I am having a very difficult time understanding this formula in set-builder notation.

$\forall x \forall y \exists z \forall u \left ( u \in z \Leftrightarrow \left (z=x \lor u=y \right) \right)$

$\textbf{My Attempt:}$

For two sets $x,y$ there exists a set $z$ such that for all sets $u$, $u \in z$ iff $u \in x$ or $y \in z$.

Every element of $z$ must be an element of $x$ or $z$ is just the set $y$.

This is the set that contains elements of $x$ and the set $y$. I am not so sure what to make of this.

  • $\begingroup$ Well, that's not set-builder notation, it's formal logic. And the formula is odd since it looks like it's trying to be an instance of a comprehension schema, but in that case you wouldn't expect to see $z$ to the right of the biconditional. So, I'm not surprised it looks a bit funny to you! $\endgroup$ – Malice Vidrine Feb 18 '14 at 20:11

If written as :

$\forall x \forall y \exists z \forall u (u \in z \equiv (u=x \lor u=y))$

ti is the Axiom of pairing of Zermelo-Fraenkel set thoey.

It says that the set $z$, of which the axiom asserts the existence, contains exactly the two sets $x$ and $y$.

See in Wiki Axiom of pairing:

What the axiom is really saying is that, given two sets $x$ and $y$, we can find a set $z$ whose members are precisely $x$ and $y$. [...] The essence of the axiom is: Any two sets have a pair.

Please, check the formula ...

Another possible reading of the formula is like the standard definition in Zermelo–Fraenkel ($\mathsf {ZF}$) set theory of the natural numbers; they are defined recursively by :

$0 = \{ \}$ (the empty set) and $n + 1 = n \cup \{ n \}$.

Your formula reads as : $z = x \cup \{ y \}$.

  • $\begingroup$ For pairing you want $u=x\vee u=y$, instead of $z=x\ldots$. $\endgroup$ – Malice Vidrine Feb 18 '14 at 20:16
  • $\begingroup$ You are right, thanks ! I'll correct it. $\endgroup$ – Mauro ALLEGRANZA Feb 18 '14 at 20:22

The straight read of the line is:

"For all $x,y$ there exists $z$ such that for all $u,u\in z$ if and only if $z=x$ or $u=y$."

Then we have $u\in z$ implies $u\in x$ or $u=y.$

There is almost certainly a trick or intended consequence associated with this particular statement, and one which is not immediately obvious what consequences arise.


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