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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 $\alpha_n$?

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

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  • $\begingroup$ 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.) $\endgroup$ – Andrés E. Caicedo Jun 29 '14 at 18:15
  • $\begingroup$ 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 $n$. $\endgroup$ – Andrés E. Caicedo Jun 29 '14 at 18:20
  • $\begingroup$ Sir, Please elaborate how you got A_n's. $\endgroup$ – akansha Jun 30 '14 at 17:15
  • $\begingroup$ @manisha If you mean the $A_n$ in the second comment, just use that there is an explicit bijection between $\omega$ and $\omega\times\omega$. $\endgroup$ – Andrés E. Caicedo Jun 30 '14 at 19:37

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