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Condition (a) of lemma 3.4 in the paper “Countable ranks at the first and second projective levels” [M. Carl, P. Schlicht, P. Welch] is

$\alpha^{+L} = \omega_1,$

where $\alpha$ denotes any infinite countable ordinal and $\omega_1 = \omega_1^V$. I am unable to extract the exact meaning of this statement, but I cannot seem to find the explanation in the paper, and the paper does not provide the explanation of what the $\alpha^{+L}$ notation means.

If I interpret $\alpha^{+L}$ as "the successor of $\alpha$, as seen by $L$", it would not seem to make sense because the first uncountable ordinal is a limit ordinal, not a successor, and I cannot imagine the situation where $\omega_1^V$ would be equal to the successor of some countable ordinal $\alpha > \omega$ (in $L$).

What mathematical entity does $\alpha^{+L}$ denote ?

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    $\begingroup$ That's not at all what that lemma says. The lemma says that six different conditions are equivalent for $\alpha$ an infinite countable ordinal; what you've written is just condition $(a)$. As to the meaning of the notation, "$\alpha^{+L}$" means the thing $L$ thinks is the successor cardinal of $\alpha$, that is, the unique ordinal $\beta$ such that $L\models$ "$\beta$ is the smallest cardinal with $\vert \beta\vert>\vert\alpha\vert$." For example, if $V=L$ then $\omega_1^V=\omega^{+L}$, while if there is a measurable cardinal then $\alpha^{+L}<\omega_1^V$ for all countable $\alpha$. $\endgroup$
    – Noah Schweber
    Commented Oct 21, 2022 at 3:36
  • $\begingroup$ @NoahSchweber: Thank you for the explanation. Yes, of course, $\alpha^{+L} = \omega_1$ is only one of conditions, not a separate lemma. I need to clarify one thing: you said that $\beta$ is "the unique ordinal", then "the smallest cardinal". Did you mean that $\beta$ is "the smallest initial ordinal"? I am asking because the notions of cardinals and ordinals imply different things. $\endgroup$
    – lyrically wicked
    Commented Oct 21, 2022 at 4:02
  • $\begingroup$ I meant precisely what I wrote. In reality ($V$), all we know about $\beta$ is that it is an ordinal. But in $L$ it must be a cardinal. $\endgroup$
    – Noah Schweber
    Commented Oct 21, 2022 at 4:13
  • $\begingroup$ @NoahSchweber: then "$|\beta|$" denotes "the cardinality (cardinal number) of the cardinal $\beta$"? $\endgroup$
    – lyrically wicked
    Commented Oct 21, 2022 at 4:27
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    $\begingroup$ Also, it seems that the confusion stems from the VERY unfortunate use of some authors which write $x^+=x\cup\{x\}$, which then translates to ordinal succession in the infinite case; whereas the prevalent set theoretic use of the notation is very firmly cardinal succession. $\endgroup$
    – Asaf Karagila ♦
    Commented Oct 21, 2022 at 7:47

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The meaning of $$\alpha^{+L}$$ for $\alpha$ an infinite ordinal (countable or not) is just "The cardinal successor of $\alpha$ as seen by $L$." If that isn't clear, you may prefer the following phrasing:

"The unique ordinal $\beta$ such that $L\models$ "$\beta$ is the smallest cardinal greater than $\alpha$"."

There are a few things to note here (and the first addresses what seems to be your specific confusion):

  • The superscript-+ notation should always be understood (unless otherwise specified) as referring to the cardinal successor. Some texts use it to denote ordinal successor, and embarrassingly it is fairly standard to denote admissible ordinal successor as well, but the default interpretation should always be cardinal successor. The $L$-superscript, meanwhile, is just the usual relativization to $L$; it could be more clearly written as "$(\alpha^+)^L$."

  • All cardinals are ordinals (precisely, "cardinal" is shorthand for "initial ordinal"), so there is no type error in comparing cardinals and ordinals. That said, if you want to avoid mixing language like this, we could instead define $\alpha^{+L}$ as "The smallest ordinal $\beta$ such that $L\models$ "$\beta>\alpha$ and there is no injection from $\beta$ to $\alpha$"," which is equivalent.

  • While $L$ and $V$ may disagree about what is and is not a cardinal, they will not disagree about what is and is not an ordinal. Similarly, $L$ and $V$ will agree on ordinal comparison and ordinal arithmetic. This is why it is perfectly fine that my "unique ordinal $\beta$" clause appears outside the scope of the "$L\models$" clause.

  • There's nothing special about $L$ here; we could redo all of the above with respect to some other inner model $M$ if we liked.

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  • $\begingroup$ @AndreasBlass Whoops, good catch! Fixed. $\endgroup$
    – Noah Schweber
    Commented Oct 21, 2022 at 17:39

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