Let $X$ be a set, and let $\text{Part}(X)$ denote the collection of all partitions of $X$. For $A, B\in \text{Part}(X)$ we set $A\leq B$ if $A$ refines $B$, that is for all $a\in A$ there is $b\in B$ such that $a\subseteq b$. This relation defines a partial order on $\text{Part}(X)$.

If $X$ is an infinite set, is there a surjective order-preserving map $f:{\cal P}(X)\to \text{Part}(X)$ (where ${\cal P}(X)$ denotes the power-set of $X$, ordered by $\subseteq$)?

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    $\begingroup$ @Ivan: Maybe a shower can help... $\endgroup$
    – Asaf Karagila
    Sep 27, 2018 at 12:47
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    $\begingroup$ Don't force him @Asaf! Gerhard "Sometimes Can't Resist The Setup" Paseman, 2018.09.27. $\endgroup$ Sep 27, 2018 at 17:16
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    $\begingroup$ @Gerhard: He left me no choice. $\endgroup$
    – Asaf Karagila
    Sep 27, 2018 at 17:29
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    $\begingroup$ @HenrikRüping Right. You get that $P(X \times X) \to \text{Equiv}(X)$ preserves arbitrary sups; it can be seen as a left adjoint to the inclusion $\text{Equiv}(X) \hookrightarrow P(X \times X)$ (this follows easily from the fact that generation of equivalence relations is a closure operator). $\endgroup$
    – Todd Trimble
    Sep 27, 2018 at 18:11
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    $\begingroup$ Indeed, since the power set lattice is distributive and the equivalence relation lattice is not, the best you can hope for is an embedding, not a surjective lattice homomorphism. (Or as complete semi lattices, as Todd mentioned.) I am unsure if a lattice embedding exists in this situation when X is infinite. Gerhard "Can't Resist Those Setups Today" Paseman, 2018.09.27. $\endgroup$ Sep 27, 2018 at 18:28

2 Answers 2


It may be clarifying to work with equivalence relations $E$ on $X$ rather than partitions on $X$. The two are in natural bijection, with $E$ inducing a partitioning quotient map $q: X \to X/E$, and $X/E$ refines $X/E'$ iff $E \subseteq E'$ as subsets of $X \times X$.

Next, there is a surjective order-preserving map $P(X \times X) \to \text{Equiv}(X)$ where a general relation $R \in P(X \times X)$ is mapped to the equivalence relation $E_R$ that it generates. This is clearly surjective since an equivalence relation $E$ is mapped to itself.

Finally, if $X$ is infinite, there is a bijection $X \cong X \times X$, which induces an isomorphism of orders $P(X) \to P(X \times X)$. The composite

$$P(X) \to P(X \times X) \to \text{Equiv}(X)$$

provides what you want.

  • $\begingroup$ That's right, working with equivalence relations makes things conceptually easier - thanks for your answer! $\endgroup$ Sep 27, 2018 at 12:42

The positive solution uses an equivalent of the Axiom of Choice: for every infinite set $A$ there is a bijection $f:A\to A\times A$.

In the basic Fraenkel Model (section 4.3 in Jech's Axiom of Choice) there is an infinite set where there is no surjection as in the question.

Let $A$ be the infinite set of atoms in the model.

Step 1. Every subset of $A$ in the model is finite or cofinite. Let $X\subseteq A$ and let $E\subseteq A$ be finite such that $\operatorname{fix} E\subseteq\operatorname{sym} X$. If $X\subseteq E$ then $X$~is finite. If $X\not\subseteq E$ then $A\setminus E\subseteq X$: fix $x\in X\setminus E$ and let $y\in A\setminus E$ be arbitrary; the permutation~$\pi$ that interchanges $x$ and $y$ is in $\operatorname{fix} E$, so $\pi X=X$, but this means that $y=\pi(x)\in\pi X=X$.

Step 2. Let $P$ be partition of $A$ in the model and let $E\subseteq A$ be finite with $\operatorname{fix} E\subseteq \operatorname{sym} P$. Assume there are $a$ and $b$ in $A\setminus E$ that are distinct and $P$-equivalent. Let $c\in A\setminus E$ be arbitrary and not equal to~$a$. Let $\pi$ be the permutation that interchanges $b$ and $c$; then $\pi\in\operatorname{fix} E$, hence $\pi P=P$. Then $a,b\in X$ for some $X\in P$, and hence $a,c\in\pi X$. As $\pi X\in\pi P=P$ this shows that $a$ and $c$ are $P$-equivalent as well. It follows that either there is an element of $P$ that contains $A\setminus E$, or all elements of $A\setminus E$ determine one-element members of $P$.

Step 3. Assume $f:\mathcal{P}(A)\to\operatorname{Part}(A)$ is an order-preserving surjection. And let $E\subseteq A$ be finite such that $\operatorname{fix} E\subseteq\operatorname{sym} F$.

Assume $X$ and $Y$ are finite such that $X\cap E=Y\cap E$ and $|X\setminus E|=|Y\setminus E|$, and such that $f(X)$ and $f(Y)$ are partitions that contain $\{\{a\}:a\in A\setminus E\}$. Then $f(X)=f(Y)$.

To see this let $\pi$ be a permutation that maps $X\setminus E$ to $Y\setminus E$ and vice versa and leaves all other points in place. Then $\pi\in\operatorname{fix} E$, so that $\pi(f(X))=f(X)$ and $\pi(f(Y)=f(Y)$. The ordered pair $\langle X,f(X)\rangle$ is in $f$, hence $\langle \pi(X),\pi(f(X))\rangle$ is in~$\pi f$, but $\pi f=f$ and $\pi X=Y$ so that $\langle Y,f(X)\rangle\in f$, that is, $f(Y)=f(X)$.

A similar statement holds when $X$ and $Y$ are infinite, hence cofinite, and $X\cap E=Y\cap E$, and $|A\setminus(E\cup X)|=|A\setminus(E\cup Y)|$.

Step 4. Now consider the set $\Pi(E)$ of partitions of~$A$ that contain $\{\{a\}:a\in A\setminus E\}$; this is essentially the set of partitions of~$E$ where each is augmented with $\{\{a\}:a\in A\setminus E\}$.

The argument above also shows the following: if $X$ and $Y$ are finite such that $X\cap E=Y\cap E$, $|X\setminus E|\le|Y\setminus E|$ and $f(X),f(Y)\in\Pi(E)$ then $f(X)\le f(Y)$. This is so because we can find a permutation $\pi$ in $\operatorname{fix} E$ such that $\pi(X)\subseteq Y$.

Likewise: if $X$ and $Y$ are infinite such that $X\cap E=Y\cap E$, $|A\setminus(X\cup E)|\le|A\setminus(Y\cup E)|$ and $f(X),f(Y)\in\Pi(E)$ then $f(X)\ge f(Y)$.

Finally: if $X$ is finite and $Y$ is infinite such that $X\cap E=Y\cap E$ and $f(X),f(Y)\in\Pi(E)$ then $f(X)\ge f(Y)$.

In conclusion: each subset of $E$ determines a chain in the set~$\Pi(E)$.

Step 5. If we enlarge $E$ then $\operatorname{fix} E$ becomes smaller, so we can choose $E$ as large as we please.

Let $n$ be a natural number. The number of partitions of the set $\{1,2,\ldots,4n\}$ into four sets of size $n$ is equal to $$ \binom{4n}{n}\binom{3n}{n}\binom{2n}{n} $$ This number is larger than $$ 3^n\cdot 2^n\cdot\frac{4^n}{2n+1} = \frac{24^n}{2n+1} $$ For $n\ge9$ we have $24^n/(2n+1)>16^n$.

This gives us our contradiction: if necessary enlarge $E$ so that $|E|=4n$ for some $n\ge9$. Then, by 4 above, the map $f$ divides $\Pi(E)$ into, at most, $2^{4n}=16^n$ chains. On the other hand the partitions of $E$ into four pieces of size $n$ form an antichain of cardinality more than $24^n/(2n+1)$, which in turn is larger than $16^n$.


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