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Gelfand-Naimark theorem and compactifications
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Let $X$ be a Hausdorff space and let $C(X)$ denote the set of all continuous and bounded functions from $X$ to $mathbbC$. It is a well-known fact that $C(X)$ forms a unital and abelian $C^*$-algebra (where the addition, multiplication, and scalar multiplication are defined pointwise, $C(X)$ is endowed with the sup norm, and the $*$ operation is conjugation)
Gelfand-Naimark theorem says the following: any unital and abelian $C^*$-algebra is isomorphic to $C(mathbfX),$ where $mathbfX$ is some compact and Hausdorff space, and $C(X)$ denotes the set of all continuous and bounded functions from $mathbfX$ to $mathbbC.$
Consequently, for any Hausdorff space $X,$ there exists a compact and Hausdorff space $mathbfX$ such that $C(X)$ is isomorphic to $C(mathbfX).$
Question: Is $mathbfX$ a compactification of $X$?
By a compactification I mean the following: $mathbfX$ is a compactification of $X$ if and only if $X$ embeds as an open and dense subspace in $mathbfX$ (I know that the requirement that $X$ is open in $mathbfX$ is usually not a part of the definition of a compactification, but I would like to include it)
Bonus question 1: What conditions do $C(X)$ or $X$ need to satisfy to make sure that $mathbfX$ is a compactification of $X?$
Bonus question 2: If $C_b(X)$ denote the set of all bounded (and not necessarily continuous) functions from $X$ to $mathbbC,$ then $C_b(X)$ is a unital and abelian $C^*$ algebra. Is $mathbfX$ from the Gelfand-Naimark theorem a compactification of $X?$ Why or why not?
As always, any help will be appreciated and rewarded.
Thank you.
general-topology functional-analysis compactification gelfand-representation
asked Aug 6 at 22:44
Pawel
2,924921
add a comment |Â
up vote
0
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Let $X$ be a Hausdorff space and let $C(X)$ denote the set of all continuous and bounded functions from $X$ to $mathbbC$. It is a well-known fact that $C(X)$ forms a unital and abelian $C^*$-algebra (where the addition, multiplication, and scalar multiplication are defined pointwise, $C(X)$ is endowed with the sup norm, and the $*$ operation is conjugation)
Gelfand-Naimark theorem says the following: any unital and abelian $C^*$-algebra is isomorphic to $C(mathbfX),$ where $mathbfX$ is some compact and Hausdorff space, and $C(X)$ denotes the set of all continuous and bounded functions from $mathbfX$ to $mathbbC.$
Consequently, for any Hausdorff space $X,$ there exists a compact and Hausdorff space $mathbfX$ such that $C(X)$ is isomorphic to $C(mathbfX).$
Question: Is $mathbfX$ a compactification of $X$?
By a compactification I mean the following: $mathbfX$ is a compactification of $X$ if and only if $X$ embeds as an open and dense subspace in $mathbfX$ (I know that the requirement that $X$ is open in $mathbfX$ is usually not a part of the definition of a compactification, but I would like to include it)
Bonus question 1: What conditions do $C(X)$ or $X$ need to satisfy to make sure that $mathbfX$ is a compactification of $X?$
Bonus question 2: If $C_b(X)$ denote the set of all bounded (and not necessarily continuous) functions from $X$ to $mathbbC,$ then $C_b(X)$ is a unital and abelian $C^*$ algebra. Is $mathbfX$ from the Gelfand-Naimark theorem a compactification of $X?$ Why or why not?
As always, any help will be appreciated and rewarded.
Thank you.
general-topology functional-analysis compactification gelfand-representation
asked Aug 6 at 22:44
Pawel
2,924921
2
$X$ has to be locally compact to have a compactification in your sense.
– Henno Brandsma
Aug 6 at 22:47
@HennoBrandsma And why is that?
– Pawel
Aug 6 at 22:53
2
because an open subset of a compact Hausdorff space is locally compact and completely regular.
– Henno Brandsma
Aug 6 at 22:55
I understand. Is this a sufficient condition or just necessary?
– Pawel
Aug 6 at 23:07
1
Both necessary and sufficient.
– Henno Brandsma
Aug 6 at 23:12
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $X$ be a Hausdorff space and let $C(X)$ denote the set of all continuous and bounded functions from $X$ to $mathbbC$. It is a well-known fact that $C(X)$ forms a unital and abelian $C^*$-algebra (where the addition, multiplication, and scalar multiplication are defined pointwise, $C(X)$ is endowed with the sup norm, and the $*$ operation is conjugation)
Gelfand-Naimark theorem says the following: any unital and abelian $C^*$-algebra is isomorphic to $C(mathbfX),$ where $mathbfX$ is some compact and Hausdorff space, and $C(X)$ denotes the set of all continuous and bounded functions from $mathbfX$ to $mathbbC.$
Consequently, for any Hausdorff space $X,$ there exists a compact and Hausdorff space $mathbfX$ such that $C(X)$ is isomorphic to $C(mathbfX).$
Question: Is $mathbfX$ a compactification of $X$?
By a compactification I mean the following: $mathbfX$ is a compactification of $X$ if and only if $X$ embeds as an open and dense subspace in $mathbfX$ (I know that the requirement that $X$ is open in $mathbfX$ is usually not a part of the definition of a compactification, but I would like to include it)
Bonus question 1: What conditions do $C(X)$ or $X$ need to satisfy to make sure that $mathbfX$ is a compactification of $X?$
Bonus question 2: If $C_b(X)$ denote the set of all bounded (and not necessarily continuous) functions from $X$ to $mathbbC,$ then $C_b(X)$ is a unital and abelian $C^*$ algebra. Is $mathbfX$ from the Gelfand-Naimark theorem a compactification of $X?$ Why or why not?
As always, any help will be appreciated and rewarded.
Thank you.
general-topology functional-analysis compactification gelfand-representation
asked Aug 6 at 22:44
Pawel
2,924921
Let $X$ be a Hausdorff space and let $C(X)$ denote the set of all continuous and bounded functions from $X$ to $mathbbC$. It is a well-known fact that $C(X)$ forms a unital and abelian $C^*$-algebra (where the addition, multiplication, and scalar multiplication are defined pointwise, $C(X)$ is endowed with the sup norm, and the $*$ operation is conjugation)
Gelfand-Naimark theorem says the following: any unital and abelian $C^*$-algebra is isomorphic to $C(mathbfX),$ where $mathbfX$ is some compact and Hausdorff space, and $C(X)$ denotes the set of all continuous and bounded functions from $mathbfX$ to $mathbbC.$
Consequently, for any Hausdorff space $X,$ there exists a compact and Hausdorff space $mathbfX$ such that $C(X)$ is isomorphic to $C(mathbfX).$
Question: Is $mathbfX$ a compactification of $X$?
By a compactification I mean the following: $mathbfX$ is a compactification of $X$ if and only if $X$ embeds as an open and dense subspace in $mathbfX$ (I know that the requirement that $X$ is open in $mathbfX$ is usually not a part of the definition of a compactification, but I would like to include it)
Bonus question 1: What conditions do $C(X)$ or $X$ need to satisfy to make sure that $mathbfX$ is a compactification of $X?$
Bonus question 2: If $C_b(X)$ denote the set of all bounded (and not necessarily continuous) functions from $X$ to $mathbbC,$ then $C_b(X)$ is a unital and abelian $C^*$ algebra. Is $mathbfX$ from the Gelfand-Naimark theorem a compactification of $X?$ Why or why not?
As always, any help will be appreciated and rewarded.
Thank you.
general-topology functional-analysis compactification gelfand-representation
asked Aug 6 at 22:44
Pawel
2,924921
asked Aug 6 at 22:44
Pawel
2,924921
asked Aug 6 at 22:44
Pawel
2,924921
asked Aug 6 at 22:44
Pawel
2,924921
2,924921
2
$X$ has to be locally compact to have a compactification in your sense.
– Henno Brandsma
Aug 6 at 22:47
@HennoBrandsma And why is that?
– Pawel
Aug 6 at 22:53
2
because an open subset of a compact Hausdorff space is locally compact and completely regular.
– Henno Brandsma
Aug 6 at 22:55
I understand. Is this a sufficient condition or just necessary?
– Pawel
Aug 6 at 23:07
1
Both necessary and sufficient.
– Henno Brandsma
Aug 6 at 23:12
add a comment |Â
2
$X$ has to be locally compact to have a compactification in your sense.
– Henno Brandsma
Aug 6 at 22:47
@HennoBrandsma And why is that?
– Pawel
Aug 6 at 22:53
2
because an open subset of a compact Hausdorff space is locally compact and completely regular.
– Henno Brandsma
Aug 6 at 22:55
I understand. Is this a sufficient condition or just necessary?
– Pawel
Aug 6 at 23:07
1
Both necessary and sufficient.
– Henno Brandsma
Aug 6 at 23:12
2
2
$X$ has to be locally compact to have a compactification in your sense.
– Henno Brandsma
Aug 6 at 22:47
$X$ has to be locally compact to have a compactification in your sense.
– Henno Brandsma
Aug 6 at 22:47
@HennoBrandsma And why is that?
– Pawel
Aug 6 at 22:53
@HennoBrandsma And why is that?
– Pawel
Aug 6 at 22:53
2
2
because an open subset of a compact Hausdorff space is locally compact and completely regular.
– Henno Brandsma
Aug 6 at 22:55
because an open subset of a compact Hausdorff space is locally compact and completely regular.
– Henno Brandsma
Aug 6 at 22:55
I understand. Is this a sufficient condition or just necessary?
– Pawel
Aug 6 at 23:07
I understand. Is this a sufficient condition or just necessary?
– Pawel
Aug 6 at 23:07
1
1
Both necessary and sufficient.
– Henno Brandsma
Aug 6 at 23:12
Both necessary and sufficient.
– Henno Brandsma
Aug 6 at 23:12
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
In general (for any topological space $X$ at all), $mathbfX$ is the Stone-Cech compactification of $X$. That is, there is a canonical continuous map $i:XtomathbfX$ such that for every compact Hausdorff space $K$ and every continuous map $f:Xto K$, there is a unique continuous map $g:mathbfXto K$ such that $gi=f$. (Explicitly, $mathbfX$ is the space of $*$-homomorphisms $C(X)tomathbbC$, and $i:XtomathbfX$ takes any $xin X$ to the homomorphism given by evaluation at $x$.)
Now, the Stone-Cech compactification is not always a compactification in your sense. Specifically, it is a compactification (embedding with dense open image) iff $X$ is locally compact and Hausdorff. This condition is obviously necessary, since any open subspace of a compact Hausdorff space is locally compact Hausdorff. Conversely, the image of $i$ is always dense, and $i$ is an embedding iff $X$ is completely regular. In particular, if $X$ is locally compact Hausdorff then $i$ is a embedding with dense image and its image is therefore open since any dense locally compact subspace of a compact Hausdorff space is open.
As for what you call $C_b(X)$, it is just $C(X_d)$ where $X_d$ is $X$ with the discrete topology. So, the $mathbfX$ you obtain from this is just the Stone-Cech compactification of $X_d$. This will never be a compactification of $X$ in a canonical way unless $X$ is discrete so that $X=X_d$ (in particular, the canonical map $i:XtomathbfX$ is not continuous with respect to any topology on $X$ besides the discrete topology, since it is an embedding with respect to the discrete topology).
(Incidentally, what you call $C(X)$ is what is more commonly called $C_b(X)$. Typically $C(X)$ refers to the algebra of all (not necessarily bounded) continuous functions on $X$.)
answered Aug 6 at 23:11
Eric Wofsey
163k12190301
1
If $X$ is a locally compact Hausdorff space and $A$ is a unital $C^*$-subalgebra of $C_b(X)$ (what you called $C(X)$), then $A$ corresponds to a compactification of $X$ iff $A$ contains all functions with compact support (and every compactification can be obtained in this way). In particular, this is true of your $C'(X)$, so it gives a compactification of $X$ (as long as $X$ is locally compact).
– Eric Wofsey
Aug 6 at 23:51
1
Every function with compact support extends continuously to every compactification, by just making it $0$ at all the new points. So it is necessary for $A$ to contain all functions with compact support. Conversely, functions with compact support generate the topology on $X$ (that is, the topology on $X$ is the coarsest topology making them all continuous). So, if they all extend continuously to $mathbfX$, then the image of $X$ in $mathbfX$ must have the same topology, so it's an embedding.
– Eric Wofsey
Aug 7 at 0:03
1
Let $beta X$ denote the Stone-Cech compactification of $X$ and let $i:Xtobeta X$ be the canonical inclusion map. The inclusion homorphism $Ato C_b(X)$ can be thought of as a homomorphism $C(mathbfX)to C(beta X)$, which is given by composition with some continuous map $f:beta XtomathbfX$ (which is surjective because the inclusion homomorphism is injective). So, composing $f$ and $i$, we have a map $fi:XtomathbfX$, and the inclusion homomorphism $C(mathbfX)cong Ato C_b(X)$ is given by composition with $fi$.
– Eric Wofsey
Aug 7 at 22:05
1
This means that the elements of $A$ are exactly the continuous functions on $X$ which can be obtained by composing a continuous function on $mathbfX$ with $fi$. In other words, if you think of $fi$ as an inclusion of $X$ into $mathbfX$, they are exactly the continuous functions on $X$ which extend continuously to $mathbfX$.
– Eric Wofsey
Aug 7 at 22:07
1
That follows from the fact that $X$ is locally compact Hausdorff, and involves Urysohn's lemma.
– Eric Wofsey
Aug 8 at 19:00
 |Â
show 6 more comments
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
In general (for any topological space $X$ at all), $mathbfX$ is the Stone-Cech compactification of $X$. That is, there is a canonical continuous map $i:XtomathbfX$ such that for every compact Hausdorff space $K$ and every continuous map $f:Xto K$, there is a unique continuous map $g:mathbfXto K$ such that $gi=f$. (Explicitly, $mathbfX$ is the space of $*$-homomorphisms $C(X)tomathbbC$, and $i:XtomathbfX$ takes any $xin X$ to the homomorphism given by evaluation at $x$.)
Now, the Stone-Cech compactification is not always a compactification in your sense. Specifically, it is a compactification (embedding with dense open image) iff $X$ is locally compact and Hausdorff. This condition is obviously necessary, since any open subspace of a compact Hausdorff space is locally compact Hausdorff. Conversely, the image of $i$ is always dense, and $i$ is an embedding iff $X$ is completely regular. In particular, if $X$ is locally compact Hausdorff then $i$ is a embedding with dense image and its image is therefore open since any dense locally compact subspace of a compact Hausdorff space is open.
As for what you call $C_b(X)$, it is just $C(X_d)$ where $X_d$ is $X$ with the discrete topology. So, the $mathbfX$ you obtain from this is just the Stone-Cech compactification of $X_d$. This will never be a compactification of $X$ in a canonical way unless $X$ is discrete so that $X=X_d$ (in particular, the canonical map $i:XtomathbfX$ is not continuous with respect to any topology on $X$ besides the discrete topology, since it is an embedding with respect to the discrete topology).
(Incidentally, what you call $C(X)$ is what is more commonly called $C_b(X)$. Typically $C(X)$ refers to the algebra of all (not necessarily bounded) continuous functions on $X$.)
answered Aug 6 at 23:11
Eric Wofsey
163k12190301
1
If $X$ is a locally compact Hausdorff space and $A$ is a unital $C^*$-subalgebra of $C_b(X)$ (what you called $C(X)$), then $A$ corresponds to a compactification of $X$ iff $A$ contains all functions with compact support (and every compactification can be obtained in this way). In particular, this is true of your $C'(X)$, so it gives a compactification of $X$ (as long as $X$ is locally compact).
– Eric Wofsey
Aug 6 at 23:51
1
Every function with compact support extends continuously to every compactification, by just making it $0$ at all the new points. So it is necessary for $A$ to contain all functions with compact support. Conversely, functions with compact support generate the topology on $X$ (that is, the topology on $X$ is the coarsest topology making them all continuous). So, if they all extend continuously to $mathbfX$, then the image of $X$ in $mathbfX$ must have the same topology, so it's an embedding.
– Eric Wofsey
Aug 7 at 0:03
1
Let $beta X$ denote the Stone-Cech compactification of $X$ and let $i:Xtobeta X$ be the canonical inclusion map. The inclusion homorphism $Ato C_b(X)$ can be thought of as a homomorphism $C(mathbfX)to C(beta X)$, which is given by composition with some continuous map $f:beta XtomathbfX$ (which is surjective because the inclusion homomorphism is injective). So, composing $f$ and $i$, we have a map $fi:XtomathbfX$, and the inclusion homomorphism $C(mathbfX)cong Ato C_b(X)$ is given by composition with $fi$.
– Eric Wofsey
Aug 7 at 22:05
1
This means that the elements of $A$ are exactly the continuous functions on $X$ which can be obtained by composing a continuous function on $mathbfX$ with $fi$. In other words, if you think of $fi$ as an inclusion of $X$ into $mathbfX$, they are exactly the continuous functions on $X$ which extend continuously to $mathbfX$.
– Eric Wofsey
Aug 7 at 22:07
1
That follows from the fact that $X$ is locally compact Hausdorff, and involves Urysohn's lemma.
– Eric Wofsey
Aug 8 at 19:00
 |Â
show 6 more comments
up vote
1
down vote
accepted
In general (for any topological space $X$ at all), $mathbfX$ is the Stone-Cech compactification of $X$. That is, there is a canonical continuous map $i:XtomathbfX$ such that for every compact Hausdorff space $K$ and every continuous map $f:Xto K$, there is a unique continuous map $g:mathbfXto K$ such that $gi=f$. (Explicitly, $mathbfX$ is the space of $*$-homomorphisms $C(X)tomathbbC$, and $i:XtomathbfX$ takes any $xin X$ to the homomorphism given by evaluation at $x$.)
Now, the Stone-Cech compactification is not always a compactification in your sense. Specifically, it is a compactification (embedding with dense open image) iff $X$ is locally compact and Hausdorff. This condition is obviously necessary, since any open subspace of a compact Hausdorff space is locally compact Hausdorff. Conversely, the image of $i$ is always dense, and $i$ is an embedding iff $X$ is completely regular. In particular, if $X$ is locally compact Hausdorff then $i$ is a embedding with dense image and its image is therefore open since any dense locally compact subspace of a compact Hausdorff space is open.
As for what you call $C_b(X)$, it is just $C(X_d)$ where $X_d$ is $X$ with the discrete topology. So, the $mathbfX$ you obtain from this is just the Stone-Cech compactification of $X_d$. This will never be a compactification of $X$ in a canonical way unless $X$ is discrete so that $X=X_d$ (in particular, the canonical map $i:XtomathbfX$ is not continuous with respect to any topology on $X$ besides the discrete topology, since it is an embedding with respect to the discrete topology).
(Incidentally, what you call $C(X)$ is what is more commonly called $C_b(X)$. Typically $C(X)$ refers to the algebra of all (not necessarily bounded) continuous functions on $X$.)
answered Aug 6 at 23:11
Eric Wofsey
163k12190301
1
If $X$ is a locally compact Hausdorff space and $A$ is a unital $C^*$-subalgebra of $C_b(X)$ (what you called $C(X)$), then $A$ corresponds to a compactification of $X$ iff $A$ contains all functions with compact support (and every compactification can be obtained in this way). In particular, this is true of your $C'(X)$, so it gives a compactification of $X$ (as long as $X$ is locally compact).
– Eric Wofsey
Aug 6 at 23:51
1
Every function with compact support extends continuously to every compactification, by just making it $0$ at all the new points. So it is necessary for $A$ to contain all functions with compact support. Conversely, functions with compact support generate the topology on $X$ (that is, the topology on $X$ is the coarsest topology making them all continuous). So, if they all extend continuously to $mathbfX$, then the image of $X$ in $mathbfX$ must have the same topology, so it's an embedding.
– Eric Wofsey
Aug 7 at 0:03
1
Let $beta X$ denote the Stone-Cech compactification of $X$ and let $i:Xtobeta X$ be the canonical inclusion map. The inclusion homorphism $Ato C_b(X)$ can be thought of as a homomorphism $C(mathbfX)to C(beta X)$, which is given by composition with some continuous map $f:beta XtomathbfX$ (which is surjective because the inclusion homomorphism is injective). So, composing $f$ and $i$, we have a map $fi:XtomathbfX$, and the inclusion homomorphism $C(mathbfX)cong Ato C_b(X)$ is given by composition with $fi$.
– Eric Wofsey
Aug 7 at 22:05
1
This means that the elements of $A$ are exactly the continuous functions on $X$ which can be obtained by composing a continuous function on $mathbfX$ with $fi$. In other words, if you think of $fi$ as an inclusion of $X$ into $mathbfX$, they are exactly the continuous functions on $X$ which extend continuously to $mathbfX$.
– Eric Wofsey
Aug 7 at 22:07
1
That follows from the fact that $X$ is locally compact Hausdorff, and involves Urysohn's lemma.
– Eric Wofsey
Aug 8 at 19:00
 |Â
show 6 more comments
up vote
1
down vote
accepted
up vote
1
down vote
accepted
In general (for any topological space $X$ at all), $mathbfX$ is the Stone-Cech compactification of $X$. That is, there is a canonical continuous map $i:XtomathbfX$ such that for every compact Hausdorff space $K$ and every continuous map $f:Xto K$, there is a unique continuous map $g:mathbfXto K$ such that $gi=f$. (Explicitly, $mathbfX$ is the space of $*$-homomorphisms $C(X)tomathbbC$, and $i:XtomathbfX$ takes any $xin X$ to the homomorphism given by evaluation at $x$.)
Now, the Stone-Cech compactification is not always a compactification in your sense. Specifically, it is a compactification (embedding with dense open image) iff $X$ is locally compact and Hausdorff. This condition is obviously necessary, since any open subspace of a compact Hausdorff space is locally compact Hausdorff. Conversely, the image of $i$ is always dense, and $i$ is an embedding iff $X$ is completely regular. In particular, if $X$ is locally compact Hausdorff then $i$ is a embedding with dense image and its image is therefore open since any dense locally compact subspace of a compact Hausdorff space is open.
As for what you call $C_b(X)$, it is just $C(X_d)$ where $X_d$ is $X$ with the discrete topology. So, the $mathbfX$ you obtain from this is just the Stone-Cech compactification of $X_d$. This will never be a compactification of $X$ in a canonical way unless $X$ is discrete so that $X=X_d$ (in particular, the canonical map $i:XtomathbfX$ is not continuous with respect to any topology on $X$ besides the discrete topology, since it is an embedding with respect to the discrete topology).
(Incidentally, what you call $C(X)$ is what is more commonly called $C_b(X)$. Typically $C(X)$ refers to the algebra of all (not necessarily bounded) continuous functions on $X$.)
answered Aug 6 at 23:11
Eric Wofsey
163k12190301
In general (for any topological space $X$ at all), $mathbfX$ is the Stone-Cech compactification of $X$. That is, there is a canonical continuous map $i:XtomathbfX$ such that for every compact Hausdorff space $K$ and every continuous map $f:Xto K$, there is a unique continuous map $g:mathbfXto K$ such that $gi=f$. (Explicitly, $mathbfX$ is the space of $*$-homomorphisms $C(X)tomathbbC$, and $i:XtomathbfX$ takes any $xin X$ to the homomorphism given by evaluation at $x$.)
Now, the Stone-Cech compactification is not always a compactification in your sense. Specifically, it is a compactification (embedding with dense open image) iff $X$ is locally compact and Hausdorff. This condition is obviously necessary, since any open subspace of a compact Hausdorff space is locally compact Hausdorff. Conversely, the image of $i$ is always dense, and $i$ is an embedding iff $X$ is completely regular. In particular, if $X$ is locally compact Hausdorff then $i$ is a embedding with dense image and its image is therefore open since any dense locally compact subspace of a compact Hausdorff space is open.
As for what you call $C_b(X)$, it is just $C(X_d)$ where $X_d$ is $X$ with the discrete topology. So, the $mathbfX$ you obtain from this is just the Stone-Cech compactification of $X_d$. This will never be a compactification of $X$ in a canonical way unless $X$ is discrete so that $X=X_d$ (in particular, the canonical map $i:XtomathbfX$ is not continuous with respect to any topology on $X$ besides the discrete topology, since it is an embedding with respect to the discrete topology).
(Incidentally, what you call $C(X)$ is what is more commonly called $C_b(X)$. Typically $C(X)$ refers to the algebra of all (not necessarily bounded) continuous functions on $X$.)
answered Aug 6 at 23:11
Eric Wofsey
163k12190301
answered Aug 6 at 23:11
Eric Wofsey
163k12190301
answered Aug 6 at 23:11
Eric Wofsey
163k12190301
answered Aug 6 at 23:11
Eric Wofsey
163k12190301
163k12190301
1
If $X$ is a locally compact Hausdorff space and $A$ is a unital $C^*$-subalgebra of $C_b(X)$ (what you called $C(X)$), then $A$ corresponds to a compactification of $X$ iff $A$ contains all functions with compact support (and every compactification can be obtained in this way). In particular, this is true of your $C'(X)$, so it gives a compactification of $X$ (as long as $X$ is locally compact).
– Eric Wofsey
Aug 6 at 23:51
1
Every function with compact support extends continuously to every compactification, by just making it $0$ at all the new points. So it is necessary for $A$ to contain all functions with compact support. Conversely, functions with compact support generate the topology on $X$ (that is, the topology on $X$ is the coarsest topology making them all continuous). So, if they all extend continuously to $mathbfX$, then the image of $X$ in $mathbfX$ must have the same topology, so it's an embedding.
– Eric Wofsey
Aug 7 at 0:03
1
Let $beta X$ denote the Stone-Cech compactification of $X$ and let $i:Xtobeta X$ be the canonical inclusion map. The inclusion homorphism $Ato C_b(X)$ can be thought of as a homomorphism $C(mathbfX)to C(beta X)$, which is given by composition with some continuous map $f:beta XtomathbfX$ (which is surjective because the inclusion homomorphism is injective). So, composing $f$ and $i$, we have a map $fi:XtomathbfX$, and the inclusion homomorphism $C(mathbfX)cong Ato C_b(X)$ is given by composition with $fi$.
– Eric Wofsey
Aug 7 at 22:05
1
This means that the elements of $A$ are exactly the continuous functions on $X$ which can be obtained by composing a continuous function on $mathbfX$ with $fi$. In other words, if you think of $fi$ as an inclusion of $X$ into $mathbfX$, they are exactly the continuous functions on $X$ which extend continuously to $mathbfX$.
– Eric Wofsey
Aug 7 at 22:07
1
That follows from the fact that $X$ is locally compact Hausdorff, and involves Urysohn's lemma.
– Eric Wofsey
Aug 8 at 19:00
 |Â
show 6 more comments
1
If $X$ is a locally compact Hausdorff space and $A$ is a unital $C^*$-subalgebra of $C_b(X)$ (what you called $C(X)$), then $A$ corresponds to a compactification of $X$ iff $A$ contains all functions with compact support (and every compactification can be obtained in this way). In particular, this is true of your $C'(X)$, so it gives a compactification of $X$ (as long as $X$ is locally compact).
– Eric Wofsey
Aug 6 at 23:51
1
Every function with compact support extends continuously to every compactification, by just making it $0$ at all the new points. So it is necessary for $A$ to contain all functions with compact support. Conversely, functions with compact support generate the topology on $X$ (that is, the topology on $X$ is the coarsest topology making them all continuous). So, if they all extend continuously to $mathbfX$, then the image of $X$ in $mathbfX$ must have the same topology, so it's an embedding.
– Eric Wofsey
Aug 7 at 0:03
1
Let $beta X$ denote the Stone-Cech compactification of $X$ and let $i:Xtobeta X$ be the canonical inclusion map. The inclusion homorphism $Ato C_b(X)$ can be thought of as a homomorphism $C(mathbfX)to C(beta X)$, which is given by composition with some continuous map $f:beta XtomathbfX$ (which is surjective because the inclusion homomorphism is injective). So, composing $f$ and $i$, we have a map $fi:XtomathbfX$, and the inclusion homomorphism $C(mathbfX)cong Ato C_b(X)$ is given by composition with $fi$.
– Eric Wofsey
Aug 7 at 22:05
1
This means that the elements of $A$ are exactly the continuous functions on $X$ which can be obtained by composing a continuous function on $mathbfX$ with $fi$. In other words, if you think of $fi$ as an inclusion of $X$ into $mathbfX$, they are exactly the continuous functions on $X$ which extend continuously to $mathbfX$.
– Eric Wofsey
Aug 7 at 22:07
1
That follows from the fact that $X$ is locally compact Hausdorff, and involves Urysohn's lemma.
– Eric Wofsey
Aug 8 at 19:00
1
1
If $X$ is a locally compact Hausdorff space and $A$ is a unital $C^*$-subalgebra of $C_b(X)$ (what you called $C(X)$), then $A$ corresponds to a compactification of $X$ iff $A$ contains all functions with compact support (and every compactification can be obtained in this way). In particular, this is true of your $C'(X)$, so it gives a compactification of $X$ (as long as $X$ is locally compact).
– Eric Wofsey
Aug 6 at 23:51
If $X$ is a locally compact Hausdorff space and $A$ is a unital $C^*$-subalgebra of $C_b(X)$ (what you called $C(X)$), then $A$ corresponds to a compactification of $X$ iff $A$ contains all functions with compact support (and every compactification can be obtained in this way). In particular, this is true of your $C'(X)$, so it gives a compactification of $X$ (as long as $X$ is locally compact).
– Eric Wofsey
Aug 6 at 23:51
1
1
Every function with compact support extends continuously to every compactification, by just making it $0$ at all the new points. So it is necessary for $A$ to contain all functions with compact support. Conversely, functions with compact support generate the topology on $X$ (that is, the topology on $X$ is the coarsest topology making them all continuous). So, if they all extend continuously to $mathbfX$, then the image of $X$ in $mathbfX$ must have the same topology, so it's an embedding.
– Eric Wofsey
Aug 7 at 0:03
Every function with compact support extends continuously to every compactification, by just making it $0$ at all the new points. So it is necessary for $A$ to contain all functions with compact support. Conversely, functions with compact support generate the topology on $X$ (that is, the topology on $X$ is the coarsest topology making them all continuous). So, if they all extend continuously to $mathbfX$, then the image of $X$ in $mathbfX$ must have the same topology, so it's an embedding.
– Eric Wofsey
Aug 7 at 0:03
1
1
Let $beta X$ denote the Stone-Cech compactification of $X$ and let $i:Xtobeta X$ be the canonical inclusion map. The inclusion homorphism $Ato C_b(X)$ can be thought of as a homomorphism $C(mathbfX)to C(beta X)$, which is given by composition with some continuous map $f:beta XtomathbfX$ (which is surjective because the inclusion homomorphism is injective). So, composing $f$ and $i$, we have a map $fi:XtomathbfX$, and the inclusion homomorphism $C(mathbfX)cong Ato C_b(X)$ is given by composition with $fi$.
– Eric Wofsey
Aug 7 at 22:05
Let $beta X$ denote the Stone-Cech compactification of $X$ and let $i:Xtobeta X$ be the canonical inclusion map. The inclusion homorphism $Ato C_b(X)$ can be thought of as a homomorphism $C(mathbfX)to C(beta X)$, which is given by composition with some continuous map $f:beta XtomathbfX$ (which is surjective because the inclusion homomorphism is injective). So, composing $f$ and $i$, we have a map $fi:XtomathbfX$, and the inclusion homomorphism $C(mathbfX)cong Ato C_b(X)$ is given by composition with $fi$.
– Eric Wofsey
Aug 7 at 22:05
1
1
This means that the elements of $A$ are exactly the continuous functions on $X$ which can be obtained by composing a continuous function on $mathbfX$ with $fi$. In other words, if you think of $fi$ as an inclusion of $X$ into $mathbfX$, they are exactly the continuous functions on $X$ which extend continuously to $mathbfX$.
– Eric Wofsey
Aug 7 at 22:07
This means that the elements of $A$ are exactly the continuous functions on $X$ which can be obtained by composing a continuous function on $mathbfX$ with $fi$. In other words, if you think of $fi$ as an inclusion of $X$ into $mathbfX$, they are exactly the continuous functions on $X$ which extend continuously to $mathbfX$.
– Eric Wofsey
Aug 7 at 22:07
1
1
That follows from the fact that $X$ is locally compact Hausdorff, and involves Urysohn's lemma.
– Eric Wofsey
Aug 8 at 19:00
That follows from the fact that $X$ is locally compact Hausdorff, and involves Urysohn's lemma.
– Eric Wofsey
Aug 8 at 19:00
 |Â
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2
$X$ has to be locally compact to have a compactification in your sense.
– Henno Brandsma
Aug 6 at 22:47
@HennoBrandsma And why is that?
– Pawel
Aug 6 at 22:53
2
because an open subset of a compact Hausdorff space is locally compact and completely regular.
– Henno Brandsma
Aug 6 at 22:55
I understand. Is this a sufficient condition or just necessary?
– Pawel
Aug 6 at 23:07
1
Both necessary and sufficient.
– Henno Brandsma
Aug 6 at 23:12