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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.

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    $\begingroup$ The topology deternined by a semimetric always has countable bases of neighborhoods at each point, and there are topological monoids which don't have that property; for example, the product of uncountably many discrete non-trivial finite groups. $\endgroup$ Apr 12, 2014 at 23:43
  • $\begingroup$ True. So let us restrict to "sufficiently small" topological monoids. What happens? $\endgroup$ Apr 12, 2014 at 23:46
  • $\begingroup$ Why do you expect this to be true? $\endgroup$ Apr 12, 2014 at 23:49
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    $\begingroup$ I seem to remember that Soviet topologists, in particular A.Archangielski, considered some notions of asymmetric (not necessarily symmetric) metrics. Perhaps in the context of metrization. $\endgroup$ Apr 13, 2014 at 1:24
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    $\begingroup$ To elaborate on my last comment, every poset with the Alexandrov topology is canonically semimetrizable (let $d(x,y)=0$ if $x\leq y$ and $d(x,y)=1$ otherwise). Incidentally, every semilattice is also a topological monoid and this semimetric is subinvariant. $\endgroup$ Apr 13, 2014 at 1:35

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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).

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  • $\begingroup$ No need to apologize for answering your own question; it is the correct thing to do if you learn an answer later! $\endgroup$ Apr 16, 2014 at 9:00
  • $\begingroup$ Thank you, Eric, I will keep it in mind. It was yesterday that I discussed the problem with Jacek. We didn't get an answer, but he made a cute observation on the multiplicative structure of the real field, and this morning I ended up with the above. $\endgroup$ Apr 16, 2014 at 10:34
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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.

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  • $\begingroup$ I had thought of something in the same lines. Yet, I don't see how this answers my question. Essentially, you're taking as $\mathbb M$ the (forced) unitization of a null sgrp. Then, you're claiming that any topology on $M$ is compatible with the structure of $\mathbb M$. But I'm afraid this is false, even assuming that $\tau$ is 1st-countable. To see why, let $\mathbb S$ be a null sgrp with $\mathbb S=(S, \cdot)$, and let $\tau_S$ be any 1st-countable topology on $S$ such that $S \setminus \{0\} \in \tau_S$, where $0$ is the absorbing element of $\mathbb S$ (e.g., $\tau_S$ can be [...] $\endgroup$ Apr 16, 2014 at 9:56
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    $\begingroup$ Chris is not saying that any topology on $M$ makes it a topological monoid, but that any topology on $X$ makes $M=X\coprod \{0,\infty\}$ a topological monoid. If I'm not mistaken, though, any semimetric on $X$ can be extended to a subinvariant semimetric on $M$ (just make $0$ and $\infty$ infinitely far away from $X$ and each other), so this will only give a counterexample in the case that $X$ is not semimetrizable at all. $\endgroup$ Apr 16, 2014 at 10:47
  • $\begingroup$ [...] the discrete topology on $S$, as far as $S$ is countable, or more generally the topology $\{\emptyset, S\setminus \{0\}, S\}$); it is clear that $(\mathbb S,\tau_S)$ is a topological sgrp. Next, take $\mathbb M$ to be the unitization of $\mathbb S$, so that $\mathbb =S\cup\{S\}$ (I do everything in TG) and $\cdot$ is, by abuse of notation, the unique extension of the composition law of $\mathbb S$ to a binary operation on $M$ for which $xS=Sx=x$ for all $x$. Lastly, let $\tau=\tau_S\cup\{M,M\setminus\{0\}\}$. Then $\tau$ is a 1st-countable topology on $M$; however, (...) $\endgroup$ Apr 16, 2014 at 10:50
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    $\begingroup$ My example is not very complicated. As a space M is just X with two disjoint points added. The topology has a basis given by the topology $\tau_X$ of X, together with two more open sets $\{ 0 \}$ and $\{\infty\}$. The points 0 and $\infty$ are isolated, and the multiplication is continuous for any space X. I agree with Eric, M admits a semimetric if and only if the topological space X admits a semimetric. There are spaces X which are not semimetrizable, and hence this gives a counter example to the question as you asked it. $\endgroup$ Apr 16, 2014 at 14:32
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    $\begingroup$ Perhaps you meant to ask a different question? Like: "Suppose that M is a topological monoid whose underlying space is semimetrizable. Does there always exist a subinvariant metric inducing the topology?". $\endgroup$ Apr 16, 2014 at 14:39

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