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I thought that the $L_{\alpha}$ notation denotes a set of reals that can occur on the output tape at stage $\beta$, where $\beta$ is any ordinal less than $\alpha$ and the input is an arbitrary real.

Similarly, the $L_{\alpha}[x_1, x_2, \ldots, x_{n-1}, x_n]$ notation would denote a set of reals that can occur on the output tape at stage $\beta$, where $\beta$ is any ordinal less than $\alpha$, assuming that an Infinite Time Turing Machine interprets the input as $n$ separate reals and $x_i$ denotes an $i$-th real.

But then I read that the recognizable closure of Infinite Time Turing Machines is equal to $L_{\sigma}$ (see Lemma 3.2 in the paper “Recognizable sets and Woodin cardinals: Computation beyond the constructible universe”), where $\sigma$ is a particular countable ordinal.

And here my understanding seems to contradict the intended meaning of the notation: assume that the input is a real that encodes any well-order $W$ of order-type greater than $\sigma$ and consider an Infinite Time Turing Machine that checks if $W$ is, in fact, a well-order. This machine will halt at stage $\tau$, where $\tau$ is greater than $\sigma$.

What is the meaning of $L_{\alpha}$ and $L_{\alpha}[x_1, x_2, \ldots, x_{n-1}, x_n]$ notations in relation to Infinite Time Turing Machines?

2 Answers2

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The $L_\alpha$s are levels of the $L$-hierarchy, which is defined as follows:

  • $L_0=\emptyset$.

  • For $\lambda$ a limit ordinal, $L_\lambda=\bigcup_{\beta<\lambda}L_\beta$.

  • $L_{\alpha+1}$ is the set of all $X\subseteq L_\alpha$ such that $X$ is definable (with parameters) in the structure $(L_\alpha; \in)$.

Note that this definition has nothing to do with ITTMs a priori; all interactions between the two notions are theorems.

"$L$" meanwhile refers to the class $\bigcup_{\alpha\in Ord}L_\alpha$. Note that this is somewhat similar to the $V$-hierarchy, the difference being the much more technical (although ultimately better-behaved) successor step. Also, note that elements of $L_\alpha$s are in general not reals, or sets of ordinals, or anything else particularly nice; for example, $\{\{\{\{\{\{0\}\}\}\}\}\}$ is an element of $L_8$.

It's a good easy exercise to check each of the following:

  • Whenever $\alpha<\beta$ we have both $L_\alpha\in L_\beta$ and $L_\alpha\subseteq L_\beta$.

  • Each $L_\alpha$ (and hence $L$ itself) is transitive.

  • $\alpha\subseteq L_\alpha$ for all ordinals $\alpha$.

  • We have $\vert L_\alpha\vert=\vert\alpha\vert$ for infinite $\alpha$.

Some fundamental, but not-at-all-obvious, facts about the $L$-hierarchy include:

  • Definability: the relation "$x\in L_\alpha$" (and hence the whole class $L$) is definable in the language of set theory.

  • Basic theory: the class structure $(L;\in)$ satisfies ZFC.

  • Condensation: if $M\preccurlyeq L_\beta$ for some $\beta$ then the Mostowski collapse of $M$ is equal to $L_\alpha$ for some $\alpha$. (Actually "$\preccurlyeq$" is overkill here, we just need "$\preccurlyeq_1$," but that's a bit more technical.) While opaque at first this is extremely powerful. For example, it lets us prove that $L\models GCH$ as follows:

    • Fix an infinite $L$-cardinal $\kappa$ and a set $X\in\mathcal{P}(\kappa)^L$; we want to show $X\in L_{(\kappa^+)^L}$.

    • By Lowenheim-Skolem, we get an elementary substructure $M$ of $L_\theta$ (with $\theta$ "large enough") such that $\vert M\vert=\kappa$, $L_\kappa\subseteq M$, and $X\in M$.

    • Let $m$ be the Mostowski collapse map of $M$. By Condensation we have $m: M\cong L_\beta$ for some $\beta$. Then by choice of $M$ we have $\beta<(\kappa^+)^L$ and $m(X)=X$, but this means $X\in L_{(\kappa^+)^L}$.


Similarly, for $x_1,..., x_n$ arbitrary sets (usually sets of ordinals), we can define the levels $L_\alpha[x_1,...,x_n]$ of the "relativized" $L$-hierarchy, and the "relativized" constructible universe $L[x_1,...,x_n]$, as above; the only change is in the successor step, where in order to compute $L_{\alpha+1}$ we look at the more complicated structure $$(L_\alpha; \in, x_1\cap L_\alpha,...,x_n\cap L_\alpha)$$ (that is, we can "identify $x_i$-ness").

Incidentally, we do not in general have $x\subseteq L[x]$ or $x\in L[x]$; for example, it's a good exercise to check that $L[\mathbb{R}]=L$ (basically, we can already tell whether something is a real number), so as long as $\mathbb{R}\not\subseteq L$ we'll have $\mathbb{R}\not\subseteq L[\mathbb{R}]$ and hence by transitivity of $L[\mathbb{R}]$ we won't have $\mathbb{R}\in L[\mathbb{R}]$ either. The notation "$L(x_1,...,x_n)$" refers to a different construction which on the plus side avoids this issue but on the minus(?) side does not generally result in a model of ZFC, but merely ZF. However, we do have $L[x_1,...,x_n]=L(x_1,...,x_n)\supseteq\{x_1,...,x_n\}$ whenever $x_1,...,x_n$ are each sets of ordinals.


Jech's book has a lot of good material on $L$, $L[...]$, and $L(...)$. I personally prefer Devlin's book, but that has a huge caveat - there's a serious mistake in the book one has to work around. Specifically, the development of the theory "$BS$" in Devlin is fundamentally flawed. However, this doesn't affect any of the significant results, and in particular the development of the relativized $L$-hierarchy is fine (see this Mathoverflow question for a description of the error, and how to address it).

Noah Schweber
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Its a nice question and I would like to know some details for the answer myself. So it would be better for other people to answer it in detail.

Though let me try to address a few points that I think I can address [regardless, this answer is slightly tentative, so please do point-out any mistakes in the answer so I can correct them]. It seems that one of the mistakes that you are making is definitely in the first paragraph.

I thought that the $L_\alpha$ notation denotes a set of reals that can occur on the output tape at stage $\beta$, where $\beta$ is any ordinal less than $\alpha$ and the input is an arbitrary real.

I think the basic idea is right but there are number of mistakes. First of all: (1) $L_\alpha$ doesn't denote a set of reals. (2) The basic idea you are describing is right but the last sentence is incorrect. One isn't allowed an input of an arbitrary real. But what is correct (I think) is that all sets of ordinals in $L_\alpha$ will have been generated in an OTM (starting from empty input) at a suitably large enough (ordinal) run-time.

(3) What is mentioned in previous paragraph is for OTMs. For ITTMs it will be a bit more complicated. First one needs to deal with how to encode a subset of ordinals and secondly one needs to add some qualifiers to above quote (see below).


Now one thing is that an ITTM, starting from empty input, will only ever generate codes for ordinals less than $\Sigma$ (the sup of AW-ordinals). And of course, now there must be a specific ITTM which actually generates codes for all ordinals (strictly) less than $\Sigma$ [and no other ordinal]. This follows directly from notion of simulation and the (downward) closure of AW-writeable values (i.e. they have no gaps).

Quite importantly, the previous paragraph means that there is a canonical way to associate a unique well-order (of $\mathbb{N}$) with every ordinal less than $\Sigma$. Now with this, as you can see, for any ordinal $\alpha<\Sigma$ any real can be thought of as a generating a subset of $\alpha$.

However, the previous paragraph doesn't answer an important question. Can an ITTM simulate an OTM? The answer is yes in a "certain sense". For all $\alpha<\Sigma$ an ITTM can simulate an OTM up-till the run-time $\alpha$. The details of the simulation aren't trivial but they aren't that difficult either. So while OTMs can generate all sets of ordinals in $L_\alpha$ (for arbitrary $\alpha$) eventually if we run them long enough, the ITTMs can't. However, any subsets of ordinals that OTMs generate below $\Sigma$, ITTMs can do that too by simply simulating an OTM all the way up to that point (in a certain sense, using a canonical well-order for any $\alpha<\Sigma$).


Hopefully other people can give more detailed and precise answers (this answer is merely meant to be a rough guideline and address a few specific points).

SSequence
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  • This isn't correct either - $L_\alpha$ does not only consist of sets of ordinals. – Noah Schweber Jan 29 '20 at 18:20
  • Thanks for the correction. I suspected it to be incorrect (but couldn't quite create a concrete example). I will remove that part (and replace others properly). If there is still an inaccurcy, please do let me know. – SSequence Jan 29 '20 at 18:36
  • Or maybe I should delete my answer? (not sure whether I should or shouldn't). I thought give bit of an alternative viewpoint. – SSequence Jan 29 '20 at 18:46
  • At a glance the bulk of this answer is fine, I was just pointing out that specific issue. – Noah Schweber Jan 29 '20 at 18:46
  • @NoahSchweber Can you give one such example where a set isn't a set of ordinals but belongs to $L_\alpha$ (I suspect it would be trivial but I can't quite think of it tbh). – SSequence Jan 29 '20 at 18:47
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    How about ${{{{{{0}}}}}}$? – Noah Schweber Jan 29 '20 at 18:50
  • What is meant by "one isn't allowed an input of an arbitrary real"? The term "real" here implies an infinite sequence of bits, and the notion of recognizability implies that for any (arbitrary) "real" $r$, there is only one possibility: (i) $r$ is in the recognizable closure; (ii) $r$ is not in the recognizable closure. – lyrically wicked Jan 30 '20 at 02:37
  • I just meant "isn't allowed" in the sense that if we are talking about enumeration of S.O.'s (sets of ordinals) of $L_\alpha$ (for arbitrary $\alpha$). – SSequence Jan 30 '20 at 03:41