3

Collection of all cardinality can not be a set.

Can we find $f : [0, 1] \rightarrow \text{ cardinals }$ s.t. $f$ is 1-1 ?

ASB
  • 3,999
Mudream
  • 534
  • Those two lines seem to contradict each other. Did you mean something else, or perhaps something more specific, when you said "can not be a set"? –  Jun 22 '17 at 01:49
  • 3
    Not really: the range of $f$ is a set of cardinalities. Proper classes can have subsets. – Robert Israel Jun 22 '17 at 01:50
  • 1
    @tilper: Not sure whether you got notified, but they do not contradict. "$f : S→T$" only means that the range of $f$ is contained within $T$, and many people use this notation even for classes $S,T$ that are not sets, in which case $f$ may not be a function (encoded as a set) in ZFC. Not mentioned in Noah's answer, but important to note, in ZFC Replacement essentially reifies definable functions with set domain, just as Specification essentially reifies definable predicates with set domain. – user21820 Oct 06 '19 at 15:04
  • So for example, the aleph map is not a function in ZFC, but is a definable function in ZFC, and so by Replacement restricting the aleph map to any set of ordinals immediately gives a function. – user21820 Oct 06 '19 at 15:10
  • @user21820, I don't recall being notified. It's been a while but I feel like I'd remember this. This is definitely new to me and interesting, thanks for the @. –  Oct 07 '19 at 19:24
  • @tilper: You're welcome! It has been so long since ZFC has been invented that most of the intuition for why we wanted its axioms are no longer widely known. Nowadays, the cumulative hierarchy is often put forth as the way to visualize the set-theoretic universe, but it is in fact conceptually circular if we attempt to use it as justification. In other words, the cumulative hierarchy is meaningful only after we already have ZFC-like ordinals, so something else was driving the desire for the specification and replacement schemas. – user21820 Oct 08 '19 at 04:08

1 Answers1

3

This is a great question!

First, note that any map $f$ from $[0, 1]$ into $Card$ would yield a well-ordering of $[0,1]$ (put $r\preccurlyeq s$ if $f(r)\le f(s)$); so if there is such a map, then $[0, 1]$ can be well-ordered.

Conversely, any set of ordinals is in bijection with a set of cardinals: think about the (class) function sending $\alpha$ to $\aleph_\alpha$. So if $[0, 1]$ can be well-ordered, then - since every well-ordering is isomorphic to an ordinal, and hence any well-orderable set is in bijection with a set of ordinals - it admits an injection into $[0, 1]$.

So we have:

The answer is yes iff $[0, 1]$ can be well-ordered.

In particular the axiom of choice implies that the answer is yes.

Note that there's nothing special about $[0, 1]$ here: in general, a set can be put in bijection with a set of cardinals iff it can be well-ordered.

EDIT: There's an important bit of clarification, here. In the above, by "cardinal" I mean "initial ordinal" or $\aleph$-number. In ZF alone, though, we can have non-well-orderable sets - these sets obviously aren't equinumerous with any $\aleph_\alpha$. So in the ZF context, we can also speak of cardinalities more general than just cardinals.

In this context - as Andres observes below - your question can be rephrased as:

Given a set $X$, is there a function $f$ taking each element $x$ of $X$ to some set $f(x)$, where $f(x)\equiv f(y)$ iff $x=y$?

I don't know anything, off the top of my head, about this more complicated question and its various versions.

Noah Schweber
  • 245,398
  • 1
    More interesting is to study the question in $\mathsf{ZF}$. The general version is whether for every $X$ there is a set of size $X$ whose members all have distinct cardinality. – Andrés E. Caicedo Jun 22 '17 at 05:07
  • @AndrésE.Caicedo Quite right, I've edited to bring this up. – Noah Schweber Jun 22 '17 at 05:21
  • @AndrésE.Caicedo : ​ ​ ​ That's also not it, or at least not obviously it (I suppose equivalence might be a theorem). ​ Namely, I don't see any way of proving that one can go from a set of |X| cardinals to a set of |X| sets whose cardinalities are all distinct. ​ ​ ​ ​ ​ ​ ​ ​ –  Jun 22 '17 at 14:02
  • @NoahSchweber : ​ See my above comment. ​ ​ ​ ​ –  Jun 22 '17 at 14:03
  • @Ricky Yes, there is also that issue. I find the version I specified more natural, but I do not know the answer to the version you suggest either. – Andrés E. Caicedo Jun 22 '17 at 14:06