Re: The collection of all sets and proper classes.

On 2007-03-02, Calvin <crice5@xxxxxxxxxxxxxx> wrote:
Are you saying that in NF the assumption that there is a
collection of all sets does not lead to the contradiction
known as Russell's Paradox?

Not having followed the thread I can't say what anyone's saying, but in NF
the assumption that there is a set of all sets - a theorem of NF - does not
lead to the contradiction known as Russell's paradox. NF is axiomatized by
extensionality and stratified comprehension, stating that any stratified
formula defines a set. A formula is said to be stratified if there is an
assignment of naturals to the variables in the formula, such that in any
subformula of the form "x = y" x and y are assigned the same naturals, and
in any subformula of the form "x in y" if x is assigned n y is assigned n+1.
The formula "x not in x" is not stratified, and thus the existence of the
Russell set cannot be proved in any straightforward fashion in NF.

--
Aatu Koskensilta (aatu.koskensilta@xxxxxxxxx)

"Wovon man nicht sprechen kann, daruber muss man schweigen"
- Ludwig Wittgenstein, Tractatus Logico-Philosophicus
.

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