Does a countably infinite preorder admit continuum many endomorphisms?

Question

The following assertion was made while discussing a question regarding Alexandrov spaces in topology.

Every countably infinite preorder has exactly continuum many endomorphisms.

Here, a preordered set is a set $P$ together with a relation $\le$ that is reflexive and transitive. Preordered sets form a category where the morphisms between two preorders are the monotone functions (satisfying $x\le y\implies f(x)\le f(y)$).

So assume $P$ is countably infinite. The question asks the number of monotone maps from $P$ to itself, which should be $2^{\aleph_0}$. (It cannot be more than that, since that is the number of set functions from $P$ to itself.)

Below is a partial proof. How can the last step be completed?

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Step 1: Given the preorder, there is an associated equivalence relation defined by $x\sim y$ if $x\le y$ and $y\le x$. If one of the equivalence classes, say $A$, is infinite, any map that sends $A$ to itself and fixes elements not in $A$ is monotone, and there are continuum many such maps. Otherwise, every equivalence class is finite. So the quotient structure $P/\sim$, which is a poset when given the natural induced order relation, is infinite and every monotone self-map of $P/\sim$ lifts to at least one monotone self-map of $P$. So without loss of generality we can assume $P$ is a poset.

Step 2: If the order relation on $P$ is the identity relation (= elements are only comparable to themselves), then any map from $P$ to itself is monotone, and there are countinuum many such maps. So we can assume there are at least two distinct comparable elements, say $x_0\le x_1$.

Step 3: Suppose $P$ has an infinite antichain $D$ (= no two elements are comparable). Any subset $A\subseteq D$ gives rise to an upper set $A^{\uparrow}=\{x\in P:x\ge a \text{ for some } a\in A\}$, and this upper set uniquely determines $A$ as its set of minimal elements. So there are continuum many upper sets $U$ in $P$. For each such $U$, the map sending $U$ to $x_1$ and the complement of $U$ to $x_0$ is monotone. So we have countinuum many such maps. We can now assume $P$ has no infinite antichain.

Step 4: As shown for example here, $P$ must then contain an infinite chain (= all elements comparable).

Step 5: handle the case of an infinite chain ...

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(Added later) From the comment to the answer, we also conclude:

Every infinite preorder has at least continuum many endomorphisms.

Answer 1

If $P$ contains an infinite chain, it either contains an infinite strictly increasing sequence or an infinite strictly decreasing sequence. We may assume without loss of generality that it contains a strictly increasing sequence $(a_n)_{n\in\mathbb{N}}$. If there exists $b\in P$ such that $b>a_n$ for all $n$, pick one such $b$ and call it $a_\infty$. Now define $f:P\to P$ by $f(x)=a_n$ for the least $n\in\mathbb{N}$ such that $a_n\not<x$, or $a_\infty$ if no such $n$ exists. Then $f$ is an endomorphism and the image of $f$ is isomorphic to either $\mathbb{N}$ or $\mathbb{N}\cup\{\infty\}$. In either case, the image of $f$ has continuum many endomorphisms which we can compose with $f$ to get continuum many different endomorphisms of $P$.