Gelfand-Naimark Theorem of non-unital case











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?










share|cite|improve this question




























    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?










    share|cite|improve this question


























      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?










      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






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Nov 19 at 16:54









      Aweygan

      13.3k21441




      13.3k21441










      asked Nov 19 at 15:19









      Ichiko

      423




      423






















          1 Answer
          1






          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.






          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
            Nov 19 at 19:59










          • Oops, it should be $>0$ instead of $<1$. I'll fix that.
            – Eric Wofsey
            Nov 19 at 20:00













          Your Answer





          StackExchange.ifUsing("editor", function () {
          return StackExchange.using("mathjaxEditing", function () {
          StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
          StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
          });
          });
          }, "mathjax-editing");

          StackExchange.ready(function() {
          var channelOptions = {
          tags: "".split(" "),
          id: "69"
          };
          initTagRenderer("".split(" "), "".split(" "), channelOptions);

          StackExchange.using("externalEditor", function() {
          // Have to fire editor after snippets, if snippets enabled
          if (StackExchange.settings.snippets.snippetsEnabled) {
          StackExchange.using("snippets", function() {
          createEditor();
          });
          }
          else {
          createEditor();
          }
          });

          function createEditor() {
          StackExchange.prepareEditor({
          heartbeatType: 'answer',
          convertImagesToLinks: true,
          noModals: true,
          showLowRepImageUploadWarning: true,
          reputationToPostImages: 10,
          bindNavPrevention: true,
          postfix: "",
          imageUploader: {
          brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
          contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
          allowUrls: true
          },
          noCode: true, onDemand: true,
          discardSelector: ".discard-answer"
          ,immediatelyShowMarkdownHelp:true
          });


          }
          });














          draft saved

          draft discarded


















          StackExchange.ready(
          function () {
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3005056%2fgelfand-naimark-theorem-of-non-unital-case%23new-answer', 'question_page');
          }
          );

          Post as a guest















          Required, but never shown

























          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.






          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
            Nov 19 at 19:59










          • Oops, it should be $>0$ instead of $<1$. I'll fix that.
            – Eric Wofsey
            Nov 19 at 20:00

















          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
            Nov 19 at 19:59










          • Oops, it should be $>0$ instead of $<1$. I'll fix that.
            – Eric Wofsey
            Nov 19 at 20:00















          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 Nov 19 at 20:00

























          answered Nov 19 at 17:32









          Eric Wofsey

          177k12202328




          177k12202328












          • 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
            Nov 19 at 19:59










          • Oops, it should be $>0$ instead of $<1$. I'll fix that.
            – Eric Wofsey
            Nov 19 at 20:00




















          • 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
            Nov 19 at 19:59










          • Oops, it should be $>0$ instead of $<1$. I'll fix that.
            – Eric Wofsey
            Nov 19 at 20:00


















          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
          Nov 19 at 19:59




          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
          Nov 19 at 19:59












          Oops, it should be $>0$ instead of $<1$. I'll fix that.
          – Eric Wofsey
          Nov 19 at 20:00






          Oops, it should be $>0$ instead of $<1$. I'll fix that.
          – Eric Wofsey
          Nov 19 at 20:00




















          draft saved

          draft discarded




















































          Thanks for contributing an answer to Mathematics Stack Exchange!


          • Please be sure to answer the question. Provide details and share your research!

          But avoid



          • Asking for help, clarification, or responding to other answers.

          • Making statements based on opinion; back them up with references or personal experience.


          Use MathJax to format equations. MathJax reference.


          To learn more, see our tips on writing great answers.





          Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


          Please pay close attention to the following guidance:


          • Please be sure to answer the question. Provide details and share your research!

          But avoid



          • Asking for help, clarification, or responding to other answers.

          • Making statements based on opinion; back them up with references or personal experience.


          To learn more, see our tips on writing great answers.




          draft saved


          draft discarded














          StackExchange.ready(
          function () {
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3005056%2fgelfand-naimark-theorem-of-non-unital-case%23new-answer', 'question_page');
          }
          );

          Post as a guest















          Required, but never shown





















































          Required, but never shown














          Required, but never shown












          Required, but never shown







          Required, but never shown

































          Required, but never shown














          Required, but never shown












          Required, but never shown







          Required, but never shown







          Popular posts from this blog

          Plaza Victoria

          Puebla de Zaragoza

          Musa