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?






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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







    share|cite|improve this question


























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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







      share|cite|improve this question















      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






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      edited yesterday









      Aweygan

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      12.9k21441










      asked yesterday









      Ichiko

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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.






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          • 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













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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.






          share|cite|improve this answer























          • 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

















          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.






          share|cite|improve this answer























          • 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















          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.






          share|cite|improve this answer














          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.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited yesterday

























          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




















          • 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




















           

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