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There is a symbol for the sharp sign in unicode: ♯

I am unsure how wise it might be to convert this page to use this character instead of a superscript # (octothorpe)

-- nertzy

Not everyone can see Unicode characters. This is why we do not use Unicode unless absolutely necessary. Dysprosia 07:27, 15 Jul 2004 (UTC)

More relevantly these days, Solovay's article calls this set "O^#", where the # sign has horizontal and diagonal bars. Contrast with the musical ♯ sign, which has diagonal and vertical bars. LegionMammal978 (talk) 23:42, 7 June 2024 (UTC)[reply]

I have a couple of thoughts on this. We should use wp:commonname if possible, which might be 0^♯ if that's what people call it now, even if it originally started out as O/0#. Looking over our sources, of the couple I can see without hitting a paywall I see a variety of choices, including ♯. Since we can't use # in a page title we could also use ♯ as close enough, like ♯P, if that seems appropriate. I'm also not sure that the original authors of these math papers would regard # and ♯ as distinct characters (so which character is right might be, in a sense, indeterminate) — typesetting is not always very sophisticated, and they did end up calling "0#" "zero sharp", after all!

I would have liked to look into this more but it's difficult and I ran out of time. Dingolover6969 (talk) 09:35, 29 April 2025 (UTC)[reply]

Which way is the canonical? --84.229.190.204 (talk) 04:56, 31 August 2009 (UTC)[reply]

Good question, or really, good catch. Strictly speaking there is no single canonical way — there are lots of inessentially different codings. Should be reworded; not going to tackle it right now. Feel free to take a crack at it if you like. --Trovatore (talk) 06:34, 31 August 2009 (UTC)[reply]

It was introduced by Solovay (1967), based on the ideas in Silver (1971). I'm impressed, but perhaps that does not read quite as intended. Charles Matthews (talk) 11:34, 1 January 2010 (UTC)[reply]

I think we mean order-indiscernibles. Perhaps at least the first reference could be changed? (173.206.238.58 (talk) 08:57, 11 January 2010 (UTC))[reply]

Good point — I added a parenthetical. Maybe there's a lovelier solution available, but this at least addresses the issue. --Trovatore (talk) 09:32, 11 January 2010 (UTC)[reply]

The closed unbounded class of indiscernible ordinals can be "defined" as those ordinals which cannot be defined even given constant symbols for all smaller ordinals. That is, α is such an indiscernible if and only if for every formula, φ (α), of set theory with parameters for ordinals less than α,

${\displaystyle (L\models \phi (\alpha ))\implies \exists \beta <\alpha (L\models \phi (\beta )).}$

OK? JRSpriggs (talk) 01:07, 3 April 2026 (UTC)[reply]

How can this fail? Assuming that we are in a standard model (set theory), L is absolute if our model has all the ordinals. Otherwise, it is L_γ for some ordinal γ.

• A false positive would mean that α would appear to be undefinable (and thus indiscernible), in L_γ, even though it could actually be defined in the larger model. This seems unlikely to me currently.
• A false negative would mean that α would appear to have a definition from smaller ordinals in L_γ even though it could not be defined in L of the larger model. This could only happen if α is somehow encoded in γ, e.g. if ${\displaystyle \gamma =\tau _{\alpha +1}}$, the α+1^st admissible ordinal. This would probably be incompatible with L_γ being a model of ZFC.

JRSpriggs (talk) 23:42, 3 April 2026 (UTC)[reply]

Assuming ZFC+V=L. But let us add a special symbol for the indiscernibles, ${\displaystyle \operatorname {indis} }$.

An ordinal is a transitive set of transitive sets:

${\displaystyle \forall \alpha (\operatorname {ord} (\alpha )\iff \forall x\in \alpha \forall y\in x(y\in \alpha \land \forall z\in y(z\in x))).}$

Indiscernibles are ordinals:

${\displaystyle \forall \alpha (\operatorname {indis} (\alpha )\implies \operatorname {ord} (\alpha )).}$

Indiscernibles are unbounded:

${\displaystyle \forall \alpha (\operatorname {ord} (\alpha )\implies \exists \beta (\alpha \in \beta \land \operatorname {indis} (\beta )).}$

Indiscernibles are closed:

${\displaystyle \forall \alpha (\operatorname {ord} (\alpha )\land 0\in \alpha \land \forall \beta \in \alpha \exists \gamma \in \alpha (\beta \in \gamma \land \operatorname {indis} (\gamma ))\implies \operatorname {indis} (\alpha )).}$

Indiscernibles are undefinable from smaller ordinals:

${\displaystyle \forall \alpha _{1}\ldots \forall \alpha _{n}\forall \beta (\operatorname {indis} (\beta )\land \alpha _{1}\ldots \in \alpha _{n}\in \beta \land \phi (\alpha _{1},\ldots ,\alpha _{n},\beta )\implies \exists \gamma \in \beta \,\phi (\alpha _{1},\ldots ,\alpha _{n},\gamma )).}$

Indiscernibles share the same properties:

${\displaystyle \forall \alpha _{1}\ldots \forall \alpha _{n}\forall \beta \forall \gamma (\operatorname {indis} (\beta )\land \operatorname {indis} (\gamma )\land \alpha _{1}\ldots \in \alpha _{n}\in \beta \in \gamma \land \phi (\alpha _{1},\ldots ,\alpha _{n},\beta )\implies \phi (\alpha _{1},\ldots ,\alpha _{n},\gamma )).}$

Provided that being indiscernible is not mentioned in φ nor in the axiom schemas of ZFC. OK? JRSpriggs (talk) 16:16, 6 April 2026 (UTC)[reply]

The paradoxical nature of indiscernibles is especially apparent if one asks what is the cofinality of indiscernibles?. If they are a closed unbounded class, then they must contain members of every infinite cofinality. But then they cannot share all their properties, since they do not share their cofinality. It would be OK if they were all regular, because then the cofinality is not a connection to a smaller ordinal. But clearly many of them have cofinality ω which is too small to be even the least indiscernible.

The usual solution to this paradox is that all the indiscernibles are regular in L which does not know that they are a closed unbounded class, but they have the usual variety of cofinalities in V where their properties can vary (they are not actually "indiscernible"). Of course, this requires that V≠L. JRSpriggs (talk) 17:20, 7 April 2026 (UTC)[reply]

Another possible resolution of the paradox would be to weaken club set to stationary set, and change the property to:

Indiscernibles are stationary:

${\displaystyle \forall \alpha (\operatorname {regord} (\alpha )\land 0\in \alpha \land \forall \beta \in \alpha \exists \gamma \in \alpha (\beta \in \gamma \land \operatorname {indis} (\gamma ))\implies \operatorname {indis} (\alpha )).}$

where "regord" means regular ordinal. JRSpriggs (talk) 14:12, 10 April 2026 (UTC)[reply]