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This is a follow up of Question 163246. For the reader's convenience, let me first copy&paste some basic definitions.

We let a semimetric on a set $X$ be a function $d: X \times X \to [0,\infty]$ such that $d(x,x) = 0$ and $d(x,z) \le d(x,y) + d(y,z)$ for all $x,y,z \in X$. Exactly as with "classical" metrics, $d$ induces a topology on $X$, which we call the canonical topology of $d$, having as a base the sets $\{y \in X: d(x,y) < r\}$ as $x$ ranges over $X$ and $r$ over $\mathbf R^+$. We say that a topology $\tau$ on $X$ is semimetrizable if there exists a semimetric $d$ on $X$ such that $\tau$ is the canonical topology of $d$.

Next, we take a topological monoid to be, as usual, a pair $(\mathbb M, \tau)$ consisting of a (multiplicatively written) monoid $\mathbb M = (M, \cdot)$ and a topology $\tau$ on $M$ such that $\cdot$ is continuous [...] in the expected sense.

Finally, if $\mathbb M = (M, \cdot)$ is a monoid and $d$ is a semimetric on $M$, we say that $d$ is: right $\mathbb M$-subinvariant if $d(xz,yz) \le d(x,y)$ for all $x,y,z \in M$; left $\mathbb M$-subinvariant if it is right $\mathbb M^{\rm op}$-subinvariant, where $\mathbb M^{\rm op}$ is the dual of $\mathbb M$; and $\mathbb M$-subinvariant if it is both right and left $\mathbb M$-subinvariant.

With that said, let $(\mathbb M, \tau)$ be a first-countable topological monoid, with $\mathbb M = (M, \cdot)$. It has been already observed in the other thread that $\tau$ does not need be semimetrizable (see Chris Schommer-Pries' construction here and the comments to his answer). Furthermore, $\tau$ doesn't need be induced by a left or right subinvariant semimetric even in the case when the topology is metrizable (see here); however, the construction used for proving it does "critically" depend on the fact that $\mathbb M$ has an absorbing element.

Based on the above, we then say that $\mathbb M$ is resilient if there doesn't exist any non-trivial submonoid $S$ of $\mathbb M$ [...] with an absorbing element (in particular, this means that the only idempotent element of $\mathbb M$ is the identity). Then here are my questions:

Q1. If $\mathbb M$ is resilient and $\tau$ is [semi]metrizable, is it true that $\tau$ is the canonical topology of a left (respectively, right) $\mathbb M$-subinvariant semimetric? Q2. And what about the case when $\mathbb M$ is, in addition, cancellative (i.e., $xz=yz$ or $zx=zy$ for some $x,y,z \in M$ imply $x=y$)?

Thanks in advance for any comment, hint, pointer, or whatever. (Some parts of the current formulation of the OP are based on comments by Eric Wofsey and Chris Schommer-Pries.)

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  • $\begingroup$ A monoid is resilient iff it has no idempotents besides the identity, since any idempotent $i$ is absorbing in the submonoid $\{1,i\}$ and any (locally) absorbing element is idempotent. $\endgroup$ Apr 16, 2014 at 14:19
  • $\begingroup$ Also, you really want to assume that the topology is semimetrizable, not just that it is first-countable (I don't know whether these are equivalent, but I doubt they are). Chris's construction is quite flexible and can easily be modified to be "resilient". For instance, you could take $X\coprod\mathbb{N}$, where adding elements of $X$ to anything but $0$ is the same as adding $1$. $\endgroup$ Apr 16, 2014 at 14:22
  • $\begingroup$ Or take the free commutative monoid generated by X, also resilient, and it contains X as a proper component. $\endgroup$ Apr 16, 2014 at 19:53
  • $\begingroup$ Sorry for the late reply (I was on holiday). In fact, there're 1st-countable topologies which are not semimetrizable (to be honest, I had no doubt about their existence, but it's only two days ago that I found a counterexample): this is the case, e.g., with the cofinite topology on a countably infinite set. So yes, I restated the OP to take into account your comments. $\endgroup$ Apr 20, 2014 at 17:21
  • $\begingroup$ I don't see you mentioning $1$ (or $e$). Do you consider monoids or semigroups? $\endgroup$ Apr 20, 2014 at 17:50

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