# Posets isomorphic to their endomorphism poset

Let $(P,\leq)$ be a poset. We set $$\text{End}(P)=\{f: P\to P: f\text{ is order-preserving}\}$$ and order $\text{End}(P)$ pointwise.

Is there a poset with more than 1 point such that $P\cong \text{End}(P)$?

• Not, if I understand your question right (non-strict order preserving). You always have constant maps, as many as elements in the set, plus at least the identity. – Alex Degtyarev Jul 14 '15 at 7:23
• Your argument is correct for finite posets, but you still have to look at infinite posets. – student9909 Jul 14 '15 at 7:25

If $P$ is infinite and $P\cong \textrm{End}(P)$, then the size of $P$ is restricted by the following facts.

Lemma. Assume that $P$ is an infinite poset.

1. $P$ has an infinite antichain, an infinite well-ordered subset, or an infinite inversely well-ordered subset.

2. If $|P|>2^{\lambda}$ for some infinite $\lambda$, then $P$ has an antichain, a well-ordered subset, or an inversely well-ordered subset of size $>\lambda$.

3. If $P$ has an antichain of size at least $\kappa\geq\omega$, then $\textrm{End}(P)$ has an antichain of size at least $2^{\kappa}$.

4. If $P$ has a well-ordered or inversely well-ordered subset of size at least $\kappa\geq\omega$, then $\textrm{End}(P)$ has an antichain of size at least $2^{\kappa}$.

In particular, these facts imply that if $P\cong \textrm{End}(P)$, then $|P|\geq \beth_{\omega}$. If also $|P|>2^{\lambda}$ for some infinite $\lambda$, then $|P|>2^{2^{\lambda}}$.

Sketch of proofs.

Part 1 of the Lemma follows from Ramsey's Theorem.

Part 2 of the Lemma follows from the Erdos-Rado Theorem.

For part 3 of the Lemma, if $P$ is an antichain, then $\textrm{End}(P)$ is an antichain of size $2^{|P|}$, so the claim is clear. Otherwise choose $a<b$ in $P$. Let $A\subseteq P$ be an antichain in $P$ of cardinality $\kappa$. It is possible to find a set $\mathcal X$ of pairwise incomparable subsets of $A$ such that $|\mathcal X|=2^{\kappa}$. For each $U\in \mathcal X$ define $f_U\colon P\to P$ by $f_U(x)=b$ if $x$ is in the filter generated by $U$ in $P$, else $f_U(x)=a$. The set $\{f_U | U\in \mathcal X\}$ is an antichain of size $2^{\kappa}$ in $\textrm{End}(P)$.

For part 4 of the lemma, assume that $W = \{w_{\alpha} : \alpha < \kappa\}$ is an enumeration of an increasing $\kappa$-sequence in $P$. Define a partial function $\sigma\colon P\to P$ by taking $\sigma(x)$ to be the least element $w_{\alpha}$ in $W$ such that $x\not\geq w_{\alpha}$. This $\sigma$ is a monotone total function if $W$ is cofinal in $P$, but only a partial function if there is some $q\in P$ above every $w_{\alpha}$. If such $q$ exists, fix one such, and extend the definition of $W$ to include $w_{\kappa}=q$ and extend the definition of $\sigma$ so that $\sigma(x)=q$ if $x\geq w_{\alpha}$ for all $\alpha<\kappa$. Either way, $W$ is well-ordered, of size $\kappa$, and $\sigma\in \textrm{End}(P)$ maps $P$ into $W$ and $\sigma(P)$ is well-ordered and of size $\kappa$. (Note: $\sigma(P)$ might be a proper subset of $W$.)

For every subset $V\subseteq \sigma(P)$ not containing $w_{\kappa}$ define $g\colon \sigma(P)\to \sigma(P)$ by $g(w_{\alpha})=w_{\alpha}$ if $w_{\alpha}\in \sigma(P)\setminus V$ and $g(w_{\alpha})=w_{\alpha+1}$ if $w_{\alpha}\in V$. If $V_1$ and $V_2$ are incomparable subsets of $\sigma(P)$, then $g_{V_1}$ and $g_{V_2}$ are incomparable elements of $\textrm{End}(P)$. There is a set of $2^{\kappa}$-many pairwise incomparable subsets of the $\kappa$-element set $\sigma(P)\setminus\{w_{\kappa}\}$, so $\textrm{End}(P)$ has an antichain of size at least $2^{\kappa}$. \\

• I must also mention that if $\mu$ is the strong limit cardinal with $\mu\leq|P|\leq 2^{\mu}$, then $\mu$ is not a weakly compact cardinal since if $\mu\rightarrow(\mu)^{2}_{3}$, then $|P|$ would have a well-ordered or anti-well ordered chain or an antichain of cardinality $\mu$ which is a contradiction. – Joseph Van Name Jul 15 '15 at 21:28
• For my own sanity: under GCH, what we conclude is that $|P|$ is uncountable, and is either a limit cardinal or the successor of a limit cardinal. – Tim Campion Jun 30 at 15:54

By Keith Kearnes answer, there is a strong limit cardinal $\mu$ such that $\mu\leq|P|\leq 2^{\mu}$. I claim that if $B$ is a complete Boolean algebra and $e:P\rightarrow B$ is an order preserving mapping with $e[P]$ dense in $B$, then $|B|<\mu$. Suppose to the contrary that $|B|\geq\mu$. Then by the Balcar-franek theorem, there is a subalgebra $A\subseteq B$ with $|A|=|B|$ that is freely generated by some set $F$. Let $r,s\in P,r<s$. For each $a\in F$, let $\ell_{a}(x)=r$ whenever $e(x)\leq a$ and $\ell_{a}(x)=s$ otherwise. Then $(\ell_{a})_{a\in F}$ is an antichain in $P$ of cardinality $\mu$ which contradicts Keith Kearnes' answer. In particular, the poset $P$ is not separative, nor is its dual separative. Furthermore, if $Q\subseteq P$ and $|Q|\geq\mu$, then $Q$ is not separative nor dual separative and there cannot be an order preserving mapping $e:Q\rightarrow B$ where $e[Q]$ is dense in $B$ and $|B|\geq\mu$.

Is there a poset with more than 1 point such that $$P\cong\text{End}(P)$$?