# Cardinality of all possible partitions of a infinite set

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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You should try to prove it yourself: $2^{|A|}\le |Part(A)|\le |A|^{|A|}=2^{|A|}$. –  Berci Dec 3 '12 at 11:50
@Berci: Good idea, though that's a different problem. I just think this problem should have already been covered by some textbook. –  Metta World Peace Dec 3 '12 at 11:57
Textbooks earn much less than your own effort. –  Berci Dec 3 '12 at 13:44
@Berci: Thanks, I shall remember your words. –  Metta World Peace Dec 3 '12 at 13:58

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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Thank you for your answer. I'm still confused with the first part.if $A=X-B$, then we $f(A)=f(B)=\{A,B\}$, so the cardinality should be $2^\kappa/2$. How can we define devision, when multiplication is not commutative? –  Metta World Peace Dec 3 '12 at 12:26
Was I sleeping that long? Did they define division on infinite cardinals during the morning? The calculation should be $2\cdot|P(X)/\sim|=2^\kappa$ therefore $|P(X)/\sim|=2^\kappa$. Also, note that multiplication is commutative for cardinals, even without the axiom of choice. –  Asaf Karagila Dec 3 '12 at 12:29
LOL I'm really enjoying your answer and humour. Thanks. –  Metta World Peace Dec 3 '12 at 12:37
Ah， I was confused between ordinals and cardinals. Thanks for pointing that out. –  Metta World Peace Dec 3 '12 at 12:43
I think that is a "duplicate answer" (of yours) to a duplicate question! ;-) –  amWhy Jan 11 '13 at 14:49