Re: not(CH) and cardinal inequality in the absence of LEM

Re: not(CH) and cardinal inequality in the absence of LEM

[David Roberts]

Dear all,

as an obvious-in-hindsight followup in my part 2), the idea that
Epi(Y,X) = 0 gives strictness of a cardinal inequality clearly loses
its meaning in the absence of the Cantor-Bernstein-Schröder Theorem.
What should (perhaps) be considered is the subobject Iso(Y,X) of
Epi(Y,X), and if Y injects to X and Iso(Y,X) = 0, then one can
unambiguously say Y < X. Or at least less ambiguously! There are of
course variants where one says instead that (X surjects onto Y) xor
(Y=0), or Y is a subquotient of X.

One gets of course variants, for fixed X, analogous to the case X=|N
where different notions of finiteness arise (Kuratowski finite,
subfinite etc)

Best regards,

David





On 28 October 2015 at 18:10, David Roberts
<david.roberts@adelaide.edu.au> wrote:
> Dear all,
>
> 1) not(CH)
>
> I'm reacquainting myself with the proof using toposes that not(CH) is
> relatively consistent. There's one step that is not quite as I'd like
> it, and that is as follows.
>
> Let P be a poset with the countable chain condition and the double
> negation topology (or indeed any site where all covering sieves are
> generated by families that are at most countable), then for objects
> X,Y in the base topos S that are infinite (in the sense that N x X =
> X), having Epi(Y,X) = 0 in S implies that Epi(Y^,X^) = 0 in Sh(P).
>
> It seems to me that the proof, as recounted in Mac Lane--Moerdijk or
> Johnstone's Baby Elephant for instance, uses LEM, by assuming that
> Epi(Y^,X^) =/= 0 and showing that Epi(Y,X) =/= 0.
>
> I haven't thought this through, but it seems like it might be possible
> to rework things so that the proof in fact shows that from an
> isomorphism 0--> Epi(Y,X) one can show that any two maps Epi(Y^,X^)
> --> \Omega are equal, and hence since 0 --> Epi(Y^,X^) is a subobject,
> it must be an isomorphism.
>
> Has anyone thought about this before? Of course, the previous
> paragraph may be nonsense, I admit...
>
> 2) Cardinal inequality in the absence of LEM
>
> This does raise the related question as to what the definition or
> definitions of <, strict inequality of cardinalities (i.e. just sets),
> could be in the absence of LEM and still be useful. If one were to
> adopt the following:
>
> Definition: B < A if and only if there a mono B >--> A and a
> factorisation B >--> A' >--> A such that Epi(B,A') = 0.
>
> then---modulo the issue in part 1)---the existing proof of the
> preservation of inequalities of sets seems to work without the
> assumption the base topos is boolean (which is needed in the existing
> proof to find a retract A -->> A' ).
>
> In the presence of LEM the definition of < reduces to the usual one,
> but I don't see a good reason or philosophy to single out this
> particular definition over other potential definitions except that it
> is what the proof uses.
>
> Thoughts?
>
> Best regards,
>
> David



-- 
Dr David Roberts
http://ncatlab.org/nlab/show/David+Roberts

Visiting Fellow
School of Mathematical Sciences
University of Adelaide
SA 5005
AUSTRALIA

"When I consider what people generally want in calculating, I found
that it always is a number."
-- al-Khwārizmī

CRICOS Provider Number 00123M
IMPORTANT: This message may contain confidential or legally privileged
information.  If you think it was sent to you by mistake,  please delete all
copies and advise the sender.  For the purposes of the SPAM Act 2003, this
email is authorised by The University of Adelaide.


[For admin and other information see: http://www.mta.ca/~cat-dist/ ]

not(CH) and cardinal inequality in the absence of LEM

[David Roberts]

Dear all,

1) not(CH)

I'm reacquainting myself with the proof using toposes that not(CH) is
relatively consistent. There's one step that is not quite as I'd like
it, and that is as follows.

Let P be a poset with the countable chain condition and the double
negation topology (or indeed any site where all covering sieves are
generated by families that are at most countable), then for objects
X,Y in the base topos S that are infinite (in the sense that N x X =
X), having Epi(Y,X) = 0 in S implies that Epi(Y^,X^) = 0 in Sh(P).

It seems to me that the proof, as recounted in Mac Lane--Moerdijk or
Johnstone's Baby Elephant for instance, uses LEM, by assuming that
Epi(Y^,X^) =/= 0 and showing that Epi(Y,X) =/= 0.

I haven't thought this through, but it seems like it might be possible
to rework things so that the proof in fact shows that from an
isomorphism 0--> Epi(Y,X) one can show that any two maps Epi(Y^,X^)
--> \Omega are equal, and hence since 0 --> Epi(Y^,X^) is a subobject,
it must be an isomorphism.

Has anyone thought about this before? Of course, the previous
paragraph may be nonsense, I admit...

2) Cardinal inequality in the absence of LEM

This does raise the related question as to what the definition or
definitions of <, strict inequality of cardinalities (i.e. just sets),
could be in the absence of LEM and still be useful. If one were to
adopt the following:

Definition: B < A if and only if there a mono B >--> A and a
factorisation B >--> A' >--> A such that Epi(B,A') = 0.

then---modulo the issue in part 1)---the existing proof of the
preservation of inequalities of sets seems to work without the
assumption the base topos is boolean (which is needed in the existing
proof to find a retract A -->> A' ).

In the presence of LEM the definition of < reduces to the usual one,
but I don't see a good reason or philosophy to single out this
particular definition over other potential definitions except that it
is what the proof uses.

Thoughts?

Best regards,

David


[For admin and other information see: http://www.mta.ca/~cat-dist/ ]