# Partitioning an infinite cardinal number

Can we partition an infinite cardinal greater than aleph null, into countable number of cofinal subsets? Can we have restriction on the cardinality of cofinal subsets?

For example, suppose $\alpha$ is an infinite cardinal which is the supremum of countably many cardinals $\alpha_n$ strictly less than it. Now, can we have partition into cofinal subsets $V_n$ with cardinality $\alpha_n$?

• What is the result of A. Stone you refer to? – Andrés E. Caicedo Jun 29 '14 at 17:45

Yes. Put in set $A_n$ those ordinals of the form $\alpha+n$ where $\alpha$ is $0$ or limit. One can do better, of course, since $\kappa=\kappa\times\kappa$ for any infinite cardinal $\kappa$, so we can in fact find $\kappa$ many cofinal subsets. This is explicit, since Gödel's pairing gives us an explicit bijection between $\kappa$ and $\kappa\times\kappa$.
• As for your suggestion, "Suppose $\alpha$ is an infinite cardinal which is the supremum of countably many cardinals $\alpha_n$ strictly less than it. Now, can we have partition into cofinal subsets $V_n$ with cardinality $\alpha_n$?" yes, you can, but not in a straightforward manner. The reason is that the cofinality of $\alpha$ is countable, but the $\alpha_n$ could have other cofinalities. (Cont.) – Andrés E. Caicedo Jun 29 '14 at 18:15
• A way to arrange what you suggest is to first fix a cofinal sequence into $\kappa$ with range $A$ of size $\omega$, then split it into infinitely many countable disjoint sets $A_0,A_1,\dots$ (each of which would be cofinal), then split $\kappa\setminus A$ into disjoint sets $B_0,B_1,\dots$ with $B_i$ of size $\alpha_i$, and finally set $C_n=B_n\cup A_n$ for each $n$. Then the sets $C_n$ are cofinal, and $|C_n|=\alpha_n$ for each $n$. – Andrés E. Caicedo Jun 29 '14 at 18:20
• @manisha If you mean the $A_n$ in the second comment, just use that there is an explicit bijection between $\omega$ and $\omega\times\omega$. – Andrés E. Caicedo Jun 30 '14 at 19:37