If $(\mathbb M, \tau)$ is a topological monoid, is $\tau$ always induced by a [left] subinvariant semimetric?

Question

Let me start by recalling some basic definitions (just for the sake of avoiding misunderstandings due to the vocabulary of the post).

Basically following some ideas of W. Lawvere (but not his terminology), 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 $\mathbf R^+$.

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 subinvariant (in $\mathbb M$) if $d(xz,yz) \le d(x,y)$ for all $x,y,z \in M$; left subinvariant (in $\mathbb M$) if it is right subinvariant in the dual of $\mathbb M$; and subinvariant (in $\mathbb M$) if it is both right and left subinvariant.

With all of this in mind, what is known about the following question?

Let $(\mathbb M, \tau)$ be a topological monoid, with $\mathbb M = (M, \cdot)$. Does there always exist a left (respectively, right) subinvariant semimetric $d$ on $M$ such that $\tau$ is the canonical topology of $d$? And what about a subinvariant semimetric?

Edit (based on the comments below). I don't expect this to be true in general. I would be happy with something of the form: "The answer is known to be positive for all the members of $\mathcal C_1$, and negative for all the members of $\mathcal C_2$", where $\mathcal C_1$ and $\mathcal C_2$ are "non-trivial" classes of ("sufficiently small") topological monoids.

Thanks in advance for any possible pointer.

Answer 1

I apologize for answering my own question.

Let $\mathcal K = (\mathbb K, \tau)$ be a T1 topological unital ring, with $\mathbb K = (K, +, \cdot)$, and let $\mathbb K_{(\cdot)}$ be the multiplicative monoid of $\mathbb K$, so $(\mathbb K_{(\cdot)}, \tau)$ is a T1 topological monoid.

Assume that $\tau$ is induced by a semimetric $d$ on $K$. This implies that $d(x,y) \ne 0$ for all distinct $x,y \in K$ (as mentioned in the comments to the OP, the topology induced by a semimetric is T1 iff the "distance" of two distinct points is non-zero). In particular, we have $d(0,1) \ne 0$.

Now suppose that $d$ is right (respectively, left) subinvariant in $\mathbb K_{(\cdot)}$. Then, $0 < d(0,1) \le d(0, x^n)$ for every $x \in \mathbb K^\times$, where $\mathbb K^\times$ is, as usual, the set of the units of $\mathbb K$.

This is however impossible if there exists at least one element $x \in \mathbb K^\times$ such that $0$ is a limit point, relative to $\tau$, of the $K$-valued sequence $(x^n)_{n \in \mathbf N}$, which is for instance the case when $\mathbb K$ is the real field and $\tau$ is the usual topology on $\bf R$ (just because we should then have $\lim_n d(0,x^n) = 0$).

Incidentally, the above shows that if $\tau$ is the topology induced by a non-trivial absolute value $|\cdot|$ of $\mathbb K$ and if $|x| \ne 1$ for some $x \in \mathbb K^\times$, then $\tau$ can not be the canonical topology of a left (respectively, right) $\mathbb K_{(\cdot)}$-subinvariant semimetric, in spite of being first-countable (which serves as a partial answer to one of the questions raised by @Wlodzimierz Holsztynski in the comments to the OP, insofar as $d$ has no way of being topologically equivalent to the canonical metric induced on $K$ by $|\cdot|$).

I'd like to publicly thank Jacek Jendrej for a fruitful conversation which paved the way to all of this (as far as I know, he's not a MO user, and that's why I'm providing an external link).

Answer 2

The topology of topological monoids can be arbitrarily bad. Here is an instructive example. Let X be any topological space. Then we will construct a (commutative) topological monoid M whose underlying topological space is X unioned with two disjoint points: $$M = X \cup \{ 0 \} \cup \{ \infty\}$$

The point "0" is the identity of the monoid structure. The remaining products are defined via the formula $x \cdot y = \infty$ for all $x,y \neq 0$. This gives a continuous commutative and associative multiplication for the space M.