of all numbers is the set containing all objects that we characterize as numbers. This quickly gets a bit meta-mathematical ("What is a number?", "two" and "2" are just the representatives / "shadows" of the thing we call two), you should take a look at how the natural numbers are defined.

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