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?







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  • I'd wager that @MikhailKatz could easily answer this question... (I cannot quite do so myself off-the-cuff...)
    – paul garrett
    Feb 10 at 22:47














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?







share|cite|improve this question




















  • I'd wager that @MikhailKatz could easily answer this question... (I cannot quite do so myself off-the-cuff...)
    – paul garrett
    Feb 10 at 22:47












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?







share|cite|improve this question












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?









share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










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...)
    – 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...)
    – 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...)
– 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...)
– paul garrett
Feb 10 at 22:47










1 Answer
1






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 $^*mathcalP(Xtimes X)$).






share|cite|improve this answer






















  • Ah, so $^*Phi$ doesn't satisfy property two, right?
    – 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.
    – Eric Wofsey
    Feb 11 at 0:33










  • so when you transfer statements about sets, they become statements about internal sets?
    – PyRulez
    Feb 11 at 20:09










  • Yes, basically.
    – Eric Wofsey
    Feb 11 at 20:15










  • oh, okay. I think I understand now.
    – PyRulez
    Feb 11 at 20:16










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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 $^*mathcalP(Xtimes X)$).






share|cite|improve this answer






















  • Ah, so $^*Phi$ doesn't satisfy property two, right?
    – 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.
    – Eric Wofsey
    Feb 11 at 0:33










  • so when you transfer statements about sets, they become statements about internal sets?
    – PyRulez
    Feb 11 at 20:09










  • Yes, basically.
    – Eric Wofsey
    Feb 11 at 20:15










  • oh, okay. I think I understand now.
    – PyRulez
    Feb 11 at 20:16














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 $^*mathcalP(Xtimes X)$).






share|cite|improve this answer






















  • Ah, so $^*Phi$ doesn't satisfy property two, right?
    – 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.
    – Eric Wofsey
    Feb 11 at 0:33










  • so when you transfer statements about sets, they become statements about internal sets?
    – PyRulez
    Feb 11 at 20:09










  • Yes, basically.
    – Eric Wofsey
    Feb 11 at 20:15










  • oh, okay. I think I understand now.
    – PyRulez
    Feb 11 at 20:16












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 $^*mathcalP(Xtimes X)$).






share|cite|improve this answer














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 $^*mathcalP(Xtimes X)$).







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








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?
    – 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.
    – Eric Wofsey
    Feb 11 at 0:33










  • so when you transfer statements about sets, they become statements about internal sets?
    – PyRulez
    Feb 11 at 20:09










  • Yes, basically.
    – Eric Wofsey
    Feb 11 at 20:15










  • oh, okay. I think I understand now.
    – PyRulez
    Feb 11 at 20:16
















  • Ah, so $^*Phi$ doesn't satisfy property two, right?
    – 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.
    – Eric Wofsey
    Feb 11 at 0:33










  • so when you transfer statements about sets, they become statements about internal sets?
    – PyRulez
    Feb 11 at 20:09










  • Yes, basically.
    – Eric Wofsey
    Feb 11 at 20:15










  • oh, okay. I think I understand now.
    – PyRulez
    Feb 11 at 20:16















Ah, so $^*Phi$ doesn't satisfy property two, right?
– PyRulez
Feb 11 at 0:26




Ah, so $^*Phi$ doesn't satisfy property two, right?
– 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.
– 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.
– Eric Wofsey
Feb 11 at 0:33












so when you transfer statements about sets, they become statements about internal sets?
– PyRulez
Feb 11 at 20:09




so when you transfer statements about sets, they become statements about internal sets?
– PyRulez
Feb 11 at 20:09












Yes, basically.
– Eric Wofsey
Feb 11 at 20:15




Yes, basically.
– Eric Wofsey
Feb 11 at 20:15












oh, okay. I think I understand now.
– PyRulez
Feb 11 at 20:16




oh, okay. I think I understand now.
– PyRulez
Feb 11 at 20:16












 

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