Mixing
Gelfand-Naimark Theorem of non-unital case
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Let $X$ be a non-compact, locally compact Hausdorff space. Then $C_0 (X)$, the space of complex valued continuous functions on X vanishing at infinity, is a non-unital $C^*$-algebra and $Omega (C_0 (X))$, the space of linear functionals $T$ on $C_0 (X)$ satisfying $T(fg)=T(f)T(g)$ for all $f,g in C_0 (X)$, is a locally compact Hausdorff space with respect to weak${}^*$ topology in $C_0 (X)^*$.
Let $x in X$. Define $hat{x}(f)=f(x)$ for all $f in C_0 (X)$.
Then $hat{x} in Omega (C_0 (X))$.
I know the map from $X$ to $Omega (C_0 (X))$, $x mapsto hat{x}$, is bijective and continuous. How is the proof that the map is open map?
general-topology operator-algebras
edited yesterday
Aweygan
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asked yesterday
Ichiko
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Let $X$ be a non-compact, locally compact Hausdorff space. Then $C_0 (X)$, the space of complex valued continuous functions on X vanishing at infinity, is a non-unital $C^*$-algebra and $Omega (C_0 (X))$, the space of linear functionals $T$ on $C_0 (X)$ satisfying $T(fg)=T(f)T(g)$ for all $f,g in C_0 (X)$, is a locally compact Hausdorff space with respect to weak${}^*$ topology in $C_0 (X)^*$.
Let $x in X$. Define $hat{x}(f)=f(x)$ for all $f in C_0 (X)$.
Then $hat{x} in Omega (C_0 (X))$.
I know the map from $X$ to $Omega (C_0 (X))$, $x mapsto hat{x}$, is bijective and continuous. How is the proof that the map is open map?
general-topology operator-algebras
edited yesterday
Aweygan
12.9k21441
asked yesterday
Ichiko
403
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Let $X$ be a non-compact, locally compact Hausdorff space. Then $C_0 (X)$, the space of complex valued continuous functions on X vanishing at infinity, is a non-unital $C^*$-algebra and $Omega (C_0 (X))$, the space of linear functionals $T$ on $C_0 (X)$ satisfying $T(fg)=T(f)T(g)$ for all $f,g in C_0 (X)$, is a locally compact Hausdorff space with respect to weak${}^*$ topology in $C_0 (X)^*$.
Let $x in X$. Define $hat{x}(f)=f(x)$ for all $f in C_0 (X)$.
Then $hat{x} in Omega (C_0 (X))$.
I know the map from $X$ to $Omega (C_0 (X))$, $x mapsto hat{x}$, is bijective and continuous. How is the proof that the map is open map?
general-topology operator-algebras
edited yesterday
Aweygan
12.9k21441
asked yesterday
Ichiko
403
Let $X$ be a non-compact, locally compact Hausdorff space. Then $C_0 (X)$, the space of complex valued continuous functions on X vanishing at infinity, is a non-unital $C^*$-algebra and $Omega (C_0 (X))$, the space of linear functionals $T$ on $C_0 (X)$ satisfying $T(fg)=T(f)T(g)$ for all $f,g in C_0 (X)$, is a locally compact Hausdorff space with respect to weak${}^*$ topology in $C_0 (X)^*$.
Let $x in X$. Define $hat{x}(f)=f(x)$ for all $f in C_0 (X)$.
Then $hat{x} in Omega (C_0 (X))$.
I know the map from $X$ to $Omega (C_0 (X))$, $x mapsto hat{x}$, is bijective and continuous. How is the proof that the map is open map?
general-topology operator-algebras
general-topology operator-algebras
edited yesterday
Aweygan
12.9k21441
asked yesterday
Ichiko
403
edited yesterday
Aweygan
12.9k21441
asked yesterday
Ichiko
403
edited yesterday
Aweygan
12.9k21441
edited yesterday
Aweygan
12.9k21441
edited yesterday
Aweygan
12.9k21441
12.9k21441
asked yesterday
Ichiko
403
asked yesterday
Ichiko
403
asked yesterday
Ichiko
403
403
add a comment |
add a comment |
1 Answer
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Let us write $F:Xto Omega(C_0(X))$ for the map $F(x)=hat{x}$. Let $Usubseteq X$ be open and $xin U$. Let $V$ be open such that $xin Vsubseteq overline{V}subseteq U$ and $overline{V}$ is compact. By Urysohn's lemma, there exists a continuous function $f:Xto [0,1]$ such that $f(x)=1$ and $f(y)=0$ for all $yin Xsetminus V$. Since $overline{V}$ is compact, $fin C_0(X)$. We now see that ${hat{y}inOmega(C_0(X)):hat{y}(f)neq 0}$ is an open subset of $Omega(C_0(X))$ which contains $hat{x}$ and is contained in $F(U)$. Since $xin U$ was arbitrary, this means $F(U)$ is open, and so $F$ is an open map.
answered yesterday
Eric Wofsey
175k12202326
Thank you. But does ${ hat{y} in Omega (C_0 (X)) colon |hat{y}(f)|<1 }$ really contain $hat{x}$ and is it really contained in $F(U)$? I think $hat{x}(f)=1$.
– Ichiko
yesterday
Oops, it should be $>0$ instead of $<1$. I'll fix that.
– Eric Wofsey
yesterday
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Let us write $F:Xto Omega(C_0(X))$ for the map $F(x)=hat{x}$. Let $Usubseteq X$ be open and $xin U$. Let $V$ be open such that $xin Vsubseteq overline{V}subseteq U$ and $overline{V}$ is compact. By Urysohn's lemma, there exists a continuous function $f:Xto [0,1]$ such that $f(x)=1$ and $f(y)=0$ for all $yin Xsetminus V$. Since $overline{V}$ is compact, $fin C_0(X)$. We now see that ${hat{y}inOmega(C_0(X)):hat{y}(f)neq 0}$ is an open subset of $Omega(C_0(X))$ which contains $hat{x}$ and is contained in $F(U)$. Since $xin U$ was arbitrary, this means $F(U)$ is open, and so $F$ is an open map.
answered yesterday
Eric Wofsey
175k12202326
Thank you. But does ${ hat{y} in Omega (C_0 (X)) colon |hat{y}(f)|<1 }$ really contain $hat{x}$ and is it really contained in $F(U)$? I think $hat{x}(f)=1$.
– Ichiko
yesterday
Oops, it should be $>0$ instead of $<1$. I'll fix that.
– Eric Wofsey
yesterday
add a comment |
up vote
1
down vote
accepted
Let us write $F:Xto Omega(C_0(X))$ for the map $F(x)=hat{x}$. Let $Usubseteq X$ be open and $xin U$. Let $V$ be open such that $xin Vsubseteq overline{V}subseteq U$ and $overline{V}$ is compact. By Urysohn's lemma, there exists a continuous function $f:Xto [0,1]$ such that $f(x)=1$ and $f(y)=0$ for all $yin Xsetminus V$. Since $overline{V}$ is compact, $fin C_0(X)$. We now see that ${hat{y}inOmega(C_0(X)):hat{y}(f)neq 0}$ is an open subset of $Omega(C_0(X))$ which contains $hat{x}$ and is contained in $F(U)$. Since $xin U$ was arbitrary, this means $F(U)$ is open, and so $F$ is an open map.
answered yesterday
Eric Wofsey
175k12202326
Thank you. But does ${ hat{y} in Omega (C_0 (X)) colon |hat{y}(f)|<1 }$ really contain $hat{x}$ and is it really contained in $F(U)$? I think $hat{x}(f)=1$.
– Ichiko
yesterday
Oops, it should be $>0$ instead of $<1$. I'll fix that.
– Eric Wofsey
yesterday
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Let us write $F:Xto Omega(C_0(X))$ for the map $F(x)=hat{x}$. Let $Usubseteq X$ be open and $xin U$. Let $V$ be open such that $xin Vsubseteq overline{V}subseteq U$ and $overline{V}$ is compact. By Urysohn's lemma, there exists a continuous function $f:Xto [0,1]$ such that $f(x)=1$ and $f(y)=0$ for all $yin Xsetminus V$. Since $overline{V}$ is compact, $fin C_0(X)$. We now see that ${hat{y}inOmega(C_0(X)):hat{y}(f)neq 0}$ is an open subset of $Omega(C_0(X))$ which contains $hat{x}$ and is contained in $F(U)$. Since $xin U$ was arbitrary, this means $F(U)$ is open, and so $F$ is an open map.
answered yesterday
Eric Wofsey
175k12202326
Let us write $F:Xto Omega(C_0(X))$ for the map $F(x)=hat{x}$. Let $Usubseteq X$ be open and $xin U$. Let $V$ be open such that $xin Vsubseteq overline{V}subseteq U$ and $overline{V}$ is compact. By Urysohn's lemma, there exists a continuous function $f:Xto [0,1]$ such that $f(x)=1$ and $f(y)=0$ for all $yin Xsetminus V$. Since $overline{V}$ is compact, $fin C_0(X)$. We now see that ${hat{y}inOmega(C_0(X)):hat{y}(f)neq 0}$ is an open subset of $Omega(C_0(X))$ which contains $hat{x}$ and is contained in $F(U)$. Since $xin U$ was arbitrary, this means $F(U)$ is open, and so $F$ is an open map.
answered yesterday
Eric Wofsey
175k12202326
edited yesterday
answered yesterday
Eric Wofsey
175k12202326
answered yesterday
Eric Wofsey
175k12202326
answered yesterday
Eric Wofsey
175k12202326
175k12202326
Thank you. But does ${ hat{y} in Omega (C_0 (X)) colon |hat{y}(f)|<1 }$ really contain $hat{x}$ and is it really contained in $F(U)$? I think $hat{x}(f)=1$.
– Ichiko
yesterday
Oops, it should be $>0$ instead of $<1$. I'll fix that.
– Eric Wofsey
yesterday
add a comment |
Thank you. But does ${ hat{y} in Omega (C_0 (X)) colon |hat{y}(f)|<1 }$ really contain $hat{x}$ and is it really contained in $F(U)$? I think $hat{x}(f)=1$.
– Ichiko
yesterday
Oops, it should be $>0$ instead of $<1$. I'll fix that.
– Eric Wofsey
yesterday
Thank you. But does ${ hat{y} in Omega (C_0 (X)) colon |hat{y}(f)|<1 }$ really contain $hat{x}$ and is it really contained in $F(U)$? I think $hat{x}(f)=1$.
– Ichiko
yesterday
Thank you. But does ${ hat{y} in Omega (C_0 (X)) colon |hat{y}(f)|<1 }$ really contain $hat{x}$ and is it really contained in $F(U)$? I think $hat{x}(f)=1$.
– Ichiko
yesterday
Oops, it should be $>0$ instead of $<1$. I'll fix that.
– Eric Wofsey
yesterday
Oops, it should be $>0$ instead of $<1$. I'll fix that.
– Eric Wofsey
yesterday
add a comment |
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