Re: Well ordering of reals

Arturo Magidin a écrit :
In article <0b28c041-9e2b-4e33-92f6-79920c58aa7d@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,
David R Tribble <david@xxxxxxxxxxx> wrote:
David R Tribble wrote:
The question then becomes, as an exercise for the reader,
how do we get all possible pairs that are well-orderings on
subsets of the reals?
Arturo Magidin wrote:
The question becomes, what does it mean to "get" all such possible
pairs? Since we cannot "construct" even one such pairing, but there
are a LOT of such pairings (at least as many as there are bijections
from R to itself), what do you mean by "getting" even one, let alone
"all" of them?
David R Tribble wrote:
Well, we can construct pairs for all the finite subsets
of R, as well as for all the countable infinite subsets
of R. It's those pesky uncountable subsets that are
the problem.
Arturo Magidin wrote:
And I notice that you failed to even come close to answering the
question. What does it mean to "get" even ONE pair, let alone "all
pairs"?

And, what does it mean that we can "construct pairs for all the finite
subsets of R"?
I meant we can construct well-orderings for finite subsets
of R.

Given a finite subset of R, we can certainly construct (many) well
orderings for it. We can even do it for many infinite subsets.

This, however, is a ->very<- far cry from constructing well orderings
"for ALL" the finite subsets of R, which implies that you can do it
for all finite subsets simultaneously.

This is a case in which "for any [given] finite subset" is different
from "for all finite subsets".


I may misunderstand something here, but is not the natural order a well-ordering for all the finite subsets of R (and in fact for many more)?
.

Relevant Pages

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  • Re: Well ordering of reals
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