Return to Question

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.)

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.)

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 $\mathbf R^+$.

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 the canonical topology of a left or right subinvariant semimetric. However, both of the answers there (my feeling is that Chris Schommer-Pries' construction can be made to work for a suitable topology, even if for the moment I don't know how) do "critically" rely 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 is my question:

Q. If $\mathbb M$ is resilient, is it possible for $\tau$ to be the canonical topology of a left (respectively, right) $\mathbb M$-subinvariant semimetric?

I don't think so (which means, in particular, that I don't have a counterexample), but it's well possible that I'm missing something trivial (it wouldn't be a novelty...). So thanks in advance for any comment, hint, pointer, or whatever.

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.)

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 $\mathbf R^+$.

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 the canonical topology of a left or right subinvariant semimetric. However, both of the answers there (my feeling is that Chris Schommer-Pries' construction can be made to work for a suitable topology, even if for the moment I don't know how) do "critically" rely 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).

Q. If $\mathbb M$ is resilient, is it possible for $\tau$ to be the canonical topology of a left (respectively, right) $\mathbb M$-subinvariant semimetric?

I don't think so, but it's well possible that I'm missing something trivial (it wouldn't be a novelty...). So thanks in advance for any comment, hint, pointer, or whatever.

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 $\mathbf R^+$.

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 the canonical topology of a left or right subinvariant semimetric. However, both of the answers there (my feeling is that Chris Schommer-Pries' construction can be made to work for a suitable topology, even if for the moment I don't know how) do "critically" rely 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 is my question:

Q. If $\mathbb M$ is resilient, is it possible for $\tau$ to be the canonical topology of a left (respectively, right) $\mathbb M$-subinvariant semimetric?

I don't think so (which means, in particular, that I don't have a counterexample), but it's well possible that I'm missing something trivial (it wouldn't be a novelty...). So thanks in advance for any comment, hint, pointer, or whatever.