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Aatu Koskensilta wrote:
lugita15@xxxxxxxxx wrote:
Aatu Koskensilta wrote:
My objections do not apply to the simple theory of types in any obvious
sense. For any type n the formula "all sets of type n are empty" is
false.
Then by similar reasoning, shouldn't "all sets of type 0 are empty"
also be false?
No. There is only one set of rank 0, the empty set, and it is obviously
empty. All other ranks contain non-empty sets.
Let me elaborate on a bit - perhaps it will clear up certain things to you.
In type theory, all axioms and rules of inference are typically
ambiguous, from which it trivially follows that if Phi is provable so is
Phi+. The reason this works is that the empty set appears anew at each
type, the reals appear anew at each type and so forth; the types are
basically just the same, the type directly below playing the role of
urelements. One might then postulate that, in fact, all types are
elementarily equivalent in a sense, which is formalized by the scheme
Phi <--> Phi+. It is not obvious that this should be so, and indeed it
is equivalent to the consistency of NF.
In set theory, on the other hand, new stuff appears at each level of the
cumulative hierarchy, and in general not only is there no reason to
suppose V_alpha to be elementarily equivalent to V_alpha+1 but this is
actually provably false. Thus even if we somehow make sense of the
notion of a rank of a variable in a sentence in the language of set
theory, so that we can define Phi+ for all Phi, the scheme Phi <--> Phi+
does not hold.
--
Aatu Koskensilta (aatu.koskensilta@xxxxxxxxx)
"Wovon man nicht sprechen kann, daruber muss man schweigen"
- Ludwig Wittgenstein, Tractatus Logico-Philosophicus
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