Re: Point-free affine real line?

[Vaughan Pratt <pratt@cs.stanford.edu>]

Peter is right that Q as a field contains order information absent from Q
as either a linear or affine space.  (−*x*² ≤ 0 ≤ *x*².)

> For the order to be definable from the algebraic structure, you need to
consider Q as a field

It certainly suffices, and indeed is traditional.

But is it necessary for the construction of the affine real line?  In
particular do we have to define multiplication on the rationals only to
throw it away at a later stage?

Obviously the ordered dyadic rationals would suffice.  Equally obviously
they don't form a field.

But can they be defined as an ordered affine space with the same degree of
formality and economy as we routinely define the ordered linear space of
rationals?

Claim.  The ordered algebra (D, xy, x+y, ≤) consisting of the set D  of
dyadic rationals under the two binary operations xy denoting 2y - x and x+y
denoting (x + y)/2 can be defined order-algebraically up to isomorphism as
the free ordered algebra on the ordinal 2 satisfying finitely many
equations in the two operations along with the inference rules, from x ≤ y
infer each of

y ≤ xy
yx ≤ x
x ≤ x+y
x+y ≤ y

The Dedekind cuts being cuts in the rational line, the following claim
depends on a distinct name for the corresponding notion of a cut in the
ordered affine line of dyadic rationals, which I suggest calling a dyadic
cut.  (When cutting at a dyadic rational follow a consistent convention as
to which side to associate that rational, as done with the Dedekind cuts,
and always have both sides of the cut nonempty.)

Claim.  The dyadic cuts in the ordered affine space of dyadic rationals are
in order-preserving bijection with the Dedekind cugts in the ordered linear
space of rationals, with the cuts at dyadic rationals correctly matched to
their Dedekind counterparts.

Vaughan Pratt

On Sun, Jun 10, 2018 at 6:54 AM, Peter Johnstone <ptj@dpmms.cam.ac.uk>
wrote:

> Sorry, what I wrote was a bit sloppy. Vaughan is right that the problem
> doesn't arise with the passage from considering Q as a linear space
> to considering it as an affine space, since it already has order-
> reversing linear automorphisms. For the order to be definable from
> the algebraic structure, you need to consider Q as a field, which is
> what the usual Dedekind-section construction does.
>
> Peter Johnstone
>
>
> On Thu, 7 Jun 2018, Vaughan Pratt wrote:
>
> > Not quite: the affine rational line doesn't have a definable total
>> order,
>>
>>> since it has order-reversing automorphisms, so any definition using
>>> Dedekind sections is problematic.
>>>
>>
>> Morphism-wise, since the affine transformations are just the composition
>> of
>> a linear transformation with a translation, and translation of the
>> rational
>> line preserves order, affinity can't be the problem here.
>>
>> Structure-wise, one can equip the rational line with either its linear
>> combinations or its linear order, or both.  Using both eliminates the
>> order-reversing linear transformations.   "Affine" only makes sense in the
>> context of having the linear combinations, as "affine" limits the linear
>> combinations to those whose coefficients sum to one.   If it is ok for the
>> linear combinations and the linear order to coexist, it must be even more
>> ok for the affine combinations and the linear order to coexist.
>>
>> So whether one considers the morphisms or the structure they preserve,
>> affinity (affineness?) must be a red herring here: any problem for the
>> rational line as an affine space is surely also a problem for it as a
>> vector space.
>>
>> Vaughan Pratt
>>
>>
>>

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