Consider the set $$S=\{x\mid x\in S\}.$$

For every element $x$, either $x\in S$ or $x\not\in S.$

If we know that $x$ is in fact element of $S$, then, by definition, $x\in S$ so it is true that $x\in S$.

If we know that $x\not\in S$, then, by definition, $x\not\in S$ so it is true that $x\not\in S$.

But if my question is what elements does $S$ contain, then the answer is "not certain."

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    $\begingroup$ By this set builder construction, $S$ is just $S$ ... which is vacuously true ... and indeed does not tell you what specific elements are in $S$. It's like saying $x=x$ ... true, but it doesn't tell you what $x$ is. $\endgroup$
    – Bram28
    Dec 31, 2017 at 14:18
  • $\begingroup$ In your opening line, "For every element x," element of what exactly? Of the set S, the set of real numbers or etc. ? $\endgroup$ Dec 31, 2017 at 14:18
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    $\begingroup$ @GaurangTandon We have that $\forall x (x \in S \lor x \in S^c)$. That means we have a universe of all x consisting of two disjoint sets, $S, S^c$. It's a tautology. And nothing else needs to be clarified. So the question "What elements does S contain?" becomes: "S contains all $x \in S$, i.e. all the elements that do not exist in the complement of S. $\endgroup$
    – amWhy
    Dec 31, 2017 at 14:32
  • $\begingroup$ @amWhy Thanks! I understand your point. Though, imho, saying that set S contains all the elements which do not not belong to its complement, is a kind of a recursive definition which doesn't lead us anywhere in terms of what set S actually contains. It's like saying the fruit basket has all the fruits which the refrigerator doesn't, without specifying what actually is there in either of them. Is my analogy correct? $\endgroup$ Dec 31, 2017 at 15:33

4 Answers 4


You haven't actually defined a set there. You've written down some symbols and are hoping that they define a set, but they don't.

The axiom schema which lets you define sets as "the things which satisfy some property" is the Axiom Schema of Comprehension: roughly, if $\phi$ is a predicate and $S$ is a set, then $\{x \in S: \phi(x)\}$ is a set. But you can't use that here because Comprehension selects a subset of a set which already exists; and $S$ isn't already known to be a set.

The axiom schema which lets you define sets by producing them as the image of another set is the Axiom Schema of Replacement: roughly, if $f$ is a function-class and $S$ is a set, then $\{f(x): x \in S \}$ is a set. But you can't use this here with $f = \mathrm{id}$ because $S$ isn't known a priori to be a set.


This is an interesting question.

A set is determined by the elements it contains. That's the intuitive content of the assertion that two sets are equal if and only if they have the same elements.

So the set you're asking about is just $S$. It contains what it contains.

I think you are a little confused by the set builder notation $\{ \ldots | \ldots\}$.

That notation is usually invoked to construct a set of objects that satisfy some condition. For example $$ E = \{ x | x = 2n \text{ for some integer } n\} $$ creates the set of even integers by telling you what it means for an integer to be even.

In your example, the condition "$ x \in S$" does indeed tell you only what you already know by definition: the tautology "$x$ belongs to $S$ if and only if $x$ belongs to $S$".

Edit. @PatrickSevens answer says about the same thing as this one, but more carefully and more formally.

  • $\begingroup$ I feel like we're not quite saying the same thing; we're approaching the answer in different ways. I have answered the question "why can't I say what elements $S$ contains?" by saying "you haven't defined a set in the first place", while you've answered by saying "because $S$ could be anything, assuming it's a set". I like both angles. $\endgroup$ Dec 31, 2017 at 14:28
  • $\begingroup$ @PatrickStevens I agree not quite the same thing, and both useful. Which helps the OP more depends on his/her level of mathematical maturity, $\endgroup$ Dec 31, 2017 at 14:30

There is the concept in $\text{Set Theory}$ of a 'collectivizing relation', in other words, a relation that can be used to define sets.

Starting with any set $S$,

$\tag 1 \text{There exists a set } X \text{ such that } X = \{x\mid x\in S \text{ AND } P(x) \}$

where P(x) is a statement about $x$ is collectivizing.

In particular, with

$\tag 2 P(x) \; :\; x = x$

we can create a new set

$\tag 3 A = \text{Set Defined by (1) \ (Set Builder Schema)}$

Exercise: Show that $A = S$.

The OP should have written,
Starting with a set $S$, what elements are in the set $S^{'} = \{x\mid x\in S\}$?

Ans: $S^{'} = S$.

In conclusion, you can only make sense out the expression

$\tag 4 S = \{x\mid x\in S\}$ in one way:

The expression (4) is a statement in set theory (perhaps an abuse of notation). The RHS builds a set, and the statement, $S = \text{RHS}$, is either TRUE or FALSE (and here (4) is always TRUE).

It is nonsensical to employ (4) to 'build a set $S$'.


To define an object, such as a set, means to provide a property that that object and only that object has.

In your case, however, it turns out that every set $S$ has the property that $$S=\{x\mid x\in S\}.$$ This is just because every set is the set consisting of its elements.

So you haven't successfully defined an object, you haven't picked out a particular set, since your property doesn't hold of one and only one object.


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