Post History
#5: Post edited
by
watchmaker
·
2025-11-26T22:02:00Z (10 months ago)
Tell you what I missed
- This is only half of an answer but here we go:
- First step is to get rid of those binary functions $+ : X\times X \to X$ and $* : X\times X \to X$, and with it, the need to consider product topologies.
- And to separate *algebra* from *topology*.
- Name the topology as $\mathcal T$.
- Addition and multiplication can be seen as *translation* and *scaling*.
- For *any* field $X$, let $a\in X$ and $a\neq 0$.
- Define translations and scalings by $t_a(x)=x+a$ and $s_a(x)=a\cdot x$.
- Those have to be continuous, that's to say for every open set $\mathcal O\in \mathcal T$, the preimage has to be open $t_a^{-1}(\mathcal O)\in \mathcal T$ and $s_a^{-1}(\mathcal O)\in \mathcal T$.
- With $a\neq 0$ this translates to $a+\mathcal O\in \mathcal T$ and $a\cdot \mathcal O\in \mathcal T$, even if $a$ is substituted by $-a$ or $\frac1a$.
- Usually, the zero case causes trouble, but here it does not. $t_0$ is simply the identity, and $s^{-1}_0$ gives us either $\emptyset$ or $X$, depending if the argument was $0$. Requiring that all these functions be continuous even spares us to look back at the axioms to assure that $\emptyset$ and $X$ are included.
- So, if you throw in *any* collection of sets $\mathcal A\subset \mathcal P(X)$ you want to call *open*, all of $\{a\cdot A | A\in\mathcal A\} \cup \{a+ A | A\in\mathcal A\}\cup \{\emptyset, X\}$ would have to be open and be a subbase of the topology.
- This is where my wisdom comes to an end and why I call it half of an answer.
- I have no idea if the topologies gained by this procedure would collapse to only the trivial ones or could result in interesting new ones.
- ---
- To explain a bit what is in my mind when I say "collapse":
- Start with the simple requirement that the set $\{0\}$ should be open. Then, by translation, *every* set with a single element will be open, and you end up with the discrete topology.
- Now start with $a\neq b$, $A=\{a,b\}$. Scaling and shifting gives you $\{0,1\}$ to be open, shifting by -1 and intersecting gives you $\{0\}$ to be open, and you end up with the discrete topology, again.
- Having *any* non-empty final set $A\in\mathcal A$ (assuming $A$ is not translation invariant, as can happen for final fields $X$; missed that case at first *and nobody noticed!* ) in your starting collection will make you end up with the discrete topology, over again, no matter how niftily you choose the other sets.
- This is only half of an answer but here we go:
- First step is to get rid of those binary functions $+ : X\times X \to X$ and $* : X\times X \to X$, and with it, the need to consider product topologies.
- And to separate *algebra* from *topology*.
- Name the topology as $\mathcal T$.
- Addition and multiplication can be seen as *translation* and *scaling*.
- For *any* field $X$, let $a\in X$ and $a\neq 0$.
- Define translations and scalings by $t_a(x)=x+a$ and $s_a(x)=a\cdot x$.
- Those have to be continuous, that's to say for every open set $\mathcal O\in \mathcal T$, the preimage has to be open $t_a^{-1}(\mathcal O)\in \mathcal T$ and $s_a^{-1}(\mathcal O)\in \mathcal T$.
- With $a\neq 0$ this translates to $a+\mathcal O\in \mathcal T$ and $a\cdot \mathcal O\in \mathcal T$, even if $a$ is substituted by $-a$ or $\frac1a$.
- Usually, the zero case causes trouble, but here it does not. $t_0$ is simply the identity, and $s^{-1}_0$ gives us either $\emptyset$ or $X$, depending if the argument was $0$. Requiring that all these functions be continuous even spares us to look back at the axioms to assure that $\emptyset$ and $X$ are included.
- **Trouble here:** I missed that from my assumptions above, it does not follow (or at least, I can't see it and missed to show it) that the multiplicative inverse function $a \mapsto a^{-1}$ also is continuous.
- *This is *co*-didact, which I read as "teaching each other", so what's wrong and why didn't anyone notice and tell? When I misspelled "Turing machine" as "Touring machine" the coment was promptly there.*
- So, if you throw in *any* collection of sets $\mathcal A\subset \mathcal P(X)$ you want to call *open*, all of $\{a\cdot A | A\in\mathcal A\} \cup \{a+ A | A\in\mathcal A\}\cup \{\emptyset, X\}$ would have to be open and be a subbase of the topology.
- This is where my wisdom comes to an end and why I call it half of an answer.
- I have no idea if the topologies gained by this procedure would collapse to only the trivial ones or could result in interesting new ones.
- ---
- To explain a bit what is in my mind when I say "collapse":
- Start with the simple requirement that the set $\{0\}$ should be open. Then, by translation, *every* set with a single element will be open, and you end up with the discrete topology.
- Now start with $a\neq b$, $A=\{a,b\}$. Scaling and shifting gives you $\{0,1\}$ to be open, shifting by -1 and intersecting gives you $\{0\}$ to be open, and you end up with the discrete topology, again.
- Having *any* non-empty final set $A\in\mathcal A$ (assuming $A$ is not translation invariant, as can happen for final fields $X$; missed that case at first *and nobody noticed!* ) in your starting collection will make you end up with the discrete topology, over again, no matter how niftily you choose the other sets.
#4: Post edited
by
watchmaker
·
2025-11-17T20:03:25Z (10 months ago)
missing case
- This is only half of an answer but here we go:
- First step is to get rid of those binary functions $+ : X\times X \to X$ and $* : X\times X \to X$, and with it, the need to consider product topologies.
- And to separate *algebra* from *topology*.
- Name the topology as $\mathcal T$.
- Addition and multiplication can be seen as *translation* and *scaling*.
- For *any* field $X$, let $a\in X$ and $a\neq 0$.
- Define translations and scalings by $t_a(x)=x+a$ and $s_a(x)=a\cdot x$.
- Those have to be continuous, that's to say for every open set $\mathcal O\in \mathcal T$, the preimage has to be open $t_a^{-1}(\mathcal O)\in \mathcal T$ and $s_a^{-1}(\mathcal O)\in \mathcal T$.
- With $a\neq 0$ this translates to $a+\mathcal O\in \mathcal T$ and $a\cdot \mathcal O\in \mathcal T$, even if $a$ is substituted by $-a$ or $\frac1a$.
- Usually, the zero case causes trouble, but here it does not. $t_0$ is simply the identity, and $s^{-1}_0$ gives us either $\emptyset$ or $X$, depending if the argument was $0$. Requiring that all these functions be continuous even spares us to look back at the axioms to assure that $\emptyset$ and $X$ are included.
- So, if you throw in *any* collection of sets $\mathcal A\subset \mathcal P(X)$ you want to call *open*, all of $\{a\cdot A | A\in\mathcal A\} \cup \{a+ A | A\in\mathcal A\}\cup \{\emptyset, X\}$ would have to be open and be a subbase of the topology.
- This is where my wisdom comes to an end and why I call it half of an answer.
- I have no idea if the topologies gained by this procedure would collapse to only the trivial ones or could result in interesting new ones.
- ---
- To explain a bit what is in my mind when I say "collapse":
- Start with the simple requirement that the set $\{0\}$ should be open. Then, by translation, *every* set with a single element will be open, and you end up with the discrete topology.
- Now start with $a\neq b$, $A=\{a,b\}$. Scaling and shifting gives you $\{0,1\}$ to be open, shifting by -1 and intersecting gives you $\{0\}$ to be open, and you end up with the discrete topology, again.
Having *any* non-empty final set $A\in\mathcal A$ in your starting collection will make you end up with the discrete topology, over again, no matter how niftily you choose the other sets.
- This is only half of an answer but here we go:
- First step is to get rid of those binary functions $+ : X\times X \to X$ and $* : X\times X \to X$, and with it, the need to consider product topologies.
- And to separate *algebra* from *topology*.
- Name the topology as $\mathcal T$.
- Addition and multiplication can be seen as *translation* and *scaling*.
- For *any* field $X$, let $a\in X$ and $a\neq 0$.
- Define translations and scalings by $t_a(x)=x+a$ and $s_a(x)=a\cdot x$.
- Those have to be continuous, that's to say for every open set $\mathcal O\in \mathcal T$, the preimage has to be open $t_a^{-1}(\mathcal O)\in \mathcal T$ and $s_a^{-1}(\mathcal O)\in \mathcal T$.
- With $a\neq 0$ this translates to $a+\mathcal O\in \mathcal T$ and $a\cdot \mathcal O\in \mathcal T$, even if $a$ is substituted by $-a$ or $\frac1a$.
- Usually, the zero case causes trouble, but here it does not. $t_0$ is simply the identity, and $s^{-1}_0$ gives us either $\emptyset$ or $X$, depending if the argument was $0$. Requiring that all these functions be continuous even spares us to look back at the axioms to assure that $\emptyset$ and $X$ are included.
- So, if you throw in *any* collection of sets $\mathcal A\subset \mathcal P(X)$ you want to call *open*, all of $\{a\cdot A | A\in\mathcal A\} \cup \{a+ A | A\in\mathcal A\}\cup \{\emptyset, X\}$ would have to be open and be a subbase of the topology.
- This is where my wisdom comes to an end and why I call it half of an answer.
- I have no idea if the topologies gained by this procedure would collapse to only the trivial ones or could result in interesting new ones.
- ---
- To explain a bit what is in my mind when I say "collapse":
- Start with the simple requirement that the set $\{0\}$ should be open. Then, by translation, *every* set with a single element will be open, and you end up with the discrete topology.
- Now start with $a\neq b$, $A=\{a,b\}$. Scaling and shifting gives you $\{0,1\}$ to be open, shifting by -1 and intersecting gives you $\{0\}$ to be open, and you end up with the discrete topology, again.
- Having *any* non-empty final set $A\in\mathcal A$ (assuming $A$ is not translation invariant, as can happen for final fields $X$; missed that case at first *and nobody noticed!* ) in your starting collection will make you end up with the discrete topology, over again, no matter how niftily you choose the other sets.
#3: Post edited
by
watchmaker
·
2025-11-16T02:47:31Z (10 months ago)
Typos, more or less
- This is only half of an answer but here we go:
- First step is to get rid of those binary functions $+ : X\times X \to X$ and $* : X\times X \to X$, and with it, the need to consider product topologies.
- And to separate *algebra* from *topology*.
- Name the topology as $\mathcal T$.
- Addition and multiplication can be seen as *translation* and *scaling*.
- For *any* field $X$, let $a\in X$ and $a\neq 0$.
- Define translations and scalings by $t_a(x)=x+a$ and $s_a(x)=a\cdot x$.
- Those have to be continuous, that's to say for every open set $\mathcal O\in \mathcal T$, the preimage has to be open $t_a^{-1}(\mathcal O)\in \mathcal T$ and $s_a^{-1}(\mathcal O)\in \mathcal T$.
- With $a\neq 0$ this translates to $a+\mathcal O\in \mathcal T$ and $a\cdot \mathcal O\in \mathcal T$, even if $a$ is substituted by $-a$ or $\frac1a$.
Usually, the zero case causes trouble, but here it does not. $t_0$ is simply the identity, and $s^{-1}_0$ gives us either $\emptyset$ or $X$, depending if the argument was $0$. Requiring that all this functions be continuous even spares us to look back at the axioms to assure that $\emptyset$ and $X$ are included.So, if you throw in *any* collection of sets $\mathcal A\in \mathcal P(X)$ you want to call *open*, all of $\{a\cdot A | A\in\mathcal A\} \cup \{a+ A | A\in\mathcal A\}\cup \{\emptyset, X\}$ would have to be open and be a subbase of the topology.- This is where my wisdom comes to an end and why I call it half of an answer.
- I have no idea if the topologies gained by this procedure would collapse to only the trivial ones or could result in interesting new ones.
- ---
- To explain a bit what is in my mind when I say "collapse":
- Start with the simple requirement that the set $\{0\}$ should be open. Then, by translation, *every* set with a single element will be open, and you end up with the discrete topology.
- Now start with $a\neq b$, $A=\{a,b\}$. Scaling and shifting gives you $\{0,1\}$ to be open, shifting by -1 and intersecting gives you $\{0\}$ to be open, and you end up with the discrete topology, again.
- Having *any* non-empty final set $A\in\mathcal A$ in your starting collection will make you end up with the discrete topology, over again, no matter how niftily you choose the other sets.
- This is only half of an answer but here we go:
- First step is to get rid of those binary functions $+ : X\times X \to X$ and $* : X\times X \to X$, and with it, the need to consider product topologies.
- And to separate *algebra* from *topology*.
- Name the topology as $\mathcal T$.
- Addition and multiplication can be seen as *translation* and *scaling*.
- For *any* field $X$, let $a\in X$ and $a\neq 0$.
- Define translations and scalings by $t_a(x)=x+a$ and $s_a(x)=a\cdot x$.
- Those have to be continuous, that's to say for every open set $\mathcal O\in \mathcal T$, the preimage has to be open $t_a^{-1}(\mathcal O)\in \mathcal T$ and $s_a^{-1}(\mathcal O)\in \mathcal T$.
- With $a\neq 0$ this translates to $a+\mathcal O\in \mathcal T$ and $a\cdot \mathcal O\in \mathcal T$, even if $a$ is substituted by $-a$ or $\frac1a$.
- Usually, the zero case causes trouble, but here it does not. $t_0$ is simply the identity, and $s^{-1}_0$ gives us either $\emptyset$ or $X$, depending if the argument was $0$. Requiring that all these functions be continuous even spares us to look back at the axioms to assure that $\emptyset$ and $X$ are included.
- So, if you throw in *any* collection of sets $\mathcal A\subset \mathcal P(X)$ you want to call *open*, all of $\{a\cdot A | A\in\mathcal A\} \cup \{a+ A | A\in\mathcal A\}\cup \{\emptyset, X\}$ would have to be open and be a subbase of the topology.
- This is where my wisdom comes to an end and why I call it half of an answer.
- I have no idea if the topologies gained by this procedure would collapse to only the trivial ones or could result in interesting new ones.
- ---
- To explain a bit what is in my mind when I say "collapse":
- Start with the simple requirement that the set $\{0\}$ should be open. Then, by translation, *every* set with a single element will be open, and you end up with the discrete topology.
- Now start with $a\neq b$, $A=\{a,b\}$. Scaling and shifting gives you $\{0,1\}$ to be open, shifting by -1 and intersecting gives you $\{0\}$ to be open, and you end up with the discrete topology, again.
- Having *any* non-empty final set $A\in\mathcal A$ in your starting collection will make you end up with the discrete topology, over again, no matter how niftily you choose the other sets.
#2: Post edited
by
watchmaker
·
2025-11-16T02:35:18Z (10 months ago)
Explaininng about "collapse"
- This is only half of an answer but here we go:
- First step is to get rid of those binary functions $+ : X\times X \to X$ and $* : X\times X \to X$, and with it, the need to consider product topologies.
- And to separate *algebra* from *topology*.
- Name the topology as $\mathcal T$.
- Addition and multiplication can be seen as *translation* and *scaling*.
- For *any* field $X$, let $a\in X$ and $a\neq 0$.
- Define translations and scalings by $t_a(x)=x+a$ and $s_a(x)=a\cdot x$.
- Those have to be continuous, that's to say for every open set $\mathcal O\in \mathcal T$, the preimage has to be open $t_a^{-1}(\mathcal O)\in \mathcal T$ and $s_a^{-1}(\mathcal O)\in \mathcal T$.
- With $a\neq 0$ this translates to $a+\mathcal O\in \mathcal T$ and $a\cdot \mathcal O\in \mathcal T$, even if $a$ is substituted by $-a$ or $\frac1a$.
- Usually, the zero case causes trouble, but here it does not. $t_0$ is simply the identity, and $s^{-1}_0$ gives us either $\emptyset$ or $X$, depending if the argument was $0$. Requiring that all this functions be continuous even spares us to look back at the axioms to assure that $\emptyset$ and $X$ are included.
- So, if you throw in *any* collection of sets $\mathcal A\in \mathcal P(X)$ you want to call *open*, all of $\{a\cdot A | A\in\mathcal A\} \cup \{a+ A | A\in\mathcal A\}\cup \{\emptyset, X\}$ would have to be open and be a subbase of the topology.
- This is where my wisdom comes to an end and why I call it half of an answer.
I have no idea if the topologies gained by this procedure would collapse to only the trivial ones or could result in interesting new ones.
- This is only half of an answer but here we go:
- First step is to get rid of those binary functions $+ : X\times X \to X$ and $* : X\times X \to X$, and with it, the need to consider product topologies.
- And to separate *algebra* from *topology*.
- Name the topology as $\mathcal T$.
- Addition and multiplication can be seen as *translation* and *scaling*.
- For *any* field $X$, let $a\in X$ and $a\neq 0$.
- Define translations and scalings by $t_a(x)=x+a$ and $s_a(x)=a\cdot x$.
- Those have to be continuous, that's to say for every open set $\mathcal O\in \mathcal T$, the preimage has to be open $t_a^{-1}(\mathcal O)\in \mathcal T$ and $s_a^{-1}(\mathcal O)\in \mathcal T$.
- With $a\neq 0$ this translates to $a+\mathcal O\in \mathcal T$ and $a\cdot \mathcal O\in \mathcal T$, even if $a$ is substituted by $-a$ or $\frac1a$.
- Usually, the zero case causes trouble, but here it does not. $t_0$ is simply the identity, and $s^{-1}_0$ gives us either $\emptyset$ or $X$, depending if the argument was $0$. Requiring that all this functions be continuous even spares us to look back at the axioms to assure that $\emptyset$ and $X$ are included.
- So, if you throw in *any* collection of sets $\mathcal A\in \mathcal P(X)$ you want to call *open*, all of $\{a\cdot A | A\in\mathcal A\} \cup \{a+ A | A\in\mathcal A\}\cup \{\emptyset, X\}$ would have to be open and be a subbase of the topology.
- This is where my wisdom comes to an end and why I call it half of an answer.
- I have no idea if the topologies gained by this procedure would collapse to only the trivial ones or could result in interesting new ones.
- ---
- To explain a bit what is in my mind when I say "collapse":
- Start with the simple requirement that the set $\{0\}$ should be open. Then, by translation, *every* set with a single element will be open, and you end up with the discrete topology.
- Now start with $a\neq b$, $A=\{a,b\}$. Scaling and shifting gives you $\{0,1\}$ to be open, shifting by -1 and intersecting gives you $\{0\}$ to be open, and you end up with the discrete topology, again.
- Having *any* non-empty final set $A\in\mathcal A$ in your starting collection will make you end up with the discrete topology, over again, no matter how niftily you choose the other sets.
#1: Initial revision
by
watchmaker
·
2025-11-15T08:57:18Z (10 months ago)
This is only half of an answer but here we go:
First step is to get rid of those binary functions $+ : X\times X \to X$ and $* : X\times X \to X$, and with it, the need to consider product topologies.
And to separate *algebra* from *topology*.
Name the topology as $\mathcal T$.
Addition and multiplication can be seen as *translation* and *scaling*.
For *any* field $X$, let $a\in X$ and $a\neq 0$.
Define translations and scalings by $t_a(x)=x+a$ and $s_a(x)=a\cdot x$.
Those have to be continuous, that's to say for every open set $\mathcal O\in \mathcal T$, the preimage has to be open $t_a^{-1}(\mathcal O)\in \mathcal T$ and $s_a^{-1}(\mathcal O)\in \mathcal T$.
With $a\neq 0$ this translates to $a+\mathcal O\in \mathcal T$ and $a\cdot \mathcal O\in \mathcal T$, even if $a$ is substituted by $-a$ or $\frac1a$.
Usually, the zero case causes trouble, but here it does not. $t_0$ is simply the identity, and $s^{-1}_0$ gives us either $\emptyset$ or $X$, depending if the argument was $0$. Requiring that all this functions be continuous even spares us to look back at the axioms to assure that $\emptyset$ and $X$ are included.
So, if you throw in *any* collection of sets $\mathcal A\in \mathcal P(X)$ you want to call *open*, all of $\{a\cdot A | A\in\mathcal A\} \cup \{a+ A | A\in\mathcal A\}\cup \{\emptyset, X\}$ would have to be open and be a subbase of the topology.
This is where my wisdom comes to an end and why I call it half of an answer.
I have no idea if the topologies gained by this procedure would collapse to only the trivial ones or could result in interesting new ones.