Talk:Inaccessible cardinal

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[edit] Is aleph-null inaccessible?

Anonymous Coward: Hey, it sure seems to me that aleph-null fits the description of an inaccessible cardinal. Am I missing something?

No. \aleph_0 would satisfy the requirements for an inaccessible cardinal; that's why the definition specifically talks only about kardinals κ > \aleph_0. -- Schnee 05:59, 30 Sep 2004 (UTC)
This is a usage question. A few authors count \aleph_0 as an inaccessible, and even as a measurable. I think Drake does; don't have a copy here to check. --Trovatore 01:55, 14 July 2005 (UTC)
Suppes explicitly counts \aleph_0 as an inaccessible (Suppes, Patrick [1960] (1972). "§ 8.3 Axioms Which Imply the Axiom of Choice", Axiomatic Set Theory, unabridged and corrected republication, Dover Publications, 251. ISBN 0-486-61630-4. ). — Tobias Bergemann 18:30, 26 June 2006 (UTC)

[edit] Rename or merge

By default, inaccessible means strongly inaccessible. The current page should be moved to "weakly inaccessible", and "strongly inaccessible" should be put here. Or better, the pages should be merged. --Trovatore 01:40, 14 July 2005 (UTC)

Merge complete. Result could still use some polishing--the two paragraphs on consistency strength could be usefully turned into one. --Trovatore 02:58, 14 July 2005 (UTC)

[edit] Consistency with ZFC

In fact, it cannot even be proven that the existence of strongly inaccessible cardinals is consistent with ZFC (as the existence of a model of ZFC + "there exists a strongly inaccessible cardinal" can be used to prove the consistency of ZFC)

I find this confusing. The existence of a model of ZFC + anything can be used to prove the consistency of ZFC, since such a model is also a model of ZFC. Josh Cherry 14:13, 31 July 2005 (UTC)

Good point. See fix. --Trovatore 15:03, 31 July 2005 (UTC)
OK, thanks. I'm not saying this is wrong, but I still don't get the argument. Proving that the existence of inaccessible cardinals is consistent with ZFC would, it seems from this, only show that the consistency of ZFC is consistent with ZFC, i.e., that if ZFC is consistent then ZFC + consis(ZFC) is consistent. Can't we prove that? Josh Cherry 15:14, 31 July 2005 (UTC)
No. If you could, then ZFC would prove that ZFC is consistent, contradicting Second Incompleteness. --Trovatore 15:31, 31 July 2005 (UTC)
Um, so I was slightly off here. If you could prove that Con(ZFC)-->Con(ZFC+Con(ZFC)), then ZFC+Con(ZFC) would prove its own consistency (and therefore be inconsistent). That's only minutely better or more plausible than ZFC itself being inconsistent. --Trovatore 16:01, 31 July 2005 (UTC)

[edit] Other descriptions

Is the first weakly inaccessible cardinal the same as \aleph_{\aleph_\ddots}? I've seen references to this, but it seems singular. Jim Apple 11:18, 18 August 2005 (UTC)

Well, depends on what you mean by that notation. Every weakly inaccessible cardinal is a fixed point of the ℵ function. I.e. if κ is weakly inaccessible then κ = ℵκ. But it's not the first fixed point, which is the limit of the ω-sequence
\aleph_0, \aleph_{\aleph_0}, \aleph_{\aleph_{\aleph_0}},\ldots
You're quite right that this limit has cofinality ω --Trovatore 21:28, 18 August 2005 (UTC)
In fact, the first weakly inaccesible cardinal τ isn't the κ'th fixed point of \aleph for any κ < τ, right? Jim Apple 06:05, 19 August 2005 (UTC)
Right again. --Trovatore 17:43, 20 August 2005 (UTC)

Why isn't the first fixed point of \aleph inaccesible? Which requirement is missing?--lion 21.2.2006

It's not a regular cardinal; its cofinality is ω, as you can see by the existence of the ω-sequence above that approaches it. --Trovatore 20:55, 21 February 2006 (UTC)

[edit] recursive definition of α-inaccessibility incomplete?

The article defines α-inaccessibility to mean that the set of smaller β-inaccessibles is cofinal, for all β<α. This recursive definition seems incomplete to me. Don't we also need to define a base for the recursion? I would guess we should define 0-inaccessible to mean regular limit cardinal? -lethe talk + 00:55, 21 May 2006 (UTC)

In my definition, "For any ordinal α, a cardinal κ is α-inaccessible iff κ is inaccessible and for every ordinal β < α, the set of β-inaccessibles less than κ is unbounded in κ (and thus of cardinality κ, since κ is regular).", the base is "κ is inaccessible", i.e. an uncountable regular strong limit cardinal. When α=0, the other part of the right-hand side, which refers to β, contributes nothing. So 0-inaccessible is the same as inaccessible. JRSpriggs 08:20, 21 May 2006 (UTC)
I think I somehow failed to see that phrase "κ is inaccessible". It makes sense now, thank you for explaining. -lethe talk + 13:47, 21 May 2006 (UTC)

You are welcome. At the risk of belaboring the obvious, let me expand my explanation. α+1-inaccessible means an α-inaccessible limit of α-inaccessibles. And when λ is a limit, λ-inaccessible means α-inaccessible for all α < λ. JRSpriggs 06:13, 22 May 2006 (UTC)

On the contrary, your expansions are quite welcome, because it's an unfamiliar concept that I'm still digesting. Stated loosely, a number is 1-inaccessible if there are an inaccessible number of inaccessible numbers below it. And so on up the recursive ladder. -lethe talk + 15:01, 22 May 2006 (UTC)
Yes. One could make a weak analogy between α-inaccessibles and multiples of \omega^{\omega^{\alpha}}. JRSpriggs 09:26, 28 May 2006 (UTC)

[edit] what does ∞ mean?

In the sentence "This is a relatively weak large cardinal axiom since it amounts to saying that ∞ is 1-inaccessible in the language of the next section.", what does the symbol ∞ mean? -lethe talk + 04:22, 4 June 2006 (UTC)

"∞" is a symbol for the class of all ordinals. That is, it refers to the least ordinal which is not an element of your model, i.e. it is κ if your model is Vκ. JRSpriggs 02:08, 5 June 2006 (UTC)
I see. Would you object to adding that to the article? -lethe talk + 02:26, 5 June 2006 (UTC)

By the way, you might be interested to read the discussion about that section which I had last week with User:Trovatore at his talk page. -lethe talk + 02:28, 5 June 2006 (UTC)

Done. And I already read that talk when you guys posted it. JRSpriggs 02:36, 5 June 2006 (UTC)


[edit] what does ω mean?

It's used in the second sentence and not defined or linked to. --Michael C. Price talk 13:39, 26 June 2006 (UTC)

I have removed the inline definitions of limit cardinal, regular cardinal and uncountable, linking to the respective pages instead. I actually think this makes the article more readable. I also replaced the reference to ω (the first infinite ordinal number) with references to ℵ0 (alef-null), the first infinite cardinal number. —Tobias Bergemann 15:04, 26 June 2006 (UTC)
Thanks. BTW the aleph symbol displays as a square on my browser, so I changed it to the form used at the beginning of this talk page: \aleph --Michael C. Price talk 15:12, 26 June 2006 (UTC)
Thanks. The corresponding HTML entity &alefsym; is apparently not widely supported. —Tobias Bergemann 15:15, 26 June 2006 (UTC)

[edit] can a cardinal k be k-hyper-inaccessible?

can a cardinal k be k-hyper-inaccessible? —The preceding unsigned comment was added by 80.41.84.38 (talk) 17:44, 6 April 2007 (UTC).

Yes, or rather, it is not known to be inconsistent that such cardinals exist. See also Mahlo cardinal#Example: showing that Mahlo cardinals are hyper-inaccessible. But, of course, κ cannot be κ+1-hyper-inaccessible. JRSpriggs 08:20, 7 April 2007 (UTC)

[edit] What is a standard model?

  • "Vκ is a standard model of ZFC which contains no strong inaccessibles."
  • "Similarly, either V contains no weak inaccessible or, taking κ to be the smallest ordinal which is weakly inaccessible relative to any standard sub-model of V, then Lκ is a standard model of ZFC which contains no weak inaccessibles."

These two sentences are from the Models and consistency section of this article. What is meant by a 'standard model' here? I can't find the definition at model theory article, and it seems obvious that it is not the physical standard model. --Acepectif (talk) 06:50, 15 January 2008 (UTC)

It probably means a model whose membership relation is the true membership relation. I'm not sure there's a completely *ahem* standard meaning for the term, though. Some people might use it to mean a model that's a Vα; others might use it to mean any model whose membership relation is wellfounded. From a model whose membership relation is wellfounded you can recover an isomorphic one whose membership relation is the true one, so the first and third definitions aren't very different -- this is by the Mostowski collapsing lemma; do we have an article on that? --Trovatore (talk) 07:25, 15 January 2008 (UTC)
It's at Mostowski collapse lemma; I'll redirect the other name. I also was concerned about the terminology "a standard model" at some point. I accept "the standard model" as commonly understood terminology, but I would prefer to say "well-founded transitive" if that is what is meant by "a standard model". But I have never had strong enough feelings to follow up on it. — Carl (CBM · talk) 20:14, 15 January 2008 (UTC)
Thanks for replies. --Acepectif (talk) 01:03, 16 January 2008 (UTC)