[FOM] Intuitionists and excluded-middle
Lew Gordeew
legor at gmx.de
Sun Oct 23 10:02:43 EDT 2005
On Fri, 21 Oct 2005 10:45:56 -0700, "Dana Scott" wrote:
> So, constructively, we ask: Can there be a CONTINUOUS function f
> such that if a < b and both are irrational, then
>
> a < f(a,b) < b,
>
> and f(a,b) is always irrational as well?
> [...]
> (Surely you cannot map the whole plane into the rationals in a
> non-trivial way. But could you map the irrational plane non-trivially
> to the irrationals? We did map the rational plane nicely to the
> irrationals. OK, by "plane" here I mean the HALF-plane where a < b.)
Strictly speaking, these are two different questions. The existence of a
constructive (hence continuous) function f in question that is defined on
the WHOLE real half-plane H = {(a,b) in R^2: a < b} corresponds to standard
constructive interpretation of the sentence
A: (forall reals a < b)(thereis real c)(a, b irrationals -> a < c < b and c
irrational)
whereas a constructive mapping of the irrational half-plane HI = {(a,b) in
(R-Q)^2: a < b} to the irrationals corresponds to standard constructive
interpretation of this sentence
B: (forall irrationals a < b)(thereis irrational c)(a < c < b)
Now A is stronger than B. Constructive solution to B is easy (see my and
Harvey Friedman's FOM replies), but it does not infer A, since the set of
(ir)rational numbers is not decidable in the constructive real universe. In
other words, there is no trivial constructive extension of a given function
f: HI -> R-Q, a < f(a,b) < b,
to a desired function
f: H -> R-Q, a < f(a,b) < b,
Regards,
LG
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