Hartogg's aleph function
Stephen J. Herschkorn
Let A(x) = sup{a: a is an ordinal and there exists an injection from
a to x}. In ZF - Foundation, show that
A(x) < A(P(P(P(x)))), where P denotes the power set.
Kunen marks this as a "more difficult" exercise and notes that it is
easier to show that A(x) < A([P^4](x)).
I am working through these exercises for my own edification (i.e., this
is not homework), and I have been scratching my head about this one
(even the "easy" part) for some time. Can anyone offer a hint? Of
course, I refer to the case where x is infinite.
My meagre thoughts on the "easy" part so far:
- A(x) is a cardinal. In fact, A(x) is the set whereof it is the
supremum.
- It suffices to show that there is an injection from A(x) to [P^4](x) .
- Might there be a way to go from x to [P^2](x) and compose this
result with itself?
- I have played with the equivalence classes of P(x) determined by
bijections.
--
Stephen J. Herschkorn hers...@rutcor.rutgers.edu
Fred Galvin
> An exercise from Chapter I of Kunen:
>
> Let A(x) = sup{a: a is an ordinal and there exists an injection
> from a to x}. In ZF - Foundation, show that
> A(x) < A(P(P(P(x)))), where P denotes the power set.
>
> Kunen marks this as a "more difficult" exercise and notes that it is
> easier to show that A(x) < A([P^4](x)).
>
> I am working through these exercises for my own edification (i.e.,
> this is not homework), and I have been scratching my head about this
> one (even the "easy" part) for some time. Can anyone offer a hint?
> Of course, I refer to the case where x is infinite.
Hartogg's is an unusual misspelling. The usual misspelling is
Hartog's. The correct spelling is Hartogs' or maybe Hartogs's, I'm not
sure; the guy's name was Hartogs.
All right, here's a hint. Forget about [P^3](x) and [P^4](x). Just
focus on injecting A(x) into *something* built up from x. After you've
done that it will be time to count iterations of the power set
operation. If it's more than you want, try to do what you did more
economically.
Another hint: forget about Von Neumann ordinals. Ordinal number are
the isomorphism types of well-ordered sets; like any other isomorphism
types, they are equivalence classes. (Give me that old time set
theory, it's good enough for me.)
Herman Rubin
>Let A(x) = sup{a: a is an ordinal and there exists an injection from
>a to x}. In ZF - Foundation, show that
>A(x) < A(P(P(P(x)))), where P denotes the power set.
This should read
A(x) <= (P(P(P(x))))
This is the original proof of Hartogs, and uses the representation
of a reflexive antisymmetric transitive relation R as
{a: for some x, a = {y: yRx}}.
This precedes the use of ordered pairs to represent relations.
Using ordered pairs, the inequality would read
A(x) <= (P(P(x) x P(x))).
Getting rid of the "=" in these relations is harder. It follows
because on can get, instead of A(x), P(A(x)) with the =.
--
This address is for information only. I do not claim that these views
are those of the Statistics Department or of Purdue University.
Herman Rubin, Department of Statistics, Purdue University
hru...@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558
Stephen J. Herschkorn
>Hartogg's is an unusual misspelling. The usual misspelling is
>Hartog's. The correct spelling is Hartogs' or maybe Hartogs's, I'm not
>sure; the guy's name was Hartogs.
>
I got "Hartogg's" from Kunen (p. 44). Thanks for the correction and the
hint; I will consider them in my efforts.
Stephen J. Herschkorn
In article <gABtc.64390$cz5.26...@news4.srv.hcvlny.cv.net>, Stephen J. Herschkorn <hers...@rutcor.rutgers.edu> wrote:An exercise from Chapter I of Kunen:Let A(x) = sup{a: a is an ordinal and there exists an injection from a to x}. In ZF - Foundation, show that A(x) < A(P(P(P(x)))), where P denotes the power set.This should read A(x) <= (P(P(P(x))))
A(x) <= A(P(P(P(x)))).
Or did you mean the curly inequality, i.e., an injection into [P^3](x)? Kunen definitely has strict inequality. Thank you for the hints; I will give them consideration to see which inequality I can get.