# Can we generalize Aleph numbers to non integer values? [duplicate]

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I'm really new to those kind of arguments so don't call me mad but I was wondering if there is a way to define an infinite set which cardinality is an Aleph-number like: $\aleph_{\frac 12}$. I have a sort of passion to generalization of math objects and I've been really astonished when I learned about fractals, whose dimensions are not an integer number so I was thinking if there is some way to build infinite sets (maybe with an unusual list of axioms) whose cardinality is not an Aleph-number with integer index and how those numbers and sets would work. Any help or reference is appreciated :)

## marked as duplicate by Jack D'Aurizio, zarathustra, Andrés E. Caicedo, Asaf Karagila♦ set-theory StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Jun 28 '15 at 16:16

• Assuming the Axiom of Choice, every infinite set has cardinality some aleph. – Hayden Jun 28 '15 at 10:40
• So maybe those bizarre Aleph numbers with non integer index exist but we don't know them ? – Renato Faraone Jun 28 '15 at 10:45
• If you're expecting them to represent some cardinality different that those alephs indexed by the ordinals, and also assuming Choice, then they provably don't exist. If you don't assume Choice, it is consistent that there exist sets which are not in bijection with an aleph. But you haven't given any properties you'd want these "non-ordinal indexed alephs" to satisfy (nor do I know of any material that deals with the idea), so I couldn't say whether such sets can be made into something you want. – Hayden Jun 28 '15 at 10:54
• This question makes no sense. See Hartogs function: $\aleph_1$ is defined as the least ordinal strictly greater than $\aleph_0$, so there's no place for $\aleph_{\frac{1}{2}}$ in ZFC, as well as there is no point in reindexing the usual alephs. – Jack D'Aurizio Jun 28 '15 at 12:06
• So this is more or less as asking if there is a way to give $\frac{1}{2}$ the dignity of a natural number. – Jack D'Aurizio Jun 28 '15 at 12:15

Beware of over-generalizations! (I loved them, but I was quite wrong on that one.)

I think that your question has a nice interpretation. I'll take your phrase

if there is a way to define an infinite set which cardinality is an Aleph-number like: $\aleph_{\frac{1}{2}}$

to mean if it is reasonable to assume that infinite cardinalities may have the ordering of the rational numbers. Written as such, this is in contradiction to the Axiom of Choice (AC), as stated in the comments. In particular, AC implies that cardinalities are well-ordered.

But perhaps you may object: “Ok, but AC is overkill. I only need that cardinalities are totally ordered; that's enough for having a neat notion of size of sets, so I can compare any two of them”.

This may be called the Hypothesis of Cardinal Comparability. Now the punchline is, cardinal comparability implies AC! So, any universe of set theory (i.e., satisfying ZF) where cardinalities are totally ordered, they must be well-ordered.

So, if your function $r \mapsto \aleph_r$ is (strictly) increasing, its domain must be wellordered.

A beautiful introduction to all this stuff is this book.

• The axiom of choice is completely unnecessary here. The $\aleph$ cardinals are by definition well-ordered. – Asaf Karagila Jun 28 '15 at 18:08
• @AsafKaragila Agree, but I guess the OP is using $\aleph$ naively as a symbol for "increasing enumeration of cardinals". I am not sure if he is fully aware of its definition (nor that its arguments are the ordinals, for that matter). Don't you think so? Or I'm missing something? – Pedro Sánchez Terraf Jun 28 '15 at 18:58
• Instead of reinforcing half-understandings, it might be a good idea to give a full understanding instead. :-) – Asaf Karagila Jun 28 '15 at 19:01
• @AsafKaragila -1 your first comment, +1 to this last one. Thanks! – Pedro Sánchez Terraf Jun 28 '15 at 19:07

This is not a real answer, but you could try to see if you can extend the $\aleph$ function from the ordinal surreal numbers onto the cardinal surreal numbers to the whole collection of positive surreal numbers with nice properties, and then study what $\widetilde{\aleph}_{\frac{1}{2}}$ is (for instance, is it an ordinal surreal number?). But with what properties, how, and what for?

• I don't think there is any natural way to extend $\aleph$ to non-ordinal inputs in the surreals. Since by definition $\aleph$ makes a sort of huge jump with each successor, I have no idea what, if any, would be an appropriate way to interpolate. Certainly you could interpolate linearly or any other way you want, but I doubt this can lead anywhere satisfying. – Mark S. Jul 20 '15 at 3:00
• I have no idea either. The definition would be somewhat non-homogeneous with respect to the property of being an ordinal, because between the ordinal $\alpha$ and its successor, there would have to be $\aleph_{\alpha}$ points sent to an ordinal number, even though the intervals $[\alpha;\alpha+1]$ are of same "diameter". But why not? – nombre Jul 21 '15 at 11:20

Although the answers above are completely correct, I think the answerers are possibly missing part of the point of the OP's question, which was motivated by learning about fractals, which have fractional dimension. The crucial part of that context is that to define the dimension of a fractal, you need an essentially new definition, one which agrees with the more "standard" notion of dimension for more "standard" shapes.

So while it is true that, as long as we are interested in sets and cardinality, there is no need to consider $\aleph_r$ for $r \notin \mathbb N$, I think what the OP is wondering about is:

Is there some generalization of set theory in which we consider "set-like" objects, for which there is a corresponding "cardinality-like" notion, with the properties that (a) for a genuine "set", the cardinality-like notion is equivalent to ordinary cardinality, but (b) for things that are not really sets but only "set-like", the cardinality-like notion might require "pseudocardinalities" that lie strictly between the alephs?

As far as I know no such generalization exists, but that doesn't mean it can't exist. In particular I think it's worth noting that as far as cardinality goes, even a finite set has to have a cardinality within $\mathbb N$, so from that point of view one does not need non-natural numbers at all, let alone alephs with a rational index -- but of course we have plenty of other uses for rational and negative numbers besides answering the question "How many elements?"

• As far as generalizations of cardinality go, I wrote my opinion here. The point is that if every set already has cardinality, then either the extension is not used (like saying that $x\mapsto x^2$ is a function into $\Bbb R$, but in reality it is into $\Bbb R^{\geq 0}$); or you had to work in a context where the objects are not sets. Or, you know, just define a new measurement of size. Like how measure theory gives a different notion of "size" for sets. – Asaf Karagila Jun 28 '15 at 18:12
• "A context where the objects are not sets" is precisely what I meant by "set-like objects". – mweiss Jun 28 '15 at 18:42