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What is the topology associated with the algebras for the ultrafilter monad?
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern)Is there a way to make tangent bundle a monad?Coproducts and pushouts of Boolean algebras and Heyting algebrasProof of theorems in the field of banach-and $c^*$-algebras in a categorial languageEpimorphisms of locally compact spacesThis is just the Eilenberg-Moore category, right?Which spaces can be used as “test spaces” for the Stone-Čech compactification?What is the opposite category of $operatornameTop$?Finitely presentable objects and the Kleisli categoryultrafilter convergence versus non-standard topologyMorita theory for algebras for a monad $T$
$begingroup$
It is easy to find references stating that the category of compact Hausdorff spaces $mathbfCompHaus$ is equivalent to the category of algebras for the ultrafilter monad, $mathbfbeta Alg$. After doing some digging, the $mathbfCompHausto mathbfbeta Alg$ half of the equivalence is simple enough, but I haven't been able to find a description of the $mathbfbeta Algto mathbfCompHaus$ half of this equivalence. I have tried to work it out, but I have little experience with topological spaces and am not sure what the associated topology ought to look like.
I'm wondering if anyone has a good reference that describes the $mathbfbeta Algto mathbfCompHaus$ half of the equivalence, or can describe it here.
general-topology category-theory
asked 2 hours ago
Malice VidrineMalice Vidrine
6,34121123
$endgroup$
add a comment |
$begingroup$
It is easy to find references stating that the category of compact Hausdorff spaces $mathbfCompHaus$ is equivalent to the category of algebras for the ultrafilter monad, $mathbfbeta Alg$. After doing some digging, the $mathbfCompHausto mathbfbeta Alg$ half of the equivalence is simple enough, but I haven't been able to find a description of the $mathbfbeta Algto mathbfCompHaus$ half of this equivalence. I have tried to work it out, but I have little experience with topological spaces and am not sure what the associated topology ought to look like.
I'm wondering if anyone has a good reference that describes the $mathbfbeta Algto mathbfCompHaus$ half of the equivalence, or can describe it here.
general-topology category-theory
asked 2 hours ago
Malice VidrineMalice Vidrine
6,34121123
$endgroup$
add a comment |
$begingroup$
It is easy to find references stating that the category of compact Hausdorff spaces $mathbfCompHaus$ is equivalent to the category of algebras for the ultrafilter monad, $mathbfbeta Alg$. After doing some digging, the $mathbfCompHausto mathbfbeta Alg$ half of the equivalence is simple enough, but I haven't been able to find a description of the $mathbfbeta Algto mathbfCompHaus$ half of this equivalence. I have tried to work it out, but I have little experience with topological spaces and am not sure what the associated topology ought to look like.
I'm wondering if anyone has a good reference that describes the $mathbfbeta Algto mathbfCompHaus$ half of the equivalence, or can describe it here.
general-topology category-theory
asked 2 hours ago
Malice VidrineMalice Vidrine
6,34121123
$endgroup$
It is easy to find references stating that the category of compact Hausdorff spaces $mathbfCompHaus$ is equivalent to the category of algebras for the ultrafilter monad, $mathbfbeta Alg$. After doing some digging, the $mathbfCompHausto mathbfbeta Alg$ half of the equivalence is simple enough, but I haven't been able to find a description of the $mathbfbeta Algto mathbfCompHaus$ half of this equivalence. I have tried to work it out, but I have little experience with topological spaces and am not sure what the associated topology ought to look like.
I'm wondering if anyone has a good reference that describes the $mathbfbeta Algto mathbfCompHaus$ half of the equivalence, or can describe it here.
general-topology category-theory
general-topology category-theory
asked 2 hours ago
Malice VidrineMalice Vidrine
6,34121123
asked 2 hours ago
Malice VidrineMalice Vidrine
6,34121123
asked 2 hours ago
Malice VidrineMalice Vidrine
6,34121123
asked 2 hours ago
Malice VidrineMalice Vidrine
6,34121123
asked 2 hours ago
Malice VidrineMalice Vidrine
6,34121123
6,34121123
add a comment |
add a comment |
1 Answer
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$begingroup$
The other half of the equivalence is described on the nLab page on ultrafilters.
Given an algebra structure $xicolon beta X to X$, we define the topology on $X$ by declaring that a subset $Usubseteq X$ is open if and only if for every point $xin U$ and every ultrafilter $Fin beta X$ such that $xi(F) = x$, we have $Uin F$.
answered 2 hours ago
Alex KruckmanAlex Kruckman
28.8k32758
$endgroup$
$begingroup$
Of course I looked at precisely the wrong nLab pages to answer my question :P Thanks!
$endgroup$
– Malice Vidrine
59 mins ago
add a comment |
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$begingroup$
The other half of the equivalence is described on the nLab page on ultrafilters.
Given an algebra structure $xicolon beta X to X$, we define the topology on $X$ by declaring that a subset $Usubseteq X$ is open if and only if for every point $xin U$ and every ultrafilter $Fin beta X$ such that $xi(F) = x$, we have $Uin F$.
answered 2 hours ago
Alex KruckmanAlex Kruckman
28.8k32758
$endgroup$
$begingroup$
Of course I looked at precisely the wrong nLab pages to answer my question :P Thanks!
$endgroup$
– Malice Vidrine
59 mins ago
add a comment |
$begingroup$
The other half of the equivalence is described on the nLab page on ultrafilters.
Given an algebra structure $xicolon beta X to X$, we define the topology on $X$ by declaring that a subset $Usubseteq X$ is open if and only if for every point $xin U$ and every ultrafilter $Fin beta X$ such that $xi(F) = x$, we have $Uin F$.
answered 2 hours ago
Alex KruckmanAlex Kruckman
28.8k32758
$endgroup$
$begingroup$
Of course I looked at precisely the wrong nLab pages to answer my question :P Thanks!
$endgroup$
– Malice Vidrine
59 mins ago
add a comment |
$begingroup$
The other half of the equivalence is described on the nLab page on ultrafilters.
Given an algebra structure $xicolon beta X to X$, we define the topology on $X$ by declaring that a subset $Usubseteq X$ is open if and only if for every point $xin U$ and every ultrafilter $Fin beta X$ such that $xi(F) = x$, we have $Uin F$.
answered 2 hours ago
Alex KruckmanAlex Kruckman
28.8k32758
$endgroup$
The other half of the equivalence is described on the nLab page on ultrafilters.
Given an algebra structure $xicolon beta X to X$, we define the topology on $X$ by declaring that a subset $Usubseteq X$ is open if and only if for every point $xin U$ and every ultrafilter $Fin beta X$ such that $xi(F) = x$, we have $Uin F$.
answered 2 hours ago
Alex KruckmanAlex Kruckman
28.8k32758
answered 2 hours ago
Alex KruckmanAlex Kruckman
28.8k32758
answered 2 hours ago
Alex KruckmanAlex Kruckman
28.8k32758
answered 2 hours ago
Alex KruckmanAlex Kruckman
28.8k32758
28.8k32758
$begingroup$
Of course I looked at precisely the wrong nLab pages to answer my question :P Thanks!
$endgroup$
– Malice Vidrine
59 mins ago
add a comment |
$begingroup$
Of course I looked at precisely the wrong nLab pages to answer my question :P Thanks!
$endgroup$
– Malice Vidrine
59 mins ago
$begingroup$
Of course I looked at precisely the wrong nLab pages to answer my question :P Thanks!
$endgroup$
– Malice Vidrine
59 mins ago
$begingroup$
Of course I looked at precisely the wrong nLab pages to answer my question :P Thanks!
$endgroup$
– Malice Vidrine
59 mins ago
add a comment |
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