The usual definition of a regular cardinal is "$\kappa$ is regular if $cf(\kappa) = \kappa$", which, assuming the axiom of choice, is equivalent to this definition: "$\kappa$ is regular iff it cannot be expressed as a union of a less than $\kappa$ sets, all of which are of cardinality less than $\kappa$."

Now consider Definition 2: a cardinal $\kappa$ is regular if for every subset $X \subset \kappa$ such that $|X| < \kappa$ we have $\bigcup X < \kappa$.

I think these definitions are not equivalent in ZF, because $X$ and $\bigcup X$ can be well-ordered without invoking the axiom of choice. In particular, I guess one could prove in ZF that with Definition 2 all successor cardinals are regular, which is not true with usual definition.

Questions: Is my intuition correct? How can I show the non-equivalence?

  • $\begingroup$ I should add that proving non-equivalence is usually very hard, because it requires constructing a model where the axiom of choice fails in a very particular way. Doing that is very often very difficult. $\endgroup$ – Asaf Karagila May 15 '13 at 12:13
  • $\begingroup$ @AsafKaragila: Well, but I was thinking perhaps one can appeal to some already-constructed examples to show this. $\endgroup$ – yfyf May 15 '13 at 21:54
  • $\begingroup$ Yes, that is often the case, that previously constructed models witness something like that. But as I remark in my answer, when we talk about cofinality we talk about ordinals, and therefore the union of $X$ is a subset of an ordinal and therefore can be well-ordered. So there is no such model to separate the definitions in this case. $\endgroup$ – Asaf Karagila May 15 '13 at 22:05

Let me rewrite the answer completely.

You wrote three definitions for a regular cardinal, and you made two false claims. First let me write the definitions, so we'll be clear on those.

Let $\kappa$ be an infinite initial ordinal (i.e. a cardinal in the context of $\sf ZFC$).

  1. $\kappa$ is $1$-regular if every unbounded subset of $\kappa$ has order type $\kappa$.
  2. $\kappa$ is $2$-regular if every union of less than $\kappa$ subsets of $\kappa$, each of cardinality less than $\kappa$, is itself of cardinality less than $\kappa$.
  3. $\kappa$ is $3$-regular if every $X\subseteq\kappa$ such that $|X|<\kappa$ has the property that $\bigcup X<\kappa$.

You also claimed that $1$ and $2$ are equivalent if we assume the axiom of choice, and you wanted to show that $3$ is also inequivalent to those without the axiom of choice. But the truth is that the axiom of choice is not needed in order to show that all three are equivalent.

If $\kappa$ is $1$-regular, then whenever $X\subseteq\kappa$ and $|X|<\kappa$ the order type of $X$ (as a well-ordered set) has to be less than $\kappa$, so $\bigcup X=\sup X<\kappa$ by the assumption that it is $1$-regular.

If $\kappa$ is $3$-regular, let $A$ be an unbounded subset of $\kappa$ then $\bigcup A=\sup A=\kappa$, then by the definition $3$-regular we have to have that $|A|=\kappa$, but since $\kappa$ is an initial ordinal it does not have subsets of size $\kappa$ whose order type is not $\kappa$ itself. Therefore $A$ has order type $\kappa$.

This establishes the equivalence between $1$ and $3$. There was absolutely no use of the axiom of choice. $\bigcup X$ is an ordinal because it is the union of ordinals.

Now to show that these are equivalent to $2$-regularity.

If $\kappa$ is $2$-regular, let $X\subseteq\kappa$ such that $|X|<\kappa$. Since $\kappa$ is an initial ordinal, all the members of $X$ have cardinality strictly less than that of $\kappa$. Therefore we take union of less than $\kappa$ sets of size less than $\kappa$, and by $2$-regularity $|\bigcup X|<\kappa$. But this means that $\sup X<\kappa$, as wanted.

If $\kappa$ is $3$-regular, let $P$ be a set of less than $\kappa$ subsets of $\kappa$, each of cardinality $\kappa$. Then for every $A\in P$ we have that $\bigcup A=\sup A<\kappa$ by the definition $3$-regular. Let $X=\{\sup A\mid A\in P\}$, then $|P|\leq|X|<\kappa$ and therefore $\bigcup X<\kappa$. It follows that $\bigcup P<\kappa$ as well, because there is some $\alpha>\bigcup X$ and that means that $\alpha\notin A$ for every $A\in P$. Therefore $\kappa$ is $2$-regular as wanted. $\square$

In the comments you remarked that your teacher challenged with the following task, if $X\subseteq\aleph_{\alpha+1}$ and $|X|<\aleph_{\alpha+1}$ then you are to find an injection from $\bigcup X$ into $\aleph_\alpha\times\aleph_\alpha$.

But that means that you have to assume that $\aleph_{\alpha+1}$ is regular. Whereas in $\sf ZFC$ every successor cardinal is indeed regular, it is consistent with $\sf ZF$ that successors are not regular. In fact, assuming very large cardinals are consistent, we can construct a model of $\sf ZF$ in which there are no regular cardinals except $\aleph_0$.

So in order to find that injection one has to assume that $\aleph_{\alpha+1}$ is regular, which is an assumption we cannot prove without the help of the axiom of choice, and therefore we cannot write an explicit injection from $\sup X$ into $\aleph_\alpha\times\aleph_\alpha$.

  • $\begingroup$ I was thinking along the same lines of your answer, but then I was challenged by my lecturer to provide an explicit injection from $\bigcup X$ to $\aleph_\alpha \times \aleph_\alpha (= \aleph_\alpha)$, where $\kappa = \aleph_{\alpha+1}$, $X$ are as in Definition 2. I thought it would be trivial to do, but I keep bumping into problems. Could you maybe sketch how would one do that? $\endgroup$ – yfyf May 16 '13 at 7:17
  • $\begingroup$ yfyf, it is not true in $\sf ZF$ that successor cardinals are regular. You do need the axiom of choice for that. $\endgroup$ – Asaf Karagila May 16 '13 at 10:26
  • $\begingroup$ Even with Definition 2? $\endgroup$ – yfyf May 16 '13 at 19:14
  • $\begingroup$ They are equivalent in $\sf ZF$, as my answer points out, so no. $\endgroup$ – Asaf Karagila May 16 '13 at 19:22
  • 1
    $\begingroup$ Amazing! Thank you very much for the effort, it's crystal clear now. $\endgroup$ – yfyf May 17 '13 at 15:08

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