4
$\begingroup$

This question is related to the one I asked yesterday here in that it's related to another one of the Zermelo-Fraenkel Axioms. After looking over the notation used to describe the axiom, that is:

$$ \forall \space x \space \forall \space y \space \forall \space z \space [\varphi (x,y,p) \wedge \varphi(x,z,p) \Rightarrow y = z] \Rightarrow \forall \space X \space \exists \space Y \space \forall \space y \space [y \in Y \equiv (\exists \space x \in X) \varphi(x, y, p) ] $$

I believe I understand most of it, but I'm unsure of why we need to involve the variable z, so I thought I'd just write how I'm interpreting this and have someone correct me where it starts to get fuzzy.

Current Interpretation: For all the elements of the three sets $X, Y, Z,$ if the property $\varphi$ holds under some parameter $p$ for $x, y $ and $x, z$ conjointly implies that $y$ equals $z$ then for any set X there exists a set Y such that for any element of $Y$ there exists an element of X such that property $\varphi$ holds under both the element of $Y$ and the chosen element of $X$ for that property $p$.

What I'm confused about is the purpose of the extra parameter $p$ and the set $Z$ why couldn't you just say something like this:

$$ \forall \space x \space \forall \space y \space \varphi (x,y) \Rightarrow \forall \space X \space \exists \space Y \space \forall \space y \space [y \in Y \equiv (\exists \space x \in X) \varphi(x, y, p) ] $$

What am I missing here? Also if someone could clear up my interpretation that would be awesome.

$\endgroup$

1 Answer 1

Reset to default
2
$\begingroup$

Use $x=x$ for $\varphi(x,y)$. Then your version of the axiom gives us a universal set $Y$: every $y$ is in $Y$. So it leads to an inconsistent theory.

The part that says that $\varphi(x,y,p) \land \varphi(x,z,p)$ implies $y=z$ says that the relation $\varphi$ is "function-like." It is a very strong set construction principle, but does not (one hopes) lead to inconsistency.

$\endgroup$
4
  • $\begingroup$ Hmm the second part makes sense to me. I can see how if that part held it would be function-like, but I don't quite understand the first-part where you say let $x = x$ . Does this mean you just replace the $Y$ and $y$ with $X$ and $x$? Sorry if its a dumb question, trying to teach myself a few topics over the summer $\endgroup$
    – Amateur Math Guy
    Commented May 18, 2013 at 1:27
  • $\begingroup$ I am saying let $\varphi$ be any formula which is always true. Instead of $x=x$ we could even use something that does not "mention" $x$ and $y$ at all, such as "there exists an empty set." Then if you trace through the meaning of your axiom for that choice of $\varphi$, it turns out that it says there is a $Y$ such that every (set) $y$ is an element of $Y$. From this we can deduce the Russell paradox. $\endgroup$
    – André Nicolas
    Commented May 18, 2013 at 1:32
  • $\begingroup$ Alright this is starting to make more sense to me, thanks! $\endgroup$
    – Amateur Math Guy
    Commented May 18, 2013 at 1:38
  • $\begingroup$ You are welcome. The traditional "Axiom of Separation" produced only subsets of already constructed sets. Replacement goes beyond that, but cautiously. It says that if we have an already constructed set $X$, and a "function-like" relation $\varphi$, then the "range" of $\varphi$, as $x$ varies over $X$, is a set. $\endgroup$
    – André Nicolas
    Commented May 18, 2013 at 1:43

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .