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• CommentRowNumber1.
• CommentAuthorTodd_Trimble
• CommentTimeJan 25th 2016

Added a bit to Hartogs number. Including the curiosity that GCH implies AC. :-)

• CommentRowNumber2.
• CommentAuthorMike Shulman
• CommentTimeJan 26th 2016

How are you phrasing the GCH in the absence of AC to make that true? I usually see GCH phrased as something lke $2^{\aleph_n} = \aleph_{n+1}$, and in the absence of AC usually the $\aleph$s are only the well-orderable cardinalities; so that doesn’t seem sufficient for your argument which applies GCH to $P(P(P(X)))$ when $X$ is not known to be well-orderable. Unless I’m missing something?

• CommentRowNumber3.
• CommentAuthorTodd_Trimble
• CommentTimeJan 26th 2016
• (edited Jan 26th 2016)

Can’t we say $\forall_{X, Y} \neg (|X| \lt |Y| \lt |P X|)$? Which is probably more or less what Cantor would have said.

• CommentRowNumber4.
• CommentAuthorMike Shulman
• CommentTimeJan 26th 2016

How do you get from that to $\aleph(X)$ being bijective to $P(X)$, $P^2(X)$, or $P^3(X)$?

• CommentRowNumber5.
• CommentAuthorTodd_Trimble
• CommentTimeJan 26th 2016

Hm, maybe I unwittingly let trichotomy sneak into my thinking. (The result ZF + GCH implies AC happens to be true, but maybe this route through Hartogs is not really the way to do it.) Let me think on it more (and thanks).

• CommentRowNumber6.
• CommentAuthorTodd_Trimble
• CommentTimeJan 26th 2016
• (edited Jan 26th 2016)

Well, here is a more responsible demonstration, which does in fact use the Hartogs numbers. The key result seems to be lemma 3 (page 552).

• CommentRowNumber7.
• CommentAuthorTodd_Trimble
• CommentTimeJan 26th 2016
• (edited Jan 26th 2016)

Okay, I’ve written up what I think is a tight proof of GCH implies AC at Hartogs number. It’s a rendition of the Gillman article cited in my last comment (I did spot a little oversight in his proof).

I learned of this fact last night while I was idly leafing through Eric Wofsey’s old blog Ultrawaffle (or whatever he calls it); it’s one of his series “Fun Little Math Problem of the Day”; see here. I think the “little” made me underestimate the amount of argumentation that is actually required, but the proof does wind up being fun (and the statement a little surprising at first, as Gillman says: what could GCH possibly have to do with AC?).

• CommentRowNumber8.
• CommentAuthorMike Shulman
• CommentTimeJan 27th 2016

Nice, thanks!

Maybe it’s too late at night, but I need help with the easy exercise that ${|P|} = {|2P|}$ if $P$ is an infinite power set. I can see how to do it if $P=P(Y)$ where $Y$ is Dedekind-infinite, but in the general case I’m stuck.

• CommentRowNumber9.
• CommentAuthorMike Shulman
• CommentTimeJan 27th 2016

I do feel like the “recall the GCH” comment merits some more discussion, since it appears to be only this particular way of phrasing GCH that implies AC, right? If we state GCH as $2^{\aleph_n} = \aleph_{n+1}$, which is equivalent in the presence of AC (and is how the article continuum hypothesis states it), then it doesn’t imply AC.

• CommentRowNumber10.
• CommentAuthorTodd_Trimble
• CommentTimeJan 27th 2016

I can see how to do it if $P=P(Y)$ where $Y$ is Dedekind-infinite, but in the general case I’m stuck.

Oh! I may have elided over this point.

So in ZF with classical logic, there is a distinction between infinite and Dedekind-infinite? I just assumed that in that context, $X$ infinite is the same as existence of a bijection $1+X \cong X$, which is what I had in mind.

• CommentRowNumber11.
• CommentAuthorTodd_Trimble
• CommentTimeJan 27th 2016

As for #9: I don’t know the history, but maybe someone should check on Sierpinski’s article. You again raise an interesting point.

• CommentRowNumber12.
• CommentAuthorTodd_Trimble
• CommentTimeJan 27th 2016

Okay, I just consulted Wikipedia, and yes you’re right that Dedekind-infinite and infinite are distinct in ZF in classical logic. (Live and learn.) However, the patch is easy: just embed $Y$ into a Dedekind-infinite set like $\mathbb{N} + Y$, and take the power set $X$ of that. The proof then goes through. I’ll put the patch in now.

• CommentRowNumber13.
• CommentAuthorTodd_Trimble
• CommentTimeJan 27th 2016

And finally (Mike), I added some remarks to continuum hypothesis to cover the point you brought up in #9. Sierpinski of course uses the stronger form of GCH.

• CommentRowNumber14.
• CommentAuthorMike Shulman
• CommentTimeJan 27th 2016

Thanks!! Dedekind-infiniteness is definitely an unexpected gotcha: you expect that constructive mathematics may have trouble defining “infinite”, but it’s surprising (to me) that even in ZF there’s some ambiguity left in what you mean by “infinite”.

Morally, I feel like $2^{\aleph_n}=\aleph_{n+1}$ and $\forall X \forall Y ({|X|}\le {|Y|}\le {|P(X)|} \to {|Y|}={|X|}\vee {|Y|}={|P(X)|}$ ought to have two different names, like “weak GCH” and “strong GCH”. Then the theorem would be that strong GCH is equivalent to the conjunction of weak GCH and AC.

• CommentRowNumber15.
• CommentAuthorDavidRoberts
• CommentTimeJan 27th 2016

The latter should be ’global CH’, since it applies to all sets, and the traditional ’G(eneralised )CH’ can be saved for the version with only alephs.

• CommentRowNumber16.
• CommentAuthorDavidRoberts
• CommentTimeJan 27th 2016

Also: is the full strength of ZF even used? The argument looks like it could be done in BZ.

• CommentRowNumber17.
• CommentAuthorTodd_Trimble
• CommentTimeJan 27th 2016
• (edited Jan 28th 2016)

As for the nomenclature, I have no strong opinions. I’m fine with implementing the proposed theorem in #14 if agreement is reached.

I’m pretty sure BZ suffices. Each subset of $X$ equipped with a well-ordering is uniquely specified by its set of principal ideals or downsets, i.e., an element of $P P(X)$. An equivalence class of well-orderings is then a subset of $P P(X)$ or an element of $P P P(X)$, so the set of equivalence classes $\aleph(X)$ is a subset of $P P P(X)$ as advertised. There is no funny business with unbounded quantifiers or replacement that I can see anywhere; everything is locally definable. This is probably worth a remark at Hartogs number.

• CommentRowNumber18.
• CommentAuthorMike Shulman
• CommentTimeJan 31st 2016

“global” to me implies that the other one would be “local” in some way, which doesn’t seem to be the case.

• CommentRowNumber19.
• CommentAuthorDavidRoberts
• CommentTimeJan 31st 2016

Localised to alephs? Of course, ’local’ has a fairly established meaning in set theory as well, which is better suited to specific instances of CH at a given set/cardinal.

• CommentRowNumber20.
• CommentAuthorTodd_Trimble
• CommentTimeMar 21st 2016

I added a few more examples (equivalent forms of AC) to Hartogs number, to illustrate the kinds of things you can do with it.

• CommentRowNumber21.
• CommentAuthorTodd_Trimble
• CommentTimeAug 9th 2018

Added the proof that the Hartogs of $S$ doesn’t inject into $S$.

• CommentRowNumber22.
• CommentAuthorDavidRoberts
• CommentTimeApr 23rd 2019

I’ve been looking around and can’t find the following. Given a set $X$, a “least” set $Y$ such that $X$ does not surject onto $Y$. The Hartogs number $Y=\aleph(X)$ doesn’t quite work, because it’s a least set such that $Y$ does not inject into $X$. With AC, given a surjection $X\to \aleph(X)$, take a section, hence an injection $\aleph(X)\to X$, which can’t exist. But without AC it’s not clear, let alone without EM.

The library lost its copy of Practical Foundation of Mathematics, and I suspect something like it could be in there (it is buying another copy).

• CommentRowNumber23.
• CommentAuthorDavidRoberts
• CommentTimeApr 23rd 2019
• (edited Apr 23rd 2019)

Hmm, this MO question tells me that one can have $X$ surjecting onto $\aleph(X)$ in the absence of AC. So it would have to be some other type of construction. I was thinking about $X=\mathbb{N}$ and $\aleph_1$, and how constructive the proof is that $\aleph_1$ has the properties it has, from different points of view. Certainly $\aleph_1 \nleq \aleph_0$ seems ok, but was thinking about the $\leq^*$ ordering, which is the one that the diagonal argument uses to show $X$ is smaller than $P(X)$.

• CommentRowNumber24.
• CommentAuthorDavidRoberts
• CommentTimeApr 23rd 2019

And this MO question discusses the dual notion

$h^*(A)$ is defined as the least ordinal $a$ such that there exists no surjection $f \colon B \to a$ with $B$ a subset of $A$.

however this uses the well-ordering of $ORD$ and seems rather impredicative. For the case of $\mathbb{N}$ there’s no problem, as the question indicates that $h^*$ returns the usual Hartogs number given a well-ordered set, but conceptually it’s messy.

Maybe something like taking the supremum of all well-orderings of subquotients of my given set? This seems better.

• CommentRowNumber25.
• CommentAuthorMike Shulman
• CommentTimeApr 24th 2019

How constructive/predicative are you trying to be? LEM (no AC) suffices for $ORD$ to be classically well-ordered.

Taking the supremum of all well-ordered quotients or subquotients does seem an obvious dual of the Hartogs construction, but I haven’t thought about whether it works.

• CommentRowNumber26.
• CommentAuthorDavidRoberts
• CommentTimeApr 24th 2019

Actually, I found one reference that defines a “surjective Hartogs number” $\aleph^*(X)$ of a set $x$ as being the set of ordinals admitting a surjection from $x$. With LEM this would be the same as the set of ordinals that are subquotients, so it’s reasonable. And $\aleph(x)\leq\aleph^*(x)\leq\aleph(P(x))$ for all sets, with $\aleph(x)=\aleph^*(x)$ for well-orderable $x$.

• CommentRowNumber27.
• CommentAuthorDavidRoberts
• CommentTimeApr 24th 2019

To answer your question, Mike, I think I’m after something that would work at least in any topos with nno. Defining the thing shouldn’t be a problem, it’s proving it has the desired properties that makes me pause.

And something that is more predicative would be good. I mean, what sort of ’cardinals’ can one construct in an arithmetic universe, for instance? Or less ambitiously, in a $\Pi$-pretopos with nno?

• CommentRowNumber28.
• CommentAuthorMike Shulman
• CommentTimeApr 24th 2019

I expect that that $\aleph^*(x)$ gives an ordinal definable in any topos with the property that $x$ doesn’t surject onto it. However, it’s not clear to me that one will be able to do very much with it beyond that; even the ordinary Hartogs number doesn’t seem to be very useful without LEM, since its defining property is purely negative.

• CommentRowNumber29.
• CommentAuthorDavidRoberts
• CommentTimeApr 24th 2019

Yes, I was trying to think if we can parallel the diagonal argument a bit more closely, in that given a function $\mathbb{N}\to \aleph^*(\mathbb{N})$ we can construct an element not in its image (and analogously for other cases). But the proofs all work by assuming surjectivity, then arriving at a contradiction. The case for the ordinary Hartogs number really does require Choice as well, since it’s possible for a set to surject onto its Hartogs number under the negation of AC.

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