Let I be a set which could be finite (includes the empty set) or infinite. How would I rewrite the set I? I included some examples for possibilities but an open to other notations as well. I know that 3 is a vacuous truth. I have always been hesitant using set builder notation, because the way you rewrite it out in words is the set of all of x such that ... . If someone knows the reason for using such that instead of where, I think that would help me with my confusion as well.

1) $I=\left \{ x\mid x\in I \right \}$

2) $I=\left \{ x\mid \exists x\in I \right \}$

3) $I=\left \{ x\mid \forall x\in I \right \}$

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    $\begingroup$ $I$ is... $I$. If we have neither a finite list of elements nor a "condition" (a formula) specifying it, what does it mean to write $I= \{ x \mid x \in I \}$ or similar ? $\endgroup$ May 31, 2017 at 15:13
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    $\begingroup$ In any case, 2) and 3) are wrong. $\endgroup$ May 31, 2017 at 15:13
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    $\begingroup$ The first suggestion is a tautology. You might just as well write $I$. The second and third make little sense. If you are concerned about using set builder notation correctly in a particular case, edit the question to show us the case that bothers you. $\endgroup$ May 31, 2017 at 15:14
  • $\begingroup$ Do you know the difference between such that and where in this context too? $\endgroup$
    – W. G.
    May 31, 2017 at 15:14
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    $\begingroup$ Obviously $I$ is "the set of those elements that belongs to $I$"... so what ? It is "formally" correct but what does it mean ? What have we achieved in rewriting it that way ? $\endgroup$ May 31, 2017 at 15:20

1 Answer 1


$\{x|....\}$ means the set of all and only those $x$ for which "..." is true.

$\exists x\in I$ means "something belongs to $I$". Note that if I write $x=2\land \exists x\in I$ then I did not say that $2\in I$. But is bad writing because the first and second appearances of the letter $x$ are different things, and it is better to write $x=2\land \exists y\in I$.

$\forall x\in I$ means "for each, any, & every thing that belongs to $I$" (whether there is any or not). Note that it is not a grammatically complete sentence, but a part of a sentence, unlike $\exists x\in I.$

$I=\{x|x\in I\}$ asserts that $I$ is the same thing as the set of all those and only those things that belong to $I$. In Set Theory this is the Axiom of Extensionality: The set $I$ and the set $J$ are equal iff they have the same members.... (There are alternate formulations of set theory in which there are sets and atoms: An atom is not the empty set but it has no members.)

So $1)$ is correct....

2) says that $I$ is the set of all and only those objects for which the sentence "Something belongs to $I$" is true, so 2) can be true only if $I$ is the empty set or if everything belongs to $I$.....

3) says that $I$ is the set of all and only those objects for which "For every thing that belongs to $I$" is true, which is meaningless.

Much modern notation is in convenient short forms for words or short sequences of words, and can be rendered literally (in the sense of a word-for-word translation) into words. Much of it is not very old. For comparison, try to read some algebra from about 400 years ago.

  • $\begingroup$ 3) NO; it says: $I$ is the set of all and only those objects for which the sentence "Everything belongs to $I$" is true, which is true only of the "universal set", if any. The syntax of the set-builder notation is $\{ x \mid \varphi(x) \}$: in the "condition" $\varphi(x)$ the var $x$ must be free, otherwise it is not the "same" $x$ of $\{ x \mid \ldots \}$. $\endgroup$ Jun 1, 2017 at 11:51
  • $\begingroup$ @MauroALLEGRANZA,,Agreed. That occurred to me today. $\endgroup$ Jun 2, 2017 at 2:27
  • $\begingroup$ @DanielWainfleet What you wrote about sentence (3) is correct: sentence (3) is meaningless and the condition inside that set is not a sentence and—contrary to Mauro's comment—does not read "Everything belongs to $I$". $\quad$ However, on the same note, "$∃x{∈}I$" too isn't a sentence, does not mean "something belongs to $I$", but rather "for some $x$ that belongs to $I\ldots$". $\endgroup$
    – ryang
    Sep 9, 2023 at 17:32

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