I have a partially ordered set $(X, \preceq)$ with the following properties:

  • $X$ has a minimum. I'll name it $1$.
  • For every $x \in X$, the principal filter ${\uparrow} x$ is order isomorphic to $X$.

I am wondering if it is possible to give $X$ the structure of an ordered monoid with $1$ as the identity. It seems like the map $\varphi_x: X \xrightarrow\cong {\uparrow} x \subseteq X$ could be thought of as an action of $x$ on $X$, but I am not sure how to justify that $\varphi_x \circ \varphi_y$ is of the form $\varphi_z$. Each $\varphi_x$ may not even be unique(?) Is it possible to choose $\varphi_x$'s in a consistent way?

Does it help if $X$ is an upward directed set, a join semilattice or a lattice? Would it be helpful if $X$ is well-founded?

Origin of the Problem:

I'll describe how this question comes up. I start with an ordered monoid $X$ whose identity $1$ is the minimum. The monoid operation and the order $\preceq$ satisfy:

$$ x \preceq y \text{ if and only if there exists a unique } z \text{ with } zx = y. \tag{1} $$

If $x \preceq y$, we can write $\frac{y}{x}$ to refer to the unique element $z$ satisfying $zx = y$. Because $1$ is the minimum, $X$ is right cancellative: for $xz = yz$, $z \preceq yz$, and so $x = \frac{yz}z = y$. $X$ is not necessarily left cancellative, but $\frac{xy}x$ always exists because $1 \preceq y$, and so $x \preceq xy$. One potential problem is I don't know that $\frac{xy}x = y$. (My fraction notation here is not symmetric. The denominator is actually to the "right" of the numerator. This asymmetry is due to $(1)$ being asymmetric.) However, that is not the most important part of the question. It might be related though.

The main question is: Can we recover the monoid operation from $\preceq$? Of course $\preceq$ must come from the monoid operation, so it will have many properties that may be useful. The question formulated at the beginning is based only on some properties of $\preceq$. My guess is that there are other potentially useful properties that I have not found.

Also, if there are interesting special cases that would make the answer positive, I would be happy to hear about them. Well-foundedness might seem natural, but it is rather strong, and does not apply to $\mathbb R_{\ge 0}$.

  • $\begingroup$ When you talk about an ordered monoid $(M, \leqslant)$, you assume that $x \leqslant y$ implies $zx \leqslant zy$ and $xz \leqslant yz$, right? $\endgroup$ – J.-E. Pin Jan 29 '14 at 18:43
  • $\begingroup$ @J.-E.Pin Yes, that is what I meant. It seems to me that the answer to this question is negative although I haven't been able to come up with a counterexample. $\endgroup$ – Tunococ Feb 10 '14 at 2:40

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