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I suppose this problem should be a commonplace, but I only find this one, in which notations are kind of idiosyncratic, along with a glaring defect.

My question is where I can find a more formal proof?

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    @Berci: Thanks, I shall remember your words.2012-12-03

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I will assume choice, otherwise things get extra messy.

Let $X$ be of cardinality $\kappa$.

  • You have at least $2^\kappa$ partitions:

    Define $\sim$ on $P(X)$ by $A\sim B\iff A=B\lor A=X\setminus B$. Clearly every equivalence class has two members so there are $2^\kappa$ of them, and all but the class of $\{\varnothing,X\}$ define a proper and distinct partition.

  • You don't have more than $2^\kappa$ partitions:

    Since every partition has at most $\kappa$ parts, we can think about it as a surjection from $\kappa$ onto itself. Of course many functions can define the same partition, but the point is that there are at most $\kappa^\kappa=2^\kappa$ functions from $X$ to itself, including those which are not surjective. Therefore there cannot be more than $2^\kappa$ partitions.

We have, if so, $2^\kappa$ partitions of $X$ whenever $|X|=\kappa$.

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    I completely agree! :-)2013-01-11