Is $approx$ actually an entourage?

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I was looking at applying the ideas in the paper On Nonstandard Topology to Uniform spaces. Given a uniform space $(X,Phi)$, we can define the relation $approx$ on $^*X$ as follows $$approx , = bigcap ^*U:U in Phi $$
For example, if $(X,Phi)$ is the real numbers with their ordinary uniform structure, then $x approx y$ iff $x-y$ is an infinitesimal in the hyperreal numbers. If we use the metric $|x^3 - y^3|$ instead of the real numbers ordinary $|x - y|$ metric, then $H + frac 1H not approx H$ for infinite $H$ (although $H + frac 1H^3 approx H$).
In fact, it appears that we can prove that $approx$ is an entourage (i.e. $approx , in ^*Phi$). Take the following statements
- The statement "$V in Phi$"
- The statement "$V subseteq U$" for each $U in Phi$.
By the third property of entourages, any finite number of these statements is satisfiable. Therefore, by the Saturation property, there is nonstandard $V$ that satisfies the transfer of all these statements. In particular, $V in ^* Phi$. Additionally, $V subseteq , approx$. By the second property of entourages, $approx$ is therefore an entourage.
Is my proof correct though? The issue is that, going back to the real line with its usual uniform structure, we can assign each entourage a $V$ "diameter", defined as the supremum of the distances between points $x$ and $y$ such that $(x,y) in V$. Therefore, we should be able to assign either a hyperreal number of $infty$ to $approx$ like this, but we can't really. It is less than every positive non-infinitesimal (so in particular is not $infty$), but greater than every infinitesimal.
So, what's going on? Is $approx$ actually an entourage?
general-topology proof-verification paradoxes nonstandard-analysis uniform-spaces
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up vote
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down vote
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I was looking at applying the ideas in the paper On Nonstandard Topology to Uniform spaces. Given a uniform space $(X,Phi)$, we can define the relation $approx$ on $^*X$ as follows $$approx , = bigcap ^*U:U in Phi $$
For example, if $(X,Phi)$ is the real numbers with their ordinary uniform structure, then $x approx y$ iff $x-y$ is an infinitesimal in the hyperreal numbers. If we use the metric $|x^3 - y^3|$ instead of the real numbers ordinary $|x - y|$ metric, then $H + frac 1H not approx H$ for infinite $H$ (although $H + frac 1H^3 approx H$).
In fact, it appears that we can prove that $approx$ is an entourage (i.e. $approx , in ^*Phi$). Take the following statements
- The statement "$V in Phi$"
- The statement "$V subseteq U$" for each $U in Phi$.
By the third property of entourages, any finite number of these statements is satisfiable. Therefore, by the Saturation property, there is nonstandard $V$ that satisfies the transfer of all these statements. In particular, $V in ^* Phi$. Additionally, $V subseteq , approx$. By the second property of entourages, $approx$ is therefore an entourage.
Is my proof correct though? The issue is that, going back to the real line with its usual uniform structure, we can assign each entourage a $V$ "diameter", defined as the supremum of the distances between points $x$ and $y$ such that $(x,y) in V$. Therefore, we should be able to assign either a hyperreal number of $infty$ to $approx$ like this, but we can't really. It is less than every positive non-infinitesimal (so in particular is not $infty$), but greater than every infinitesimal.
So, what's going on? Is $approx$ actually an entourage?
general-topology proof-verification paradoxes nonstandard-analysis uniform-spaces
I'd wager that @MikhailKatz could easily answer this question... (I cannot quite do so myself off-the-cuff...)
â paul garrett
Feb 10 at 22:47
add a comment |Â
up vote
4
down vote
favorite
up vote
4
down vote
favorite
I was looking at applying the ideas in the paper On Nonstandard Topology to Uniform spaces. Given a uniform space $(X,Phi)$, we can define the relation $approx$ on $^*X$ as follows $$approx , = bigcap ^*U:U in Phi $$
For example, if $(X,Phi)$ is the real numbers with their ordinary uniform structure, then $x approx y$ iff $x-y$ is an infinitesimal in the hyperreal numbers. If we use the metric $|x^3 - y^3|$ instead of the real numbers ordinary $|x - y|$ metric, then $H + frac 1H not approx H$ for infinite $H$ (although $H + frac 1H^3 approx H$).
In fact, it appears that we can prove that $approx$ is an entourage (i.e. $approx , in ^*Phi$). Take the following statements
- The statement "$V in Phi$"
- The statement "$V subseteq U$" for each $U in Phi$.
By the third property of entourages, any finite number of these statements is satisfiable. Therefore, by the Saturation property, there is nonstandard $V$ that satisfies the transfer of all these statements. In particular, $V in ^* Phi$. Additionally, $V subseteq , approx$. By the second property of entourages, $approx$ is therefore an entourage.
Is my proof correct though? The issue is that, going back to the real line with its usual uniform structure, we can assign each entourage a $V$ "diameter", defined as the supremum of the distances between points $x$ and $y$ such that $(x,y) in V$. Therefore, we should be able to assign either a hyperreal number of $infty$ to $approx$ like this, but we can't really. It is less than every positive non-infinitesimal (so in particular is not $infty$), but greater than every infinitesimal.
So, what's going on? Is $approx$ actually an entourage?
general-topology proof-verification paradoxes nonstandard-analysis uniform-spaces
I was looking at applying the ideas in the paper On Nonstandard Topology to Uniform spaces. Given a uniform space $(X,Phi)$, we can define the relation $approx$ on $^*X$ as follows $$approx , = bigcap ^*U:U in Phi $$
For example, if $(X,Phi)$ is the real numbers with their ordinary uniform structure, then $x approx y$ iff $x-y$ is an infinitesimal in the hyperreal numbers. If we use the metric $|x^3 - y^3|$ instead of the real numbers ordinary $|x - y|$ metric, then $H + frac 1H not approx H$ for infinite $H$ (although $H + frac 1H^3 approx H$).
In fact, it appears that we can prove that $approx$ is an entourage (i.e. $approx , in ^*Phi$). Take the following statements
- The statement "$V in Phi$"
- The statement "$V subseteq U$" for each $U in Phi$.
By the third property of entourages, any finite number of these statements is satisfiable. Therefore, by the Saturation property, there is nonstandard $V$ that satisfies the transfer of all these statements. In particular, $V in ^* Phi$. Additionally, $V subseteq , approx$. By the second property of entourages, $approx$ is therefore an entourage.
Is my proof correct though? The issue is that, going back to the real line with its usual uniform structure, we can assign each entourage a $V$ "diameter", defined as the supremum of the distances between points $x$ and $y$ such that $(x,y) in V$. Therefore, we should be able to assign either a hyperreal number of $infty$ to $approx$ like this, but we can't really. It is less than every positive non-infinitesimal (so in particular is not $infty$), but greater than every infinitesimal.
So, what's going on? Is $approx$ actually an entourage?
general-topology proof-verification paradoxes nonstandard-analysis uniform-spaces
asked Feb 10 at 22:31
PyRulez
4,53822168
4,53822168
I'd wager that @MikhailKatz could easily answer this question... (I cannot quite do so myself off-the-cuff...)
â paul garrett
Feb 10 at 22:47
add a comment |Â
I'd wager that @MikhailKatz could easily answer this question... (I cannot quite do so myself off-the-cuff...)
â paul garrett
Feb 10 at 22:47
I'd wager that @MikhailKatz could easily answer this question... (I cannot quite do so myself off-the-cuff...)
â paul garrett
Feb 10 at 22:47
I'd wager that @MikhailKatz could easily answer this question... (I cannot quite do so myself off-the-cuff...)
â paul garrett
Feb 10 at 22:47
add a comment |Â
1 Answer
1
active
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up vote
2
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accepted
Your proof would be correct if you knew that $approx$ is an internal set. However, it usually won't be, and if it is not an internal set, it certainly can't be an element of $^*Phi$. Indeed the contradiction you reached shows that for the usual uniformity on the real line, $approx$ cannot be internal.
To be clear, $^*Phi$ is not a uniformity in the most literal set-theoretic sense. Rather, it satisfies the version of the definition of a uniformity you get by applying Transfer to the usual definition. In particular, the usual definition involves quantification over all subsets of $Xtimes X$ (i.e., elements of $mathcalP(Xtimes X)$), and this will turn into a quantification over all internal subsets of $^*Xtimes ^*X$ (since those are the elements of $^*mathcalP(Xtimes X)$).
Ah, so $^*Phi$ doesn't satisfy property two, right?
â PyRulez
Feb 11 at 0:26
If "property two" is "any subset of $Xtimes X$ containing an entourage is an entourage", then that's correct.
â Eric Wofsey
Feb 11 at 0:33
so when you transfer statements about sets, they become statements about internal sets?
â PyRulez
Feb 11 at 20:09
Yes, basically.
â Eric Wofsey
Feb 11 at 20:15
oh, okay. I think I understand now.
â PyRulez
Feb 11 at 20:16
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Your proof would be correct if you knew that $approx$ is an internal set. However, it usually won't be, and if it is not an internal set, it certainly can't be an element of $^*Phi$. Indeed the contradiction you reached shows that for the usual uniformity on the real line, $approx$ cannot be internal.
To be clear, $^*Phi$ is not a uniformity in the most literal set-theoretic sense. Rather, it satisfies the version of the definition of a uniformity you get by applying Transfer to the usual definition. In particular, the usual definition involves quantification over all subsets of $Xtimes X$ (i.e., elements of $mathcalP(Xtimes X)$), and this will turn into a quantification over all internal subsets of $^*Xtimes ^*X$ (since those are the elements of $^*mathcalP(Xtimes X)$).
Ah, so $^*Phi$ doesn't satisfy property two, right?
â PyRulez
Feb 11 at 0:26
If "property two" is "any subset of $Xtimes X$ containing an entourage is an entourage", then that's correct.
â Eric Wofsey
Feb 11 at 0:33
so when you transfer statements about sets, they become statements about internal sets?
â PyRulez
Feb 11 at 20:09
Yes, basically.
â Eric Wofsey
Feb 11 at 20:15
oh, okay. I think I understand now.
â PyRulez
Feb 11 at 20:16
add a comment |Â
up vote
2
down vote
accepted
Your proof would be correct if you knew that $approx$ is an internal set. However, it usually won't be, and if it is not an internal set, it certainly can't be an element of $^*Phi$. Indeed the contradiction you reached shows that for the usual uniformity on the real line, $approx$ cannot be internal.
To be clear, $^*Phi$ is not a uniformity in the most literal set-theoretic sense. Rather, it satisfies the version of the definition of a uniformity you get by applying Transfer to the usual definition. In particular, the usual definition involves quantification over all subsets of $Xtimes X$ (i.e., elements of $mathcalP(Xtimes X)$), and this will turn into a quantification over all internal subsets of $^*Xtimes ^*X$ (since those are the elements of $^*mathcalP(Xtimes X)$).
Ah, so $^*Phi$ doesn't satisfy property two, right?
â PyRulez
Feb 11 at 0:26
If "property two" is "any subset of $Xtimes X$ containing an entourage is an entourage", then that's correct.
â Eric Wofsey
Feb 11 at 0:33
so when you transfer statements about sets, they become statements about internal sets?
â PyRulez
Feb 11 at 20:09
Yes, basically.
â Eric Wofsey
Feb 11 at 20:15
oh, okay. I think I understand now.
â PyRulez
Feb 11 at 20:16
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Your proof would be correct if you knew that $approx$ is an internal set. However, it usually won't be, and if it is not an internal set, it certainly can't be an element of $^*Phi$. Indeed the contradiction you reached shows that for the usual uniformity on the real line, $approx$ cannot be internal.
To be clear, $^*Phi$ is not a uniformity in the most literal set-theoretic sense. Rather, it satisfies the version of the definition of a uniformity you get by applying Transfer to the usual definition. In particular, the usual definition involves quantification over all subsets of $Xtimes X$ (i.e., elements of $mathcalP(Xtimes X)$), and this will turn into a quantification over all internal subsets of $^*Xtimes ^*X$ (since those are the elements of $^*mathcalP(Xtimes X)$).
Your proof would be correct if you knew that $approx$ is an internal set. However, it usually won't be, and if it is not an internal set, it certainly can't be an element of $^*Phi$. Indeed the contradiction you reached shows that for the usual uniformity on the real line, $approx$ cannot be internal.
To be clear, $^*Phi$ is not a uniformity in the most literal set-theoretic sense. Rather, it satisfies the version of the definition of a uniformity you get by applying Transfer to the usual definition. In particular, the usual definition involves quantification over all subsets of $Xtimes X$ (i.e., elements of $mathcalP(Xtimes X)$), and this will turn into a quantification over all internal subsets of $^*Xtimes ^*X$ (since those are the elements of $^*mathcalP(Xtimes X)$).
edited Aug 20 at 4:38
answered Feb 10 at 23:57
Eric Wofsey
164k12191305
164k12191305
Ah, so $^*Phi$ doesn't satisfy property two, right?
â PyRulez
Feb 11 at 0:26
If "property two" is "any subset of $Xtimes X$ containing an entourage is an entourage", then that's correct.
â Eric Wofsey
Feb 11 at 0:33
so when you transfer statements about sets, they become statements about internal sets?
â PyRulez
Feb 11 at 20:09
Yes, basically.
â Eric Wofsey
Feb 11 at 20:15
oh, okay. I think I understand now.
â PyRulez
Feb 11 at 20:16
add a comment |Â
Ah, so $^*Phi$ doesn't satisfy property two, right?
â PyRulez
Feb 11 at 0:26
If "property two" is "any subset of $Xtimes X$ containing an entourage is an entourage", then that's correct.
â Eric Wofsey
Feb 11 at 0:33
so when you transfer statements about sets, they become statements about internal sets?
â PyRulez
Feb 11 at 20:09
Yes, basically.
â Eric Wofsey
Feb 11 at 20:15
oh, okay. I think I understand now.
â PyRulez
Feb 11 at 20:16
Ah, so $^*Phi$ doesn't satisfy property two, right?
â PyRulez
Feb 11 at 0:26
Ah, so $^*Phi$ doesn't satisfy property two, right?
â PyRulez
Feb 11 at 0:26
If "property two" is "any subset of $Xtimes X$ containing an entourage is an entourage", then that's correct.
â Eric Wofsey
Feb 11 at 0:33
If "property two" is "any subset of $Xtimes X$ containing an entourage is an entourage", then that's correct.
â Eric Wofsey
Feb 11 at 0:33
so when you transfer statements about sets, they become statements about internal sets?
â PyRulez
Feb 11 at 20:09
so when you transfer statements about sets, they become statements about internal sets?
â PyRulez
Feb 11 at 20:09
Yes, basically.
â Eric Wofsey
Feb 11 at 20:15
Yes, basically.
â Eric Wofsey
Feb 11 at 20:15
oh, okay. I think I understand now.
â PyRulez
Feb 11 at 20:16
oh, okay. I think I understand now.
â PyRulez
Feb 11 at 20:16
add a comment |Â
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I'd wager that @MikhailKatz could easily answer this question... (I cannot quite do so myself off-the-cuff...)
â paul garrett
Feb 10 at 22:47