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Salmon's 'proof' for the existence of the empty set

  1. Nov 10, 2014 #1
    Hi guys,

    I stumbled upon this lovely quote from the philosopher of science Wesley Salmon: "The fool hath said in his heart that there is no null set. But if that were so, then the set of all such sets would be empty, and hence, it would be the null set. Q.E.D." (in Martin Gardner, Mathematical Magic Show, 1989, p.33)

    This proof was obviously meant by Salmon as having to be taken with a grain of salt. Nevertheless, I don't believe he took it to be nothing but a joke. What I would like to know is if this proof holds up in any of the formalized systems of set theory. In other words: is it possible to formalize Salmon's proof?

    The problem with that, I think, is that Salmon presupposes a principle of comprehension. He seems to reason as follows: if there is a property E such that Ex means "x is an empty set", then there must be a set S containing all objects of which E is true. Hence if -∃x(Ex), then S must be an empty set.

    The problem, of course, is that comprehension leads to paradoxes in set theory and is therefore replaced by extensionality as a way to identify sets. One of these paradoxes is that comprension can lead to self-inclusion which can lead to Russell's paradox. In Salmon's proof, too, there seems to be paradoxical self-inclusion, since S being the set of all empty sets must be an element of itself if it is an empty set, in which case it wouldn't be empty...

    I would like to hear what you think? Is Salmon's proof complete nonsense from a formal point of view? Or are there formel systems in which it could be expressed? And if so, how?

    Thank you
    P.
     
  2. jcsd
  3. Nov 10, 2014 #2

    Erland

    User Avatar
    Science Advisor

    I agree with you. This cannot be formalized since it assumes the existence of a set which only contains only empty sets. Without a general principle of comprehension, which leads to Russell's paradox etc, it is not clear how to prove that such a set exists, using only the argument in this "proof".

    The simplest way to prove that there exists an empty set (and there is then only one, by the axiom of extensionality), is to use the limited principle of comprehension called the axiom of subsets (although it is actually a theorem in ZF):

    ##\forall x\exists y \forall z (z\in y \leftrightarrow P(z)\land z\in x)##

    where ##P(z)## is a predicate logical formula with ##z## as its only free variable.

    This says that given any expressible (with predicate logic) condition ##P(z)## and any set ##x##, there is a set ##y## which consists of those elements in ##z\in x## which satisfy ##P(z)##.

    Now, we can apply this to an arbitrary ##x## and some contradictory condition ##P(z)##, for example ##z\in z\land z\notin z##.
    ##P(z)\land z\in x## is then always false, no matter what ##x## is, so the condition
    ##z\in y \leftrightarrow P(z)\land z\in x## is equivalent to ##z\notin y##, and hence we obtain
    ##\exists y\forall z(z \notin y)##, which says that there exists an empty set.
     
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