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What is the topology associated with the algebras for the ultrafilter monad?
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It is easy to find references stating that the category of compact Hausdorff spaces $mathbf{CompHaus}$ is equivalent to the category of algebras for the ultrafilter monad, $mathbf{beta Alg}$. After doing some digging, the $mathbf{CompHaus}to mathbf{beta Alg}$ half of the equivalence is simple enough, but I haven't been able to find a description of the $mathbf{beta Alg}to mathbf{CompHaus}$ 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 $mathbf{beta Alg}to mathbf{CompHaus}$ 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 $mathbf{CompHaus}$ is equivalent to the category of algebras for the ultrafilter monad, $mathbf{beta Alg}$. After doing some digging, the $mathbf{CompHaus}to mathbf{beta Alg}$ half of the equivalence is simple enough, but I haven't been able to find a description of the $mathbf{beta Alg}to mathbf{CompHaus}$ 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 $mathbf{beta Alg}to mathbf{CompHaus}$ 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 $mathbf{CompHaus}$ is equivalent to the category of algebras for the ultrafilter monad, $mathbf{beta Alg}$. After doing some digging, the $mathbf{CompHaus}to mathbf{beta Alg}$ half of the equivalence is simple enough, but I haven't been able to find a description of the $mathbf{beta Alg}to mathbf{CompHaus}$ 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 $mathbf{beta Alg}to mathbf{CompHaus}$ 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 $mathbf{CompHaus}$ is equivalent to the category of algebras for the ultrafilter monad, $mathbf{beta Alg}$. After doing some digging, the $mathbf{CompHaus}to mathbf{beta Alg}$ half of the equivalence is simple enough, but I haven't been able to find a description of the $mathbf{beta Alg}to mathbf{CompHaus}$ 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 $mathbf{beta Alg}to mathbf{CompHaus}$ 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
1 hour ago
add a comment |
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1 Answer
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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
1 hour 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
1 hour 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
1 hour ago
add a comment |
$begingroup$
Of course I looked at precisely the wrong nLab pages to answer my question :P Thanks!
$endgroup$
– Malice Vidrine
1 hour ago
$begingroup$
Of course I looked at precisely the wrong nLab pages to answer my question :P Thanks!
$endgroup$
– Malice Vidrine
1 hour ago
$begingroup$
Of course I looked at precisely the wrong nLab pages to answer my question :P Thanks!
$endgroup$
– Malice Vidrine
1 hour ago
add a comment |
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