Ultrafilters as a double dual
$begingroup$
Given a set $X$, let $beta X$ denote the set of ultrafilters. The following theorems are known:
$X$ canonically embeds into $beta X$ (by taking principal ultrafilters);- If $X$ is finite, then there are no non-principal ultrafilters, so $beta X = X$.
- If $X$ is infinite, then (assuming choice) we have $|beta X| = 2^{2^{|X|}}$.
These are reminiscent of similar claims that can be made about vector spaces and double duals:
$V$ canonically embeds into $V^{star star}$;- If $V$ is finite-dimensional, then we have $V = V^{star star}$;
- If $V$ is infinite-dimensional, then (assuming choice) we have $dim(V^{star star}) = 2^{2^{dim(V)}}$.
This suggests that the operation of taking the collection of ultrafilters on a set can be viewed as a double iterate of some 'duality' of sets. Can this be made precise: that is to say, is there some notion of a 'dual' of a set $X$, $delta X$, such that the following are true?
- The double dual $delta delta X$ is (canonically isomorphic to) the set $beta X$ of ultrafilters on $X$;
- If $X$ is finite, then $|delta X| = |X|$ (but not canonically so);
- If $X$ is infinite, then (assuming choice) $|delta X| = 2^{|X|}$.
Apart from the tempting analogy between $beta X$ and $V^{star star}$, further evidence for this conjecture is that $beta$ can be given the structure of a monad (the 'ultrafilter monad'), and monads can be obtained from a pair of adjunctions.
ct.category-theory lo.logic gn.general-topology ultrafilters
edited 7 hours ago
YCor
28.1k483136
asked 8 hours ago
Adam P. GoucherAdam P. Goucher
6,56322857
$endgroup$
add a comment |
$begingroup$
Given a set $X$, let $beta X$ denote the set of ultrafilters. The following theorems are known:
$X$ canonically embeds into $beta X$ (by taking principal ultrafilters);- If $X$ is finite, then there are no non-principal ultrafilters, so $beta X = X$.
- If $X$ is infinite, then (assuming choice) we have $|beta X| = 2^{2^{|X|}}$.
These are reminiscent of similar claims that can be made about vector spaces and double duals:
$V$ canonically embeds into $V^{star star}$;- If $V$ is finite-dimensional, then we have $V = V^{star star}$;
- If $V$ is infinite-dimensional, then (assuming choice) we have $dim(V^{star star}) = 2^{2^{dim(V)}}$.
This suggests that the operation of taking the collection of ultrafilters on a set can be viewed as a double iterate of some 'duality' of sets. Can this be made precise: that is to say, is there some notion of a 'dual' of a set $X$, $delta X$, such that the following are true?
- The double dual $delta delta X$ is (canonically isomorphic to) the set $beta X$ of ultrafilters on $X$;
- If $X$ is finite, then $|delta X| = |X|$ (but not canonically so);
- If $X$ is infinite, then (assuming choice) $|delta X| = 2^{|X|}$.
Apart from the tempting analogy between $beta X$ and $V^{star star}$, further evidence for this conjecture is that $beta$ can be given the structure of a monad (the 'ultrafilter monad'), and monads can be obtained from a pair of adjunctions.
ct.category-theory lo.logic gn.general-topology ultrafilters
edited 7 hours ago
YCor
28.1k483136
asked 8 hours ago
Adam P. GoucherAdam P. Goucher
6,56322857
$endgroup$
19
$begingroup$
Well, one of the very first things that comes to mind that's sort of in this vein is that $beta X = hom_{text{Bool}}(hom_{text{Set}}(X, 2), 2)$. But if you want to pursue your analogy at a deeper level, try golem.ph.utexas.edu/category/2012/09/…, where both the ultrafilter monad and the double dualization monad are reckoned to be codensity monads induced by the full inclusions of finitary objects.
$endgroup$
– Todd Trimble♦
8 hours ago
2
$begingroup$
Close to Todd's comment, I'd view $beta X$ as $F(X)=mathrm{hom}_{mathrm{Bool}}(mathrm{hom}_{mathrm{Top}}(X,mathbf{Z}/2mathbf{Z}))$. In general, I guess that for a topological space $X$, the map $Xto F(X)$ is the initial object for the category of continuous maps from $X$ to compact Hausdorff totally disconnected topological spaces. A difference with taking biduals is that $F(F(X))=F(X)$ by Stone duality.
$endgroup$
– YCor
7 hours ago
1
$begingroup$
Well, one major difference is that without choice it is always the case that $X$ embeds into $beta X$, it's just not provable that the embedding is not surjective; whereas $V^*$ might be trivial, let alone $V^{**}$, even though $V$ isn't.
$endgroup$
– Asaf Karagila
7 hours ago
add a comment |
$begingroup$
Given a set $X$, let $beta X$ denote the set of ultrafilters. The following theorems are known:
$X$ canonically embeds into $beta X$ (by taking principal ultrafilters);- If $X$ is finite, then there are no non-principal ultrafilters, so $beta X = X$.
- If $X$ is infinite, then (assuming choice) we have $|beta X| = 2^{2^{|X|}}$.
These are reminiscent of similar claims that can be made about vector spaces and double duals:
$V$ canonically embeds into $V^{star star}$;- If $V$ is finite-dimensional, then we have $V = V^{star star}$;
- If $V$ is infinite-dimensional, then (assuming choice) we have $dim(V^{star star}) = 2^{2^{dim(V)}}$.
This suggests that the operation of taking the collection of ultrafilters on a set can be viewed as a double iterate of some 'duality' of sets. Can this be made precise: that is to say, is there some notion of a 'dual' of a set $X$, $delta X$, such that the following are true?
- The double dual $delta delta X$ is (canonically isomorphic to) the set $beta X$ of ultrafilters on $X$;
- If $X$ is finite, then $|delta X| = |X|$ (but not canonically so);
- If $X$ is infinite, then (assuming choice) $|delta X| = 2^{|X|}$.
Apart from the tempting analogy between $beta X$ and $V^{star star}$, further evidence for this conjecture is that $beta$ can be given the structure of a monad (the 'ultrafilter monad'), and monads can be obtained from a pair of adjunctions.
ct.category-theory lo.logic gn.general-topology ultrafilters
edited 7 hours ago
YCor
28.1k483136
asked 8 hours ago
Adam P. GoucherAdam P. Goucher
6,56322857
$endgroup$
Given a set $X$, let $beta X$ denote the set of ultrafilters. The following theorems are known:
$X$ canonically embeds into $beta X$ (by taking principal ultrafilters);- If $X$ is finite, then there are no non-principal ultrafilters, so $beta X = X$.
- If $X$ is infinite, then (assuming choice) we have $|beta X| = 2^{2^{|X|}}$.
These are reminiscent of similar claims that can be made about vector spaces and double duals:
$V$ canonically embeds into $V^{star star}$;- If $V$ is finite-dimensional, then we have $V = V^{star star}$;
- If $V$ is infinite-dimensional, then (assuming choice) we have $dim(V^{star star}) = 2^{2^{dim(V)}}$.
This suggests that the operation of taking the collection of ultrafilters on a set can be viewed as a double iterate of some 'duality' of sets. Can this be made precise: that is to say, is there some notion of a 'dual' of a set $X$, $delta X$, such that the following are true?
- The double dual $delta delta X$ is (canonically isomorphic to) the set $beta X$ of ultrafilters on $X$;
- If $X$ is finite, then $|delta X| = |X|$ (but not canonically so);
- If $X$ is infinite, then (assuming choice) $|delta X| = 2^{|X|}$.
Apart from the tempting analogy between $beta X$ and $V^{star star}$, further evidence for this conjecture is that $beta$ can be given the structure of a monad (the 'ultrafilter monad'), and monads can be obtained from a pair of adjunctions.
ct.category-theory lo.logic gn.general-topology ultrafilters
ct.category-theory lo.logic gn.general-topology ultrafilters
edited 7 hours ago
YCor
28.1k483136
asked 8 hours ago
Adam P. GoucherAdam P. Goucher
6,56322857
edited 7 hours ago
YCor
28.1k483136
asked 8 hours ago
Adam P. GoucherAdam P. Goucher
6,56322857
edited 7 hours ago
YCor
28.1k483136
edited 7 hours ago
YCor
28.1k483136
edited 7 hours ago
YCor
28.1k483136
28.1k483136
asked 8 hours ago
Adam P. GoucherAdam P. Goucher
6,56322857
asked 8 hours ago
Adam P. GoucherAdam P. Goucher
6,56322857
asked 8 hours ago
Adam P. GoucherAdam P. Goucher
6,56322857
6,56322857
19
$begingroup$
Well, one of the very first things that comes to mind that's sort of in this vein is that $beta X = hom_{text{Bool}}(hom_{text{Set}}(X, 2), 2)$. But if you want to pursue your analogy at a deeper level, try golem.ph.utexas.edu/category/2012/09/…, where both the ultrafilter monad and the double dualization monad are reckoned to be codensity monads induced by the full inclusions of finitary objects.
$endgroup$
– Todd Trimble♦
8 hours ago
2
$begingroup$
Close to Todd's comment, I'd view $beta X$ as $F(X)=mathrm{hom}_{mathrm{Bool}}(mathrm{hom}_{mathrm{Top}}(X,mathbf{Z}/2mathbf{Z}))$. In general, I guess that for a topological space $X$, the map $Xto F(X)$ is the initial object for the category of continuous maps from $X$ to compact Hausdorff totally disconnected topological spaces. A difference with taking biduals is that $F(F(X))=F(X)$ by Stone duality.
$endgroup$
– YCor
7 hours ago
1
$begingroup$
Well, one major difference is that without choice it is always the case that $X$ embeds into $beta X$, it's just not provable that the embedding is not surjective; whereas $V^*$ might be trivial, let alone $V^{**}$, even though $V$ isn't.
$endgroup$
– Asaf Karagila
7 hours ago
add a comment |
19
$begingroup$
Well, one of the very first things that comes to mind that's sort of in this vein is that $beta X = hom_{text{Bool}}(hom_{text{Set}}(X, 2), 2)$. But if you want to pursue your analogy at a deeper level, try golem.ph.utexas.edu/category/2012/09/…, where both the ultrafilter monad and the double dualization monad are reckoned to be codensity monads induced by the full inclusions of finitary objects.
$endgroup$
– Todd Trimble♦
8 hours ago
2
$begingroup$
Close to Todd's comment, I'd view $beta X$ as $F(X)=mathrm{hom}_{mathrm{Bool}}(mathrm{hom}_{mathrm{Top}}(X,mathbf{Z}/2mathbf{Z}))$. In general, I guess that for a topological space $X$, the map $Xto F(X)$ is the initial object for the category of continuous maps from $X$ to compact Hausdorff totally disconnected topological spaces. A difference with taking biduals is that $F(F(X))=F(X)$ by Stone duality.
$endgroup$
– YCor
7 hours ago
1
$begingroup$
Well, one major difference is that without choice it is always the case that $X$ embeds into $beta X$, it's just not provable that the embedding is not surjective; whereas $V^*$ might be trivial, let alone $V^{**}$, even though $V$ isn't.
$endgroup$
– Asaf Karagila
7 hours ago
19
19
$begingroup$
Well, one of the very first things that comes to mind that's sort of in this vein is that $beta X = hom_{text{Bool}}(hom_{text{Set}}(X, 2), 2)$. But if you want to pursue your analogy at a deeper level, try golem.ph.utexas.edu/category/2012/09/…, where both the ultrafilter monad and the double dualization monad are reckoned to be codensity monads induced by the full inclusions of finitary objects.
$endgroup$
– Todd Trimble♦
8 hours ago
$begingroup$
Well, one of the very first things that comes to mind that's sort of in this vein is that $beta X = hom_{text{Bool}}(hom_{text{Set}}(X, 2), 2)$. But if you want to pursue your analogy at a deeper level, try golem.ph.utexas.edu/category/2012/09/…, where both the ultrafilter monad and the double dualization monad are reckoned to be codensity monads induced by the full inclusions of finitary objects.
$endgroup$
– Todd Trimble♦
8 hours ago
2
2
$begingroup$
Close to Todd's comment, I'd view $beta X$ as $F(X)=mathrm{hom}_{mathrm{Bool}}(mathrm{hom}_{mathrm{Top}}(X,mathbf{Z}/2mathbf{Z}))$. In general, I guess that for a topological space $X$, the map $Xto F(X)$ is the initial object for the category of continuous maps from $X$ to compact Hausdorff totally disconnected topological spaces. A difference with taking biduals is that $F(F(X))=F(X)$ by Stone duality.
$endgroup$
– YCor
7 hours ago
$begingroup$
Close to Todd's comment, I'd view $beta X$ as $F(X)=mathrm{hom}_{mathrm{Bool}}(mathrm{hom}_{mathrm{Top}}(X,mathbf{Z}/2mathbf{Z}))$. In general, I guess that for a topological space $X$, the map $Xto F(X)$ is the initial object for the category of continuous maps from $X$ to compact Hausdorff totally disconnected topological spaces. A difference with taking biduals is that $F(F(X))=F(X)$ by Stone duality.
$endgroup$
– YCor
7 hours ago
1
1
$begingroup$
Well, one major difference is that without choice it is always the case that $X$ embeds into $beta X$, it's just not provable that the embedding is not surjective; whereas $V^*$ might be trivial, let alone $V^{**}$, even though $V$ isn't.
$endgroup$
– Asaf Karagila
7 hours ago
$begingroup$
Well, one major difference is that without choice it is always the case that $X$ embeds into $beta X$, it's just not provable that the embedding is not surjective; whereas $V^*$ might be trivial, let alone $V^{**}$, even though $V$ isn't.
$endgroup$
– Asaf Karagila
7 hours ago
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
This is a quite standard idea in functional analysis. Let $X$ be any set and let $c_0(X)$ be the space of all functions from $X$ to $mathbb{C}$ which go to zero at infinity. Then the algebra homomorphisms from $c_0(X)$ to $mathbb{C}$ are precisely the point evaluations at elements of $X$, i.e., the spectrum of $c_0(X)$ is naturally identified with $X$.
Going to the second dual we get $l^infty(X)$, the space of all bounded functions from $X$ to $mathbb{C}$, whose spectrum is naturally identified with $beta X$.
By the way, this generalizes to locally compact Hausdorff spaces. If $X$ is such a space, then the spectrum of $C_0(X)$ (the continuous functions on $X$ which vanish at infinity), equipped with its natural topology, is homeomorphic to $X$, and the spectrum of $C_0(X)^{**}$ is homeomorphic to $beta X$. This is bread-and-butter C*-algebra stuff.
answered 7 hours ago
Nik WeaverNik Weaver
21.4k148131
$endgroup$
2
$begingroup$
What does it mean for a function from a set $X$ to $mathbb{C}$ to "go to zero at infinity"?
$endgroup$
– Alex Kruckman
4 hours ago
4
$begingroup$
For any $epsilon > 0$, there is a finite subset of $X$ off of which $|f(x)| leq epsilon$.
$endgroup$
– Nik Weaver
4 hours ago
3
$begingroup$
Equivalently, the extension of $f$ to the one point compactification of $X$ (with the discrete topology) which sets $f(infty) = 0$ is continuous. Hence "goes to zero at infinity".
$endgroup$
– Nik Weaver
4 hours ago
add a comment |
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$begingroup$
This is a quite standard idea in functional analysis. Let $X$ be any set and let $c_0(X)$ be the space of all functions from $X$ to $mathbb{C}$ which go to zero at infinity. Then the algebra homomorphisms from $c_0(X)$ to $mathbb{C}$ are precisely the point evaluations at elements of $X$, i.e., the spectrum of $c_0(X)$ is naturally identified with $X$.
Going to the second dual we get $l^infty(X)$, the space of all bounded functions from $X$ to $mathbb{C}$, whose spectrum is naturally identified with $beta X$.
By the way, this generalizes to locally compact Hausdorff spaces. If $X$ is such a space, then the spectrum of $C_0(X)$ (the continuous functions on $X$ which vanish at infinity), equipped with its natural topology, is homeomorphic to $X$, and the spectrum of $C_0(X)^{**}$ is homeomorphic to $beta X$. This is bread-and-butter C*-algebra stuff.
answered 7 hours ago
Nik WeaverNik Weaver
21.4k148131
$endgroup$
2
$begingroup$
What does it mean for a function from a set $X$ to $mathbb{C}$ to "go to zero at infinity"?
$endgroup$
– Alex Kruckman
4 hours ago
4
$begingroup$
For any $epsilon > 0$, there is a finite subset of $X$ off of which $|f(x)| leq epsilon$.
$endgroup$
– Nik Weaver
4 hours ago
3
$begingroup$
Equivalently, the extension of $f$ to the one point compactification of $X$ (with the discrete topology) which sets $f(infty) = 0$ is continuous. Hence "goes to zero at infinity".
$endgroup$
– Nik Weaver
4 hours ago
add a comment |
$begingroup$
This is a quite standard idea in functional analysis. Let $X$ be any set and let $c_0(X)$ be the space of all functions from $X$ to $mathbb{C}$ which go to zero at infinity. Then the algebra homomorphisms from $c_0(X)$ to $mathbb{C}$ are precisely the point evaluations at elements of $X$, i.e., the spectrum of $c_0(X)$ is naturally identified with $X$.
Going to the second dual we get $l^infty(X)$, the space of all bounded functions from $X$ to $mathbb{C}$, whose spectrum is naturally identified with $beta X$.
By the way, this generalizes to locally compact Hausdorff spaces. If $X$ is such a space, then the spectrum of $C_0(X)$ (the continuous functions on $X$ which vanish at infinity), equipped with its natural topology, is homeomorphic to $X$, and the spectrum of $C_0(X)^{**}$ is homeomorphic to $beta X$. This is bread-and-butter C*-algebra stuff.
answered 7 hours ago
Nik WeaverNik Weaver
21.4k148131
$endgroup$
2
$begingroup$
What does it mean for a function from a set $X$ to $mathbb{C}$ to "go to zero at infinity"?
$endgroup$
– Alex Kruckman
4 hours ago
4
$begingroup$
For any $epsilon > 0$, there is a finite subset of $X$ off of which $|f(x)| leq epsilon$.
$endgroup$
– Nik Weaver
4 hours ago
3
$begingroup$
Equivalently, the extension of $f$ to the one point compactification of $X$ (with the discrete topology) which sets $f(infty) = 0$ is continuous. Hence "goes to zero at infinity".
$endgroup$
– Nik Weaver
4 hours ago
add a comment |
$begingroup$
This is a quite standard idea in functional analysis. Let $X$ be any set and let $c_0(X)$ be the space of all functions from $X$ to $mathbb{C}$ which go to zero at infinity. Then the algebra homomorphisms from $c_0(X)$ to $mathbb{C}$ are precisely the point evaluations at elements of $X$, i.e., the spectrum of $c_0(X)$ is naturally identified with $X$.
Going to the second dual we get $l^infty(X)$, the space of all bounded functions from $X$ to $mathbb{C}$, whose spectrum is naturally identified with $beta X$.
By the way, this generalizes to locally compact Hausdorff spaces. If $X$ is such a space, then the spectrum of $C_0(X)$ (the continuous functions on $X$ which vanish at infinity), equipped with its natural topology, is homeomorphic to $X$, and the spectrum of $C_0(X)^{**}$ is homeomorphic to $beta X$. This is bread-and-butter C*-algebra stuff.
answered 7 hours ago
Nik WeaverNik Weaver
21.4k148131
$endgroup$
This is a quite standard idea in functional analysis. Let $X$ be any set and let $c_0(X)$ be the space of all functions from $X$ to $mathbb{C}$ which go to zero at infinity. Then the algebra homomorphisms from $c_0(X)$ to $mathbb{C}$ are precisely the point evaluations at elements of $X$, i.e., the spectrum of $c_0(X)$ is naturally identified with $X$.
Going to the second dual we get $l^infty(X)$, the space of all bounded functions from $X$ to $mathbb{C}$, whose spectrum is naturally identified with $beta X$.
By the way, this generalizes to locally compact Hausdorff spaces. If $X$ is such a space, then the spectrum of $C_0(X)$ (the continuous functions on $X$ which vanish at infinity), equipped with its natural topology, is homeomorphic to $X$, and the spectrum of $C_0(X)^{**}$ is homeomorphic to $beta X$. This is bread-and-butter C*-algebra stuff.
answered 7 hours ago
Nik WeaverNik Weaver
21.4k148131
edited 1 hour ago
answered 7 hours ago
Nik WeaverNik Weaver
21.4k148131
answered 7 hours ago
Nik WeaverNik Weaver
21.4k148131
answered 7 hours ago
Nik WeaverNik Weaver
21.4k148131
21.4k148131
2
$begingroup$
What does it mean for a function from a set $X$ to $mathbb{C}$ to "go to zero at infinity"?
$endgroup$
– Alex Kruckman
4 hours ago
4
$begingroup$
For any $epsilon > 0$, there is a finite subset of $X$ off of which $|f(x)| leq epsilon$.
$endgroup$
– Nik Weaver
4 hours ago
3
$begingroup$
Equivalently, the extension of $f$ to the one point compactification of $X$ (with the discrete topology) which sets $f(infty) = 0$ is continuous. Hence "goes to zero at infinity".
$endgroup$
– Nik Weaver
4 hours ago
add a comment |
2
$begingroup$
What does it mean for a function from a set $X$ to $mathbb{C}$ to "go to zero at infinity"?
$endgroup$
– Alex Kruckman
4 hours ago
4
$begingroup$
For any $epsilon > 0$, there is a finite subset of $X$ off of which $|f(x)| leq epsilon$.
$endgroup$
– Nik Weaver
4 hours ago
3
$begingroup$
Equivalently, the extension of $f$ to the one point compactification of $X$ (with the discrete topology) which sets $f(infty) = 0$ is continuous. Hence "goes to zero at infinity".
$endgroup$
– Nik Weaver
4 hours ago
2
2
$begingroup$
What does it mean for a function from a set $X$ to $mathbb{C}$ to "go to zero at infinity"?
$endgroup$
– Alex Kruckman
4 hours ago
$begingroup$
What does it mean for a function from a set $X$ to $mathbb{C}$ to "go to zero at infinity"?
$endgroup$
– Alex Kruckman
4 hours ago
4
4
$begingroup$
For any $epsilon > 0$, there is a finite subset of $X$ off of which $|f(x)| leq epsilon$.
$endgroup$
– Nik Weaver
4 hours ago
$begingroup$
For any $epsilon > 0$, there is a finite subset of $X$ off of which $|f(x)| leq epsilon$.
$endgroup$
– Nik Weaver
4 hours ago
3
3
$begingroup$
Equivalently, the extension of $f$ to the one point compactification of $X$ (with the discrete topology) which sets $f(infty) = 0$ is continuous. Hence "goes to zero at infinity".
$endgroup$
– Nik Weaver
4 hours ago
$begingroup$
Equivalently, the extension of $f$ to the one point compactification of $X$ (with the discrete topology) which sets $f(infty) = 0$ is continuous. Hence "goes to zero at infinity".
$endgroup$
– Nik Weaver
4 hours ago
add a comment |
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Well, one of the very first things that comes to mind that's sort of in this vein is that $beta X = hom_{text{Bool}}(hom_{text{Set}}(X, 2), 2)$. But if you want to pursue your analogy at a deeper level, try golem.ph.utexas.edu/category/2012/09/…, where both the ultrafilter monad and the double dualization monad are reckoned to be codensity monads induced by the full inclusions of finitary objects.
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– Todd Trimble♦
8 hours ago
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Close to Todd's comment, I'd view $beta X$ as $F(X)=mathrm{hom}_{mathrm{Bool}}(mathrm{hom}_{mathrm{Top}}(X,mathbf{Z}/2mathbf{Z}))$. In general, I guess that for a topological space $X$, the map $Xto F(X)$ is the initial object for the category of continuous maps from $X$ to compact Hausdorff totally disconnected topological spaces. A difference with taking biduals is that $F(F(X))=F(X)$ by Stone duality.
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– YCor
7 hours ago
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Well, one major difference is that without choice it is always the case that $X$ embeds into $beta X$, it's just not provable that the embedding is not surjective; whereas $V^*$ might be trivial, let alone $V^{**}$, even though $V$ isn't.
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– Asaf Karagila
7 hours ago