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$X = \{x\ :\ P(x)\}$ is it true that $a\in X⟺P(a)$

I think it is true that $a\in X ⟹ P(a)$ but I'm not sure whether the converse is correct.

jameselmore
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naive set theory
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    By definition of the "set-builder" notation, this is true. Every predicate has this set associated with it, which is exactly the set of elements satisfying the predicate. Here "exactly" means that the left-to-right and right-to-left of your logical equivalence work. – Colm Bhandal Aug 31 '15 at 13:23
  • The converse is $P(a) \Longrightarrow a \in { x : P(x) }$. – Björn Friedrich Aug 31 '15 at 13:23
  • It may be a bit elementary, but look at this page for set builder notation: http://www.mathsisfun.com/sets/set-builder-notation.html. Because you're building a set containing all elements with that property, the converse will be true. – Colm Bhandal Aug 31 '15 at 13:26
  • Actually, if you are using a 3-valued logic with Gödel implication to handle the undefined, $a \in X \rightarrow P(a)$ would be false when $X$ is undefined (which makes $a \in X$ silly) and $P(a)$ is true, while the converse implication would never be false when $X$ is undefined. So I'd say the converse is actually less problematic, and in fact should be assertable if one considers non-false statements to be assertable. (Personally, I think in math indicative statements should be assertions of truth and subjunctive ones of non-falsity, and the subjunctive should be the default mood in math.) – Stephen A. Meigs Aug 31 '15 at 17:39
  • I should have added that the converse has problems too if one allows the possibility that $P(a)$ is silly (but $a$ is defined and not in the defined $X$). I sort of assumed I was to assume $P(a)$ was taken to be true or false, but if that assumption bothers you more than than the assumption of X being defined, then the converse might rightly seem more problematic. – Stephen A. Meigs Aug 31 '15 at 18:14

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Set builder notation has its limitations. In this case, it would help to translate $X = \{x\ :\ P(x)\}$ to $\forall x:[x\in X \iff P(x)]$.

Then, for any $a$, you obviously have $a\in X \iff P(a)$.

Dan Christensen
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