## July 8, 2021

### Large Sets 9.5

#### Posted by Mike Shulman Previously: Part 9. Next: Part 10

In the last comment thread, Tom invited me to write a post about some of the sizes of sets in between inaccessible sets and measurable sets. I’m not sure he was serious, but I’m going to take him up on it anyway. (-:

There are a lot of sizes of sets in between inaccessibles and measurables, but in this post I’ll just talk about “higher inaccessible” sets and Mahlo sets. I think these are worth thinking a bit about, especially as a followup to Tom’s very nice description of various kinds of large sets that are smaller than inaccessibles, because they can be thought of roughly as continuing the project of “making things that can’t be reached from below”. Measurable sets and their ilk feel to me like less of a straightforward continuation of that project, bringing in somewhat more exotic definitions that turn out to make them very large.

In addition, I hope to give a very fragmentary idea of how must vastly bigger than an inaccessible set a measurable set must be, by exploring just a bit of the terrain in between.

Some of the sizes of large sets that Tom introduced can be described roughly as “un-reachable from below by operations X”, where X often involves the assumed existence of smaller large sets. For instance, if beths exist (post 6), then the beth fixed points (post 7) are the sets that are un-reachable from below by constructing beths. Inaccessible sets are un-reachable from below by any of the other operations that Tom mentioned. But what about sets that are un-reachable from below by constructing inaccessibles?

In order to make “constructing inaccessibles” into an operation, consider the property that every set is smaller than some inaccessible. We then have an operation sending any set to the smallest inacessible that’s larger than it. A set is called 1-inaccessible if it’s unreachable from below by this operation. More precisely, $X$ is 1-inaccessible if it is inaccessible and for any $A\lt X$, there is an inaccessible $B$ such that $A\lt B\lt X$. Thus, essentially by definition, we have

For any 1-inaccessible set $X$, there are unboundedly many inaccessible sets $\lt X$.

If $X$ is 1-inaccessible, then the collection of sets $\lt X$ model ETCS, and also model “every set is smaller than some inaccessible”. Thus, as before we have (assuming, as always, that ETCS is consistent):

It is consistent with ETCS + (every set is smaller than some inaccessible) that there are no 1-inaccessibles.

1-inaccessible sets are somewhat relevant to the modeling of type theory. When modeling ordinary type theory, we often assume there are a countable number of inaccessibles, to model a countable hierarchy of universes. This isn’t good enough if we want to state things about all of those universes, so sometimes we introduce an $\omega$th universe. And if we want to be able to use type theory to reason about arbitrary sets, it’s natural to assume that every set is contained in some universe, i.e. every set is smaller than some inaccessible.

Of course, putting an index “1” in front immediately leads us to the next generalization. A set is 2-inaccessible if it is inaccessible and there are unboundedly many 1-inaccessibles smaller than it. And it is 3-inaccessible if it is inaccessible and there are unboundedly many 2-inaccessibles smaller than it. Ad infinitum…

What kind of infinitum? Well, we clearly get $n$-inaccessibles for all natural numbers $n$. We can then say a set is $\omega$-inaccessible if it is $n$-inaccessible for all natural numbers $n$. Then we can say it’s $(\omega+1)$-inaccessible if there are unboundedly many $\omega$-inaccessibles smaller than it. And $(\omega+2)$-inaccessible if there are unboundedly many $(\omega+1)$-inaccessibles smaller than it. Ad infinitum…

More generally, for any well-ordered set $W$, a set $X$ is $W$-inaccessible if it is inaccessible and for every well-ordered set $V\prec W$, there are unboundedly many $V$-inaccessibles smaller than $X$. So after $(\omega+n)$-inaccessibles we have $(\omega+\omega)$-inaccessibles, and so on to $\omega^2$-inaccessibles, $\omega^\omega$-inaccessibles, and so on. After exhausting the countable ordinals we have $\beth_1$-inaccessibles, $\beth_2$-inaccessibles, and so on. Then $W$-inaccessibles where $W$ is the smallest beth fixed point. Then $W$-inaccessibles where $W$ is the smallest inaccessible. Then $W$-inaccessibles where $W$ is the smallest 1-inaccessible.

If we aren’t tired yet, we can imagine a well-ordered set $W$ whose underlying set is itself $W$-inaccessible. According to wikipedia, some people call this a hyper-inaccessible; one might also call it an “inaccessibility fixed point”.

But, of course, we’re not done: we can consider an inaccessible set such that there are unboundedly many hyper-inaccessibles smaller than it, which we can call 1-hyper-inaccessible. Then we have 2-hyper-inaccessibles, which have unboundedly many 1-hyper-inaccessibles smaller than them. And so on, until a $W$-hyper-inaccessible set (for some well-ordered set $W$) has unboundedly many $V$-hyper-inaccessibles smaller than it for every well-ordered set $V\prec W$. And, of course, a hyper-hyper-inaccessible set is the underlying set of a well-ordering $W$ that is itself $W$-hyper-inaccessible. It’s natural to expect to be able to define “hyper$^W$-inaccessibles” for any well-ordered set $W$, although I haven’t seen this written down.

I already find it kind of mind-blowing to think about how big these sets are. At first it seems like an inaccessible must be very large, especially after seeing how it’s bigger than all the kinds of large sets Tom discussed in parts 1-8. Then it seems like a 1-inaccessible must be unimaginably larger than that: not only are there an inaccessible number of inaccessible sets smaller than a 1-inaccessible $X$, there are $X$-many inaccessibles smaller than it! But 1-inaccessibles are just the beginning; with limit and “diagonalization” constructions we can quickly leap into incredibly larger and larger sets.

And yet, as always, once we can describe an operation that produces large sets, that same description gives us a way to leap beyond it. Define a set $X$ to be Mahlo if it is inaccessible and given any notion of “large set” such that there are unboundedly many large sets smaller than $X$, there exists an inaccessible $Y\lt X$ such that there are unboundedly many large sets smaller than $Y$.

To show how a Mahlo set is larger than all the ones we’ve seen so far, first let $A\lt X$ and call a set “large” if it is larger than $A$. Then there are certainly unboundedly many such sets smaller than $X$, so there is an inaccessible $Y$ such that, in particular, $A\lt Y\lt X$. Varying $A$, we see there are unboundedly many inaccessibles smaller than $X$, i.e. $X$ is 1-inaccessible.

But now, we can instead call a set “large” if (for some fixed $A$) it is larger than $A$ and inaccessible. Then by what we just proved, there are unboundedly many large sets smaller than $X$, so there is an inaccessible $Y$ such that $A\lt Y\lt X$ and there are unboundedly many inaccessibles sets smaller than $Y$. Hence $Y$ is 1-inaccessible, and so (varying $A$) we’ve just proven that $X$ is 2-inaccessible. Proceeding in this way, we can show that a Mahlo set is $W$-inaccessible for any well-ordered set $W$ (Edit: as long as $W$ is smaller than it is), and indeed also $W$-hyper-inaccessible for any $W$, and so on.

Are we done? Of course not! Now we can call a set $X$ 1-Mahlo if it is inaccessible and given any notion of “large set” such that there are unboundedly many large sets smaller than $X$, there exists a Mahlo $Y\lt X$ such that there are unboundedly many large sets smaller than $Y$. Then we have 2-Mahlo sets, 3-Mahlo sets, $W$-Mahlo sets, $W$-hyper-Mahlo sets, etc. Wikipedia tells me that there is a notion of “greatly Mahlo” that’s presumably even larger than these, and so on.

And yet, despite how unimaginably large all of these sets must be, set theorists have come up with ways to describe sets that must be incomprehensibly larger than all of them. In fact, all these kinds of hyper-inaccessibles and Mahlos, as well as many sets even larger than them, are sometimes called “small large sets”. Next time Tom will tell us about measurable sets, which are the first kind of “large large set” – which are much, much larger than all of the small large sets!

Posted at July 8, 2021 4:02 AM UTC

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### Re: Large Sets 9.5

Posted by: Tom Leinster on July 8, 2021 9:39 AM | Permalink | Reply to this

### Re: Large Sets 9.5

Two small things…

A typo: where you define $W$-inaccessibility, “$V \preceq W$” should be “$V \prec W$”.

And a language question: what is the accepted meaning of “it is consistent with $P$ that $Q$”? (Here $P$ and $Q$ are collections of axioms.) I took it to mean “if $P$ is consistent then $P + Q$ is consistent”. And that’s how I used it in earlier posts. But I have a little worry in the back of my mind, because I feel like I’ve read things roughly of the form “assuming $P$ is consistent, $P$ is consistent with $Q$”, which would be redundant if my interpretation were correct. And you’ve written something along the same lines just before your second displayed result. I’m not wanting to pick holes here; just taking this opportunity to clear up my doubts over how this language is used.

Posted by: Tom Leinster on July 8, 2021 1:54 PM | Permalink | Reply to this

### Re: Large Sets 9.5

Thanks, fixed the typo.

I’ve always understood “it is consistent with P that Q” to mean just “P+Q is consistent”. In particular, note that this is equivalent to “P does not disprove Q” (whereas if P is inconsistent, then it does disprove Q — as well as proving it, of course). But maybe a logician will show up and set us straight.

Posted by: Mike Shulman on July 8, 2021 2:20 PM | Permalink | Reply to this

### Re: Large Sets 9.5

Don’t you want “If P is consistent then P+Q is consistent”?

Posted by: Noah Snyder on July 8, 2021 8:14 PM | Permalink | Reply to this

### Re: Large Sets 9.5

That’s what Tom was saying he meant, but I was saying I thought it just meant “P+Q is consistent”. Are you saying you agree with him?

Posted by: Mike Shulman on July 8, 2021 8:31 PM | Permalink | Reply to this

### Re: Large Sets 9.5

Don’t you want to be able to say “it is consistent with ZFC that Y” even though you don’t know ZFC is consistent?

Posted by: Noah Snyder on July 8, 2021 11:50 PM | Permalink | Reply to this

### Re: Large Sets 9.5

I’m not sure I know what that’s supposed to mean. If ZFC is inconsistent, then so is ZFC+Y.

Posted by: Todd Trimble on July 9, 2021 1:07 AM | Permalink | Reply to this

### Re: Large Sets 9.5

I took Noah’s point to be that one would like to be able to say things like “It is consistent with ZF that C holds”, without worrying about consistency of ZF.

Indeed, surely one would like to be able to say something vacuous like “It is consistent with ZFC that ZFC”, which, according to the definition “It is consistent with P that Q” = “P + Q is consistent”, we cannot.

Posted by: L Spice on July 10, 2021 6:08 PM | Permalink | Reply to this

### Re: Large Sets 9.5

I would argue that if ZFC is inconsistent, it is not consistent with ZFC that ZFC. In support of this, if Y is intrinsically inconsistent, wouldn’t we want to say that “it is not consistent with T that Y” for any theory T? Including the case when T=Y? So I would say the statement “it is consistent with ZFC that ZFC” is not vacuous.

Posted by: Mike Shulman on July 10, 2021 6:14 PM | Permalink | Reply to this

### Re: Large Sets 9.5

Maybe it will clarify things if we take P to be a theory that’s known to be inconsistent, like set theory with Unrestricted Comprehension.

Would you want to say “it is consistent with ZFC+UC that $0=1$”?

Would you want to say “it is consistent with ZFC+UC that $0\neq 1$”?

Note that since ZFC+UC is inconsistent, it proves both $0\neq 1$ and $0=1$. Do you want a theory to be “consistent with” a statement whose negation it proves?

Posted by: Mike Shulman on July 9, 2021 2:27 AM | Permalink | Reply to this

### Re: Large Sets 9.5

Hmm, to me this has the flavour of hard cases make bad law. Most of the time when people say “it is consistent with $P$ that $Q$”, the consistency of $P$ is unknown, right? And it’s rare that $P$ is known to be inconsistent. Maybe the usage should be whatever’s most convenient for that majority case.

I think I’m OK with things of the form

It is consistent with $0 = 1$ that $0 \neq 1$.

Anyway, I seem to be acquiring an opinion, which I was trying to avoid :-)

Posted by: Tom Leinster on July 9, 2021 11:51 AM | Permalink | Reply to this

### Re: Large Sets 9.5

But in mathematics (as opposed to law), we do try to make definitions that are correct even in degenerate cases.

Posted by: Mike Shulman on July 9, 2021 6:38 PM | Permalink | Reply to this

### Re: Large Sets 9.5

Posted by: Tom Leinster on July 9, 2021 8:30 PM | Permalink | Reply to this

### Re: Large Sets 9.5

If T is a theory and A is an additional axiom, T + A is said to be consistent relative to T (or simply that A is consistent with T) if it can be proved that if T is consistent then T + A is consistent.

On the other hand, in particular cases wikipedia does sometimes adhere to my usage instead, like at continuum hypothesis:

The existence of an inner model of ZF in which additional axioms hold shows that the additional axioms are consistent with ZF, provided ZF itself is consistent.

and large cardinal:

A necessary condition for a property of cardinal numbers to be a large cardinal property is that the existence of such a cardinal is not known to be inconsistent with ZFC and it has been proven that if ZFC is consistent, then ZFC is consistent with the statement that “no such cardinal exists.”

Posted by: Mike Shulman on July 9, 2021 4:45 AM | Permalink | Reply to this

### Re: Large Sets 9.5

I’m trying to digest the recursive operations in play here. Let me try thinking out loud.

Suppose that we have a property of sets; let’s call it being “purple”. The idea is that purpleness is some form of largeness, but I’m not going to assume anything about it apart from isomorphism-invariance.

First definition:

For a well-ordered set $W$, a set is $W$-purple if $X$ is purple and for all $V \prec W$, there are unboundedly many $V$-purple sets $\lt X$.

This is a recursive definition. It starts like this: $0$-purple $=$ purple. Then $X$ is $1$-purple if it is purple and there are unboundedly many purple sets $\lt X$. And so on, as per Mike’s explanation in the case purple $=$ inaccessible.

Clearly, if $V \prec W$ then $W$-purple implies $V$-purple. So the larger $W$ is, the harder it is to be $W$-purple.

Second definition:

A set $X$ is hyperpurple if it is $I(X)$-purple.

Here $I(X)$ is the initial well-ordered set with underlying set $X$. An equivalent definition, writing $U$ for underlying set: $X$ is hyperpurple if $X$ is purple and whenever $W$ is a well-ordered set with $U(W) \lt X$, there are unboundedly many $W$-purple sets $\lt X$.

Taking “purple” to mean “inaccessible” reproduces Mike’s definitions of $W$-inaccessible and hyperinaccessible.

Mike goes on to take purple to mean “hyperinaccessible”, then “hyperhyperinaccessible”, and so on… but let me try taking more simple meanings of purple.

What if we take “purple” to mean “a set”, i.e. all sets are purple?

I don’t even see how this most simple case works out. Here’s where I am:

• $X$ is a 1-set (i.e. 1-purple for this trivial meaning of purple) if there are unboundedly many sets $\lt X$. This means that $X$ is empty or a weak limit. Equivalently, $X$ is empty or of the form $\aleph_{\omega \cdot V}$ for some well-ordered set $V$.

• $X$ is a 2-set if there are unboundedly many (empty sets or) weak limits $\lt X$. I think this is equivalent to $X$ being either empty or of the form $\aleph_{\omega^2 \cdot V}$ for some well-ordered set $V \gt 0$.

• For an arbitrary well-ordered $W$, I guess $X$ is a $W$-set if $X$ is either empty or of the form $\aleph_{\omega^W \cdot V}$ for some well-ordered set $V \gt 0$. Here $\omega^W$ is an ordinal exponential.

• What are the “hypersets”, i.e. the hyperpurple sets when purple means “a set”? The empty set is one, but apart from that… Are they the aleph fixed points? The weakly inaccessible sets?

It might be more convenient to take purple to mean “nonempty”. This excludes the case of the empty set throughout the points above. Then the question becomes: what are the hypernonempty sets?

Posted by: Tom Leinster on July 8, 2021 9:26 PM | Permalink | Reply to this

### Re: Large Sets 9.5

If your guess about $W$-sets is right (which seems plausible), it suggests that a nonempty hyper-set (with its initial well-ordering) should satisfy $W = \aleph_{\omega^W\cdot V}$ for some $V$. In particular, any fixed point of the operation $W\mapsto \aleph_{\omega^W}$ would be one. I would expect such fixed points to exist in ZFC by the fixed-point lemma for normal functions, so in particular they needn’t be weakly inaccessible. But they don’t seem quite the same as aleph fixed points either.

Posted by: Mike Shulman on July 8, 2021 9:41 PM | Permalink | Reply to this

### Re: Large Sets 9.5

I am getting the impression that many (probably not all) large cardinal definitions are definable as fixed points of some operator. This make me wonder if the operators themselves make sense as objects in and of themselves (they are sort of functions whose domain is all cardinals).

Posted by: Ana N Mouse on July 9, 2021 2:46 AM | Permalink | Reply to this

### Re: Large Sets 9.5

You may be thinking of a normal function. However, I think the impression you’re getting doesn’t last much past the Mahlos. Larger large cardinals, like weakly compacts and especially measurables, aren’t really any kind of fixed point, as far as I know.

Posted by: Mike Shulman on July 9, 2021 3:47 AM | Permalink | Reply to this

### Re: Large Sets 9.5

I think that once one gets up to measurable cardinals, that is, up towards the large large cardinals, they can start to be defined as critical points of nontrivial elementary embeddings. And I suspect that all large cardinals above measurables can be defined like this.

Posted by: David Roberts on July 9, 2021 7:48 AM | Permalink | Reply to this

### Re: Large Sets 9.5

up towards the large large cardinals, they can start to be defined as critical points of nontrivial elementary embeddings.

It’s good that you bring this up, because I’m conscious that this kind of definition is over the horizon of what I’m covering in this series.

Right at the start of Part 1, I wrote this:

My primary interest is not actually in large cardinals themselves. What I’m really interested in is exploring the hypothesis that everying in traditional, membership-based set theory that’s relevant to the rest of mathematics can be done smoothly in categorical set theory.

One could spend entire lifetimes going through set theory research and attempting to recast it in categorical terms. (The question is not really whether everything can be done categorically, but whether things look smooth and natural.) And personally, I don’t intend to spend very much of a lifetime doing this kind of thing.

There’s always going to be something just over the horizon. As far as this series is concerned, there are two such things I’m particularly aware of. One is the style of large cardinal axiom that you mention. The other is PCF. I don’t know how either of these things look in categorical set theory.

Posted by: Tom Leinster on July 9, 2021 11:41 AM | Permalink | Reply to this

### Re: Large Sets 9.5

I don’t know of a good categorical statement of V=L, or even a categorical construction of L. In my mind this is an important open problems in structural set theory.

Posted by: Mike Shulman on July 10, 2021 3:45 PM | Permalink | Reply to this

### Re: Large Sets 9.5

Mike wrote:

Define a set $X$ to be Mahlo if it is inaccessible and given any notion of “large set” such that there are unboundedly many large sets smaller than $X$, there exists an inaccessible $Y\lt X$ such that there are unboundedly many large sets smaller than $Y$.

I’ve been trying to process this. It’s different from the definition of Mahlo in Jech’s or Kanamori’s book, or Wikipedia, or, for that matter, your “Set theory for category theory”. But presumably it’s equivalent.

I see why Mahlo in the sense of your post implies Mahlo in the “usual” sense (I mean, according to the definition given in those other sources). But why is the converse true?

(For safety’s sake, let me say that I interpret the word “smaller” in your post as “strictly smaller”, as I think you do too.)

Here’s what I believe to be a structural translation of the usual definition. For a set $X$, write

$K(X) = \{\text{iso classes of sets } \ \lt X\}.$

Then:

• $K(X)$ carries an order, cardinal inequality. A subset $C$ is unbounded if, well, it has no upper bound: whenever $Y \in K(X)$, there is some set $Z$ with $Y \lt Z \lt X$ and $[Z] \in C$. Here $[Z]$ means the iso class of $Z$.

• $K(X)$ carries a topology, the order topology. In that topology, the closure of a subset $L$ of $K(X)$ consists of the elements that can be expressed as the supremum of some nonempty subset of $L$. (At least, that’s what the closure is if $X$ is not a successor; I haven’t thought about the case where it is.)

Then $X$ is Mahlo if it is inaccessible and every closed unbounded subset of $K(X)$ contains at least one inaccessible.

If $X$ is Mike-Mahlo (I mean, Mahlo in the sense of your post) then it’s Mahlo. For take a closed unbounded subset $C$ of $K(X)$. Since $C$ is unbounded, the Mike-Mahlo property implies that there’s an inaccessible $Y \lt X$ such that there are unboundedly many elements of $C$ smaller than $Y$. Since $C$ is closed, this implies that $[Y] \in C$. So $C$ contains an inaccessible, as required.

But I’m not seeing how to prove the converse. Take a Mahlo set $X$ and an unbounded subset $L$ of $K(X)$. To prove that $X$ is Mike-Mahlo, we have to find an inaccessible $Y \lt X$ such that there are unboundedly many elements of $L$ smaller than $Y$. My first thought was to apply the definition of Mahlo to the closure of $L$ in $K(X)$. This gives us an inaccessible in $K(X)$ that’s a sup of elements of $L$. But we can’t take that as our “$Y$”, can we? E.g. it could be the smallest element of $L$. Help!

Posted by: Tom Leinster on July 15, 2021 8:11 PM | Permalink | Reply to this

### Re: Large Sets 9.5

I hope it’s equivalent. I haven’t seen it written like this before either, but I was trying to phrase it in a simple way without introducing notions like “club sets”.

How about this: let $X$ be Mahlo and $L$ unbounded in $K(X)$. Let $L' \subseteq K(X)$ consist of those $Y\lt X$ such that $L$ is unbounded below $Y$; then it suffices to show that $L'$ is closed and unbounded.

For unboundedness, if $Y\lt X$ then since $L$ is unbounded, there is a $Z_0$ with $Y\lt Z_0\lt X$ and $Z_0\in L$. By the same reasoning, there is a $Z_1$ with $Z_0 \lt Z_1\lt X$ and $Z_1\in L$. Proceed inductively and let $Z = \sup_{n\in\mathbb{N}} Z_n$; then $Z\in L'$ and $Y \lt Z\lt X$.

For closedness, suppose $Y\lt X$ is a limit point of $L'$, which is to say that $L'$ is unbounded below $Y$. We want to show $Y\in L'$, which is to say that $L$ is unbounded below $Y$. Now by assumption, for any $Z\lt Y$ there is a $W\in L'$ with $Z\lt W\lt Y$. But $W\in L'$ means $L$ is unbounded below $W$, so there is a $U\in L$ with $Z\lt U\lt W$, hence $Z\lt U\lt Y$. Since this holds for any $Z\lt Y$, it follows that $L$ is unbounded below $Y$ as desired.

Posted by: Mike Shulman on July 15, 2021 8:57 PM | Permalink | Reply to this

### Re: Large Sets 9.5

Fantastic; thanks!

• In the proof of unboundedness, where we put $Z = \sup_{n \in \mathbb{N}} Z_n$, we’re implicitly using the fact that $X$ has unbounded cofinality (which it does because it’s inaccessible, hence uncountable and regular).

• The proof that $L'$ is closed is a special case of the proof that in an arbitrary topological space, the Cantor–Bendixson derivative of any subset is closed. Presumably that’s why you chose the notation $L'$.

(In my comment to which you were replying, I messed up the definition of the closed sets in $K(X)$, though I got the closure operator right. I’ve fixed it now.)

Mike wrote:

I haven’t seen it written like this before either, but I was trying to phrase it in a simple way without introducing notions like “club sets”.

Yes, that’s great. It took me a long time to get round to learning the definition of Mahlo set, for exactly that reason.

What would happen is that I’d get the urge to learn the definition of Mahlo and I’d go to Wikipedia. The Wikipedia page on Mahlo cardinals would tell me that the definition depends on the definition of stationary set. The Wikipedia page on stationary sets would tell me that that depends on the definition of club set. The Wikipedia page on club sets would tell me that it’s just an abbreviation for “closed and unbounded”. Now “closed” and “unbounded” sound like things that should be easy enough, but the definitions of those were phrased in a von-Neumannized way that would have to be deciphered.

So: in order to learn the definition of Mahlo, I’d first have to convert the definitions of closed and unbounded into structural terms I could understand, then digest what they mean, then put them together, then read the definition of stationary set, then digest what that means, then read the definition of Mahlo cardinal, and only then, finally, start digesting that too. And the prospect of doing all that would make me say to myself “you know what? maybe I’m not so interested in Mahlo cardinals after all.”

So, having a direct definition like the one you’ve given is great. I thought of adding it to the Wikipedia page, but unless someone can find a reference, I guess it falls foul of their “no original research” rule.

Posted by: Tom Leinster on July 16, 2021 12:06 PM | Permalink | Reply to this

### Re: Large Sets 9.5

I wrote:

in an arbitrary topological space, the Cantor–Bendixson derivative of any subset is closed.

Oops, that’s false. As the Wikipedia article says, a counterexample is the two-point indiscrete space. The CB derivative of either singleton is the other singleton, which isn’t closed.

Posted by: Tom Leinster on July 21, 2021 4:15 PM | Permalink | Reply to this

### Re: Large Sets 9.5

Right, but it is true in a $T_1$ space, which I think applies here?

Posted by: Mike Shulman on July 22, 2021 5:01 AM | Permalink | Reply to this

### Re: Large Sets 9.5

Thanks for adding those clarifications; I could have been more explicit about those facts. (Your link to the CB derivative is missing a parenthesis.)

I thought of adding it to the Wikipedia page, but unless someone can find a reference, I guess it falls foul of their “no original research” rule.

The usual thing I do in a situation like that is add it to the nLab instead!

Posted by: Mike Shulman on July 16, 2021 4:22 PM | Permalink | Reply to this

### Re: Large Sets 9.5

If the definition is on the nLab, it can then be cited on WP, I feel, since the nLab seems to be considered a reliable secondary source (to the point where there is a citation template, IIRC). In any case, the nLab is going to be citing this blog post, making it a secondary source ;-)

Posted by: David Roberts on July 18, 2021 10:11 AM | Permalink | Reply to this

### Re: Large Sets 9.5

If the definition is on the nLab, it can then be cited on WP, I feel, since the nLab seems to be considered a reliable secondary source (to the point where there is a citation template, IIRC).

That’s bad form from Wikipedia, since the nLab does allow for original research: for example, the article on absorption category which the original author admitted to being original research in the discussion page, and Urs Schreiber explicitly saying original research is allowed on the nLab in the ensuing discussion here

### Re: Large Sets 9.5

a Mahlo set is $W$-inaccessible for any well-ordered set $W$

Wait, this isn’t right, is it? I think that a set $X$ can be at most $I(X)$-inaccessible, where $I(X)$ is $X$ equipped with an initial well-order. (And Wikipedia agrees, saying that a cardinal $\kappa$ can never be $(\kappa + 1)$-inaccessible.)

Proof: if not, there is a least set $X$ that is more than $I(X)$-inaccessible. Then there are unboundedly many $I(X)$-inaccessible sets $\lt X$. In particular, there is at least one $I(X)$-inaccessible set $Y \lt X$. But by minimality of $X$, the set $Y$ is at most $I(Y)$-inaccessible, so $I(X) \preceq I(Y)$, contradicting $Y \lt X$.

I agree, though, that any Mahlo set $X$ is $I(X)$-inaccessible; that is, Mahlo implies hyperinaccessible.

Posted by: Tom Leinster on October 17, 2021 1:05 PM | Permalink | Reply to this

### Re: Large Sets 9.5

Yes, that’s what I should have said. Thanks for catching it.

Posted by: Mike Shulman on October 18, 2021 1:54 AM | Permalink | Reply to this

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