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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 $P\cong \text{End}(P)$?

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    $\begingroup$ 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. $\endgroup$ – Alex Degtyarev Jul 14 '15 at 7:23
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    $\begingroup$ Your argument is correct for finite posets, but you still have to look at infinite posets. $\endgroup$ – student9909 Jul 14 '15 at 7:25
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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 $2^{\kappa}$. \\

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  • $\begingroup$ 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. $\endgroup$ – Joseph Van Name Jul 15 '15 at 21:28
  • $\begingroup$ 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. $\endgroup$ – Tim Campion Jun 30 at 15:54
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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 $|B|\geq\mu$.

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The answer to

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

is No. This follows immediately from Theorem 3 in

Roy O. Davies, Allan Hayes and George Rousseau, Complete lattices and the generalized Cantor theorem, Proc. Amer. Math. Soc. 27 (1971), 253–258.

This reference was pointed out to me by user bof in a (now deleted) comment to this post. More precisely bof gave a link to this answer of them.

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