Velleman - How to prove it - Do these two statements really mean the same thing?

Question

Hello and thanks in advance for reading!

In How to Prove it P29 Velleman writes:

" In general, the statement y ∈ { x | P(x)} means the same thing as P(y), ... "

In my understanding the first statement means that we apply the elementhood test to y and it fulfills it (makes it true), and the second statement means that we apply the same test but it can be true or false (as in we cannot know if P(y) is true or false).

So I don't see how these mean the same. What am I missing?

Thanks a lot!

Answer 1

Note that an object $y$ is an element of $\{ x : P(x) \}$ iff $P(y)$ is true.

• If $y$ is in $\{ x : P(x) \}$, then $y$ must satisfy the defining property of being an element of $\{ x : P(x) \}$, namely having property $P(\cdot)$.
• If $P(y)$ is true, then $y$ satisfies the defining property of being an element of $\{ x : P(x) \}$, and so it is an element of that collection.

This is pretty much by notational definition, but we therefore have that elementhood in $\{ x : P(x) \}$ is equivalent to having property $P(\cdot)$.

Answer 2

Two points. (1) It is certainly the case that, assuming that the relevant set exists, that

$$y \in \{x \mid P(x)\} \leftrightarrow P(y),$$

as Arthur Fischer explains.

You might wonder, then, why Velleman writes "In general" that holds, and not "Always". Well, suppose $P$ is the property of being a non-self-membered set [or of being a set which has a rank in the hierarchy, or some such]. Then, it is not true that some given $y$ is such that $P(y)$ if and only if $y \in \{x \mid P(x)\}$, taking $\{x \mid P(x)\}$ as a purported set-designator, for the simple reason that there is no such set as the Russell set of all non-self-membered sets. We do need the set $\{x \mid P(x)\}$ to exist for the equivalence to hold!

(2) Even when the relevant set does exist, I wouldn't want to put it quite Velleman's way -- i.e. I wouldn't assert that $P(y)$ actually means the same as $y \in \{x \mid P(x)\}$. For, in most cases, only one of them commits you to the existence of a set: and if their ontological commitments are distinct, they can't mean just the same. They are not outright meaning-equivalent, only equivalent inside a bit of elementary set-theory.

Answer 3

In the context of $\mathsf{ZFC}$ set theory, the reason that the statement "$y \in \{ x \mid P(x)\}$" means the same thing as $P(y)$ is that we define it to mean $P(y)$. Here the variable symbols $x$ and $y$ denote sets.

We refer to $\{ x \mid P(x)\}$ as the class of all sets $x$ having property $P$, and to say that a set $y$ is a member of this class is just a fancy way of saying that it has property $P$.

(A naive alternative is to define $\{ x \mid P(x)\}$ to be the "set" of all sets $x$ having property $P$, but this is problematic because assuming that such a set always exists will lead to Russell's paradox.)

The statement $y \in \{ x \mid P(x)\}$ can be either true or false according to whether $y$ is a member of the class $\{ x \mid P(x)\}$. Likewise, the (equivalent by definition) statement $P(y)$ can be either true or false according to whether $y$ has property $P$.

Answer 4

Set-builder notation can be confusing at times. Rewrite $p=\{x | P(x)\}$ with a quantifier as $\forall y (y\in p \iff P(y))$ and I think your answer becomes obvious.