Re: Why is the Russell Paradox necessary?
- From: "Butch Malahide" <fred.galvin@xxxxxxxxx>
- Date: 5 Mar 2007 20:15:31 -0800
On Mar 5, 9:31 pm, "Calvin" <cri...@xxxxxxxxxxxxxx> wrote:
In ordinary set theory, in which Russell's Paradox is used
to show that assuming the existence of a set of all sets
leads to that paradox, ie. a contradiction, I wonder why
the Russell Paradox method is necessary.
Did someone tell you that it is necessary?
We know that the cardinality of the set of all subsets of a
given set is greater than the cardinality of the given set.
But if we assume the existence of a set of all sets, by
definition it contains all of its subsets. Therefore the
cardinality of the set of all subsets of the set of all sets
cannot be greater than the cardinality of the set of all
sets, which is a violation of what we know about
cardinality. Hence there is no set of all sets.
This is another way of proving the nonexistence of a universal set. Of
*course*, if there are two different ways to do the same thing, then
neither one of them is "necessary", as we could get by with the other
one. It is not at all unusual to have alternative proofs of the same
theorem, e.g., the hundreds of different proofs of the theorem of
Pythagoras about right triangles. In the case you're talking about,
the two proofs are not so very different. As I understand it (could be
wrong, this is hearsay), it came about this way. The proof via
cardinalities--if there is a universal set V, then P(V) = V,
contradicting Cantor's theorem that |P(X)| > |X| -- came first.
Russell studied that proof, and *simplified* it to the one-line
argument known as "Russell's paradox". Namely, if P(V) = V, then the
identity function f(x) = x is a bijection from V to P(V). The
inequality |P(V)| > |V| is proved by constructing, for a given
function f:V-->V, the "diagonal" set D = {x in V: x is not in f(x)}.
Applying the diagonal construction to the identity function f(x) = x,
we get the so-called Russell set.
.
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