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

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

Thanks in advance for any possible pointer.

P.S.: I'm not so sure about the tags to use for this question.

Salvo Tringali
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