Mixing
For every closed neighborhood $Delta_Xsubset D$ , Is there an entourage $U$ with $Usubseteq D$?
$begingroup$
Let $(X, mathcal{U})$ be an uniform space. It is known that every entourage $Uinmathcal{U}$ is a neighborhood of $Delta_X$, but the converse is not true, in general.
What can say about closed neighborhood of $Delta_X$? Is it true that for a closed neighborhood $Dneq Delta_X$ of $Delta_X$, there is $Uinmathcal{U}$ with $Usubseteq D$?
Thanks a lot.
general-topology uniform-spaces
asked Jan 6 at 11:33
user479859user479859
756
$endgroup$
add a comment |
$begingroup$
Let $(X, mathcal{U})$ be an uniform space. It is known that every entourage $Uinmathcal{U}$ is a neighborhood of $Delta_X$, but the converse is not true, in general.
What can say about closed neighborhood of $Delta_X$? Is it true that for a closed neighborhood $Dneq Delta_X$ of $Delta_X$, there is $Uinmathcal{U}$ with $Usubseteq D$?
Thanks a lot.
general-topology uniform-spaces
asked Jan 6 at 11:33
user479859user479859
756
$endgroup$
add a comment |
$begingroup$
Let $(X, mathcal{U})$ be an uniform space. It is known that every entourage $Uinmathcal{U}$ is a neighborhood of $Delta_X$, but the converse is not true, in general.
What can say about closed neighborhood of $Delta_X$? Is it true that for a closed neighborhood $Dneq Delta_X$ of $Delta_X$, there is $Uinmathcal{U}$ with $Usubseteq D$?
Thanks a lot.
general-topology uniform-spaces
asked Jan 6 at 11:33
user479859user479859
756
$endgroup$
Let $(X, mathcal{U})$ be an uniform space. It is known that every entourage $Uinmathcal{U}$ is a neighborhood of $Delta_X$, but the converse is not true, in general.
What can say about closed neighborhood of $Delta_X$? Is it true that for a closed neighborhood $Dneq Delta_X$ of $Delta_X$, there is $Uinmathcal{U}$ with $Usubseteq D$?
Thanks a lot.
general-topology uniform-spaces
general-topology uniform-spaces
asked Jan 6 at 11:33
user479859user479859
756
asked Jan 6 at 11:33
user479859user479859
756
asked Jan 6 at 11:33
user479859user479859
756
asked Jan 6 at 11:33
user479859user479859
756
asked Jan 6 at 11:33
user479859user479859
756
756
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
No, that's not true in general. Consider $mathbb{R}$ with its standard uniform structure. Let$$D^star=left{(x,y)inmathbb{R}^2,middle|,-frac1{1+x^2}leqslant yleqslantfrac1{1+x^2}right}$$and let $D$ be what you obtain when you apply to $D^star$ a rotation of $fracpi4$ radians around the origin. Then $D$ is a closed neighborhood of $Delta_{mathbb R}$, but it contains no entourage.
answered Jan 6 at 11:44
José Carlos SantosJosé Carlos Santos
156k22126227
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add a comment |
$begingroup$
Show that in a separated uniform space: If we have a neighbourhood $U$ of $Delta_X$ that is no entourage, then $overline{U}$ is a closed neighbourhood of $Delta_X$ that contains no entourage.
answered Jan 6 at 13:19
Henno BrandsmaHenno Brandsma
107k347114
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add a comment |
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2 Answers
2
active
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2 Answers
2
active
oldest
votes
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$begingroup$
No, that's not true in general. Consider $mathbb{R}$ with its standard uniform structure. Let$$D^star=left{(x,y)inmathbb{R}^2,middle|,-frac1{1+x^2}leqslant yleqslantfrac1{1+x^2}right}$$and let $D$ be what you obtain when you apply to $D^star$ a rotation of $fracpi4$ radians around the origin. Then $D$ is a closed neighborhood of $Delta_{mathbb R}$, but it contains no entourage.
answered Jan 6 at 11:44
José Carlos SantosJosé Carlos Santos
156k22126227
$endgroup$
add a comment |
$begingroup$
No, that's not true in general. Consider $mathbb{R}$ with its standard uniform structure. Let$$D^star=left{(x,y)inmathbb{R}^2,middle|,-frac1{1+x^2}leqslant yleqslantfrac1{1+x^2}right}$$and let $D$ be what you obtain when you apply to $D^star$ a rotation of $fracpi4$ radians around the origin. Then $D$ is a closed neighborhood of $Delta_{mathbb R}$, but it contains no entourage.
answered Jan 6 at 11:44
José Carlos SantosJosé Carlos Santos
156k22126227
$endgroup$
add a comment |
$begingroup$
No, that's not true in general. Consider $mathbb{R}$ with its standard uniform structure. Let$$D^star=left{(x,y)inmathbb{R}^2,middle|,-frac1{1+x^2}leqslant yleqslantfrac1{1+x^2}right}$$and let $D$ be what you obtain when you apply to $D^star$ a rotation of $fracpi4$ radians around the origin. Then $D$ is a closed neighborhood of $Delta_{mathbb R}$, but it contains no entourage.
answered Jan 6 at 11:44
José Carlos SantosJosé Carlos Santos
156k22126227
$endgroup$
No, that's not true in general. Consider $mathbb{R}$ with its standard uniform structure. Let$$D^star=left{(x,y)inmathbb{R}^2,middle|,-frac1{1+x^2}leqslant yleqslantfrac1{1+x^2}right}$$and let $D$ be what you obtain when you apply to $D^star$ a rotation of $fracpi4$ radians around the origin. Then $D$ is a closed neighborhood of $Delta_{mathbb R}$, but it contains no entourage.
answered Jan 6 at 11:44
José Carlos SantosJosé Carlos Santos
156k22126227
answered Jan 6 at 11:44
José Carlos SantosJosé Carlos Santos
156k22126227
answered Jan 6 at 11:44
José Carlos SantosJosé Carlos Santos
156k22126227
answered Jan 6 at 11:44
José Carlos SantosJosé Carlos Santos
156k22126227
156k22126227
add a comment |
add a comment |
$begingroup$
Show that in a separated uniform space: If we have a neighbourhood $U$ of $Delta_X$ that is no entourage, then $overline{U}$ is a closed neighbourhood of $Delta_X$ that contains no entourage.
answered Jan 6 at 13:19
Henno BrandsmaHenno Brandsma
107k347114
$endgroup$
add a comment |
$begingroup$
Show that in a separated uniform space: If we have a neighbourhood $U$ of $Delta_X$ that is no entourage, then $overline{U}$ is a closed neighbourhood of $Delta_X$ that contains no entourage.
answered Jan 6 at 13:19
Henno BrandsmaHenno Brandsma
107k347114
$endgroup$
add a comment |
$begingroup$
Show that in a separated uniform space: If we have a neighbourhood $U$ of $Delta_X$ that is no entourage, then $overline{U}$ is a closed neighbourhood of $Delta_X$ that contains no entourage.
answered Jan 6 at 13:19
Henno BrandsmaHenno Brandsma
107k347114
$endgroup$
Show that in a separated uniform space: If we have a neighbourhood $U$ of $Delta_X$ that is no entourage, then $overline{U}$ is a closed neighbourhood of $Delta_X$ that contains no entourage.
answered Jan 6 at 13:19
Henno BrandsmaHenno Brandsma
107k347114
answered Jan 6 at 13:19
Henno BrandsmaHenno Brandsma
107k347114
answered Jan 6 at 13:19
Henno BrandsmaHenno Brandsma
107k347114
answered Jan 6 at 13:19
Henno BrandsmaHenno Brandsma
107k347114
107k347114
add a comment |
add a comment |
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