Timeline for If $(\mathbb M, \tau)$ is a topological monoid, is $\tau$ always induced by a [left] subinvariant semimetric?
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| Apr 23, 2014 at 14:24 | comment | added | Salvo Tringali | I'm finally convinced: there're first-countable topologies which are not semimetrizable (see the comments to mathoverflow.net/questions/163559 for details). | |
| Apr 16, 2014 at 17:56 | comment | added | Salvo Tringali | @Chris: Thanks for your clarifications, which change many things and prove that also my 2nd reading of your answer was incorrect. To be honest, it wasn't so clear (to me?) which open sets you were actually adding to $\tau_X$: there's not a unique way to do it, and the "cheapest" way to do it is to add just $M$. Further, your answer still says: "Let $X$ be any topological space. [...]" So it's not really a question of being complicated or not. Btw, there's a mistake in my last comment: it should really read as "$d(0,x):=d(1,x):=0$ for all $x\in M$ and $d(x,0):=d(x,1):=\infty$ for $x\in X$". | |
| Apr 16, 2014 at 14:39 | comment | added | Chris Schommer-Pries | 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?". | |
| Apr 16, 2014 at 14:32 | comment | added | Chris Schommer-Pries | 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. | |
| Apr 16, 2014 at 13:17 | comment | added | Salvo Tringali | If my 2nd reading of this answer (based on Eric's comments) is right, then Chris' construction doesn't however work, at least in general (and it's not clear to me if it works at all). Eric is in fact right: a semimetric $d_X$ on $X$ always extends to a semimetric $d$ on $M$ by $d(0,x):=d(x,0):=d(1,y):=d(y,1):=\infty$ for all $x,y\in M$ with $x\ne 0$ and $y\ne 1$; but the only neighborhood of both $0$ and $1$ in the canonical topology of $d$ is $M$, so $d$ induces $\tau_X\cup\{M\}$ if $d_X$ induces $\tau_X$ (the topology initially given on $X$). The rest follows from the previous comment. | |
| Apr 16, 2014 at 11:43 | comment | added | Salvo Tringali | @Eric Wofsey. I see! So I was misreading Chris' answer. He's just starting with a topology $\tau_X$ on $X$ and taking $\tau=\tau_X\cup\{M\}$ as a topology on $M$, right? That's fine! Still, how to prove that $\tau$ is not induced by a left/right $\mathbb M$-invariant semimetric? This boils down to $\tau$ not being semimetrizable at all (which is close to your remarks), since for a semimetric $d$ on $M$ we would have (for Chris' construction) that, for all $x,y,z\in M$, $d(xz,yz)=d(zx,zy)=d(0,0)\le d(x,y)$ if $z\ne 1$ or $d(xz,yz)=d(zx,zy)=d(x,y)$ if $z=1$, i.e. $d$ would be subinvariant. | |
| Apr 16, 2014 at 10:53 | comment | added | Salvo Tringali | (...) for which $\mathbb M$ is the unitization of a null sgrp, how do you think to prove that there doesn't exist any left/right $\mathbb M$-subinvariant semimetric $d$ on $M$ such that $\tau$ the canonical topology induced by $d$? That's the thing. | |
| Apr 16, 2014 at 10:53 | comment | added | Salvo Tringali | (...) $(\mathbb M,\tau)$ is not a topological monoid, unless $\{S\}\in\tau$, which is the case iff $S=\{0\}$. In fact, $M\setminus \{0\}$ is an open neighborhood of $1$ in $\tau$ (by construction), and given any open neighborhood $U$ of $(S,S)$ in the product topology induced on $M\times M$ by $\tau$ we should have that $xy\ne 0$ for all $x,y\in U$, but this is false unless we can take $U=\{(S,S)\}$, which in turn is possible iff $M\setminus \{0\}=\{S\}$, i.e. $S=\{0\}$. Do I miss something? And in any case, even assuming that $(\mathbb M,\tau)$ is a 1st-countable topological monoid (...) | |
| Apr 16, 2014 at 10:50 | comment | added | Salvo Tringali | [...] 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, (...) | |
| Apr 16, 2014 at 10:47 | comment | added | Eric Wofsey | 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. | |
| Apr 16, 2014 at 9:56 | comment | added | Salvo Tringali | 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 [...] | |
| Apr 16, 2014 at 8:29 | history | answered | Chris Schommer-Pries | CC BY-SA 3.0 |