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Given a set X with a certain cardinality, there are explicit constructions for getting a set with the "next bigger" cardinality, e.g. constructing the power set.

Does some analogous construction exist for the reverse, i.e. getting a set with the next smaller cardinality?

Martin Sleziak
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Benno
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  • Pick a single point? – Henno Brandsma Jan 13 '16 at 16:30
  • Henno: True, but I was looking for something nontrivial ;) I will edit the question to reflect this. – Benno Jan 13 '16 at 16:34
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    The power set only has the "next bigger" cardinality if we assume the continuum hypothesis. – sbares Jan 13 '16 at 16:39
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    Yes, thats why it is in quotes. I fear that I don't have the mathematical expertise to ask this question precisely, so feel free to assume anything that I have forgotten. – Benno Jan 13 '16 at 16:41
  • Perhaps a quotient set construction using an equivalence relation might work, but I can't think of any nontrivial examples that always lower the cardinality of an infinite set. (moments later) When I wrote the above, I didn't realize you wanted the "next lower cardinality" (I was distracted while writing it . . .), but I'll leave my comment here in case you're willing to settle for simply lowering the cardinality. – Dave L. Renfro Jan 13 '16 at 16:46
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    @SBareS The power set only has the next bigger cardinality if we assume the generalized continuum hypothesis. – BrianO Jan 13 '16 at 18:04
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    @Benno There may not be a "next smaller" cardinality. If $\kappa$ is a limit cardinal (e.g. $\kappa = \aleph_0$, or $\kappa = \aleph_{\omega}$) then there is no such thing. – BrianO Jan 13 '16 at 18:06

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In general no, because not every cardinality has a "next-smaller" cardinality. In particular, if $\kappa$ is a limit cardinal then it is not a successor cardinal, so there is no cardinal $\lambda$ such that $\kappa = \lambda^+$.

However, if $\kappa$ is a successor cardinal, or if for a limit cardinal you're ok with finding a subset of any smaller cardinality, the procedure is the same:

Let $\kappa$ well-order $S$, pick your desired cardinal $\lambda < \kappa$, let $x$ be the element of $S$ with order-type $\lambda$, and then let $T = \{s \in S : s < x\}$.

Dan Simon
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This is impossible as there just is no "next smaller cardinality" in general. Just consider $\mathbb{N}$, what do you want to do. It is impossible.

quid
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