How many partitions does an infinite set have?

Question

I've been looking into how many partitions a set has, mainly for infinite sets since for finite ones we have the Bell number $B_n$. I'll use the following notation:$$B(S)=\{P|P \text{ is a partition of }S \}$$

There exists a rather trivial injection $g:S \to B(S)$, but we can also find the injection $f:P(S) \to B(S)$ (assuming $3<|S|$) as follows:

$f(\emptyset)=\{S\}$ and $f(S)$ is the partition into singletons. We then pair up all remaining subsets of $S$ as $(X,S \setminus X)$ such that $|X| \geq |S \setminus X|$ and choose an $a\in X$, then: $f(X)=\{X,S \setminus X \}$ and $f(S \setminus X)=\{ S \setminus X, X \setminus \{ a \}, \{ a \} \}$ (I believe this requires the axiom of choice to pick which one is $X$). Since $B(S) \subset P(P(S))$ we have the following inequality: $$|P(S)| \leq |B(S)| \leq |P(P(S))|$$

I have two questions:

1. Is this proof correct?
2. Can we turn one of those $\leq$ into an equality or strict inequality sign?

Answer 1

For 1, it is incorrect.

For example, if $S=\mathbb N$, $X_0=S\setminus\{0\}, X_1=S\setminus\{1\}$.

Now, if accidently $f(S\setminus X_0)=\{S\setminus X_0, X_0\setminus\{1\}, \{1\}\}$ and $f(S\setminus X_1)=\{S\setminus X_1, X_0\setminus\{0\}, \{0\}\}$, then $f(S\setminus X_0)=f(S\setminus X_1)$.

To show that $|\mathcal P(S)| \leq |B(S)|$ we need to be more careful, a way to ensure it is to pick the "$a$"s more carefully, e.g.:

First let $(s_i\mid i\in |S|)$ be some ordering of $S$

If $|S\setminus X|>1$, do the same as you did, if $S\setminus X=\{x\}$, choose the $a$ to be such that $a>x$.

This is injective: It is clearly injective for elements $X$ such that $|f(X)|=2$, take $X_0,X_1$ whose image is of size $3$ and are both of cardinality $>1$, if $f(X_0)=f(X_1)$ then $\{X_0, (S\setminus X_0)\setminus\{a\},\{a\}\}=\{ X_1, (S\setminus X_1)\setminus\{a\},\{a\}\}$, from cardinality arguments $\{X_0, (S\setminus X_0)\setminus\{a\}\}=\{X_1, (S\setminus X_1)\setminus\{a\}\}$, but because $a\in S\setminus X_0,S\setminus X_1$ we get that $\{X_0, S\setminus X_0\}=\{X_1, S\setminus X_1\}$, so (by the way we arrange this kind of tuples) $X_0=X_1$.

If $X_0=\{x\},X_1=\{y\}$ for $x\ne y$, and $\{y\}∈f(X_0)$, then $y>x$ which means that $\{x\}∉f(X_1)$, so $f(X_0)≠f(X_1)$.

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As for 2: I claim that $|B(X)|=|\mathcal P(X)|$:

Again, remember that $(s_i\mid i\in |S|)$ is a well ordering of $S$, now define an injection from $B(X)$ to $S^S$:

If $π∈B(X)$, for $x\in S$ we denote $[x]_{π}$ to be the unique $ζ∈π$ such that $x\in\zeta$ (the projection of $x$ onto $π$).

Define $f_π:S\to S$ as follow: $f_π(x)=\min ([x]_{π})$.

Let $h: B(X)→S^S$ defined as $h(π)=f_π$.

$h$ is injective: if $π_0\ne π_1$, let $ζ∈π_0$ such that $ζ∉π_1$ define $η=[\min(ζ)]_{π_1}$, if $\min(η)≠\min(ζ)$, then $h(π_0)(\min(ζ))≠h(π_1)(\min(ζ))$ hence $h(π_0)≠h(π_1)$, if it is equal, take an element from $ηΔζ$, this element will witness that $h(π_0)≠h(π_1)$.

So we get $|\mathcal P(S)| \leq |B(S)|≤|S^S|=|\mathcal P(S)|$, so they are all equal.