Show that if $f:[a,b]rightarrow mathbb{R}$ is continuous and $[a,b]subseteq f([a,b])$ then there exists a...












1












$begingroup$


I have tried to prove the statement but there is a part where I am not sure if I can argue like that, I will mark it by setting it in bold



We know that $[a,b]subseteq f([a,b])$ and that $f$ is continuous



$$Longrightarrow exists_{x,yin [a,b]}f(x)=awedge f(y)=b$$



Without loss of generality assume $x<y$



$$Longrightarrow[x,y]subseteq[a,b]$$



The function $f_{|[x,y]}$ is continuous and for all $gammain [f(x),f(y)]$ there is a $zin[x,y]$ such that $f(z)= gamma$




Here is the part where I am not quite sure. Also if you have an idea how to prove it differently please share it with me:



Assume there exists finitely many points $zin [x,y]$ such that $f(z)neq z$.



Then we are already done because that means there must exist infinitely many Points for which the opposite is true, since the interval consists of infinitely many Points.



Therefore assume there exists infinitely many Points such that $zin [x,y] wedge f(z)neq z$



Then there must also exist infinitely many points for which $f(z)>z$ and $f(z')<z'$



Without loss of generality assume there are only finitely many Points for which $z<f(z)$. Then there exists a maximal element of this set and one can split the interval $[x,y]$ into $[x,z]$ and $(z,y]$. If there exists a $win (z,y]$ such that $f(w) = w$ we are already done. Therefore assume $forall_{win (z,y]} f(w)>w$. This would invoke a contradiction that $f$ is continuous in $z$.



Therefore we have showed there exists infinitely many $z,z'in [x,y]$ such that $f(z)>z$ and $f(z')<z'$



One can pick now $z,z'$ such that $f(z)>z$ and $f(z')<z'$



Without loss of generality $z<z'$



Then $[z,z']$ is a closed interval and due to the Intermediate-value-theorem



$$forall_{gammain[f(z),f(z')]}exists_{xin[z,z']}f(x)=gamma$$



If we repeat this process by Always choosing an element $l$ for the left end of the interval for which we have $f(l)>l$ and for the Right end an element for which we have $f(r)<r$. Then we get a nested interval which must converge to a fixed Point.




Is my thought process consistent? One Thing that this idea lacks is probably that there must always exist Points $l$ and $r$ like described above otherwise the proof does not work. Is it possible to prove this somehow?






real-analysis proof-verification







share|cite|improve this question











$endgroup$

















    1












    $begingroup$


    I have tried to prove the statement but there is a part where I am not sure if I can argue like that, I will mark it by setting it in bold



    We know that $[a,b]subseteq f([a,b])$ and that $f$ is continuous



    $$Longrightarrow exists_{x,yin [a,b]}f(x)=awedge f(y)=b$$



    Without loss of generality assume $x<y$



    $$Longrightarrow[x,y]subseteq[a,b]$$



    The function $f_{|[x,y]}$ is continuous and for all $gammain [f(x),f(y)]$ there is a $zin[x,y]$ such that $f(z)= gamma$




    Here is the part where I am not quite sure. Also if you have an idea how to prove it differently please share it with me:



    Assume there exists finitely many points $zin [x,y]$ such that $f(z)neq z$.



    Then we are already done because that means there must exist infinitely many Points for which the opposite is true, since the interval consists of infinitely many Points.



    Therefore assume there exists infinitely many Points such that $zin [x,y] wedge f(z)neq z$



    Then there must also exist infinitely many points for which $f(z)>z$ and $f(z')<z'$



    Without loss of generality assume there are only finitely many Points for which $z<f(z)$. Then there exists a maximal element of this set and one can split the interval $[x,y]$ into $[x,z]$ and $(z,y]$. If there exists a $win (z,y]$ such that $f(w) = w$ we are already done. Therefore assume $forall_{win (z,y]} f(w)>w$. This would invoke a contradiction that $f$ is continuous in $z$.



    Therefore we have showed there exists infinitely many $z,z'in [x,y]$ such that $f(z)>z$ and $f(z')<z'$



    One can pick now $z,z'$ such that $f(z)>z$ and $f(z')<z'$



    Without loss of generality $z<z'$



    Then $[z,z']$ is a closed interval and due to the Intermediate-value-theorem



    $$forall_{gammain[f(z),f(z')]}exists_{xin[z,z']}f(x)=gamma$$



    If we repeat this process by Always choosing an element $l$ for the left end of the interval for which we have $f(l)>l$ and for the Right end an element for which we have $f(r)<r$. Then we get a nested interval which must converge to a fixed Point.




    Is my thought process consistent? One Thing that this idea lacks is probably that there must always exist Points $l$ and $r$ like described above otherwise the proof does not work. Is it possible to prove this somehow?






    real-analysis proof-verification







    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      I have tried to prove the statement but there is a part where I am not sure if I can argue like that, I will mark it by setting it in bold



      We know that $[a,b]subseteq f([a,b])$ and that $f$ is continuous



      $$Longrightarrow exists_{x,yin [a,b]}f(x)=awedge f(y)=b$$



      Without loss of generality assume $x<y$



      $$Longrightarrow[x,y]subseteq[a,b]$$



      The function $f_{|[x,y]}$ is continuous and for all $gammain [f(x),f(y)]$ there is a $zin[x,y]$ such that $f(z)= gamma$




      Here is the part where I am not quite sure. Also if you have an idea how to prove it differently please share it with me:



      Assume there exists finitely many points $zin [x,y]$ such that $f(z)neq z$.



      Then we are already done because that means there must exist infinitely many Points for which the opposite is true, since the interval consists of infinitely many Points.



      Therefore assume there exists infinitely many Points such that $zin [x,y] wedge f(z)neq z$



      Then there must also exist infinitely many points for which $f(z)>z$ and $f(z')<z'$



      Without loss of generality assume there are only finitely many Points for which $z<f(z)$. Then there exists a maximal element of this set and one can split the interval $[x,y]$ into $[x,z]$ and $(z,y]$. If there exists a $win (z,y]$ such that $f(w) = w$ we are already done. Therefore assume $forall_{win (z,y]} f(w)>w$. This would invoke a contradiction that $f$ is continuous in $z$.



      Therefore we have showed there exists infinitely many $z,z'in [x,y]$ such that $f(z)>z$ and $f(z')<z'$



      One can pick now $z,z'$ such that $f(z)>z$ and $f(z')<z'$



      Without loss of generality $z<z'$



      Then $[z,z']$ is a closed interval and due to the Intermediate-value-theorem



      $$forall_{gammain[f(z),f(z')]}exists_{xin[z,z']}f(x)=gamma$$



      If we repeat this process by Always choosing an element $l$ for the left end of the interval for which we have $f(l)>l$ and for the Right end an element for which we have $f(r)<r$. Then we get a nested interval which must converge to a fixed Point.




      Is my thought process consistent? One Thing that this idea lacks is probably that there must always exist Points $l$ and $r$ like described above otherwise the proof does not work. Is it possible to prove this somehow?






      real-analysis proof-verification







      share|cite|improve this question











      $endgroup$




      I have tried to prove the statement but there is a part where I am not sure if I can argue like that, I will mark it by setting it in bold



      We know that $[a,b]subseteq f([a,b])$ and that $f$ is continuous



      $$Longrightarrow exists_{x,yin [a,b]}f(x)=awedge f(y)=b$$



      Without loss of generality assume $x<y$



      $$Longrightarrow[x,y]subseteq[a,b]$$



      The function $f_{|[x,y]}$ is continuous and for all $gammain [f(x),f(y)]$ there is a $zin[x,y]$ such that $f(z)= gamma$




      Here is the part where I am not quite sure. Also if you have an idea how to prove it differently please share it with me:



      Assume there exists finitely many points $zin [x,y]$ such that $f(z)neq z$.



      Then we are already done because that means there must exist infinitely many Points for which the opposite is true, since the interval consists of infinitely many Points.



      Therefore assume there exists infinitely many Points such that $zin [x,y] wedge f(z)neq z$



      Then there must also exist infinitely many points for which $f(z)>z$ and $f(z')<z'$



      Without loss of generality assume there are only finitely many Points for which $z<f(z)$. Then there exists a maximal element of this set and one can split the interval $[x,y]$ into $[x,z]$ and $(z,y]$. If there exists a $win (z,y]$ such that $f(w) = w$ we are already done. Therefore assume $forall_{win (z,y]} f(w)>w$. This would invoke a contradiction that $f$ is continuous in $z$.



      Therefore we have showed there exists infinitely many $z,z'in [x,y]$ such that $f(z)>z$ and $f(z')<z'$



      One can pick now $z,z'$ such that $f(z)>z$ and $f(z')<z'$



      Without loss of generality $z<z'$



      Then $[z,z']$ is a closed interval and due to the Intermediate-value-theorem



      $$forall_{gammain[f(z),f(z')]}exists_{xin[z,z']}f(x)=gamma$$



      If we repeat this process by Always choosing an element $l$ for the left end of the interval for which we have $f(l)>l$ and for the Right end an element for which we have $f(r)<r$. Then we get a nested interval which must converge to a fixed Point.




      Is my thought process consistent? One Thing that this idea lacks is probably that there must always exist Points $l$ and $r$ like described above otherwise the proof does not work. Is it possible to prove this somehow?






      real-analysis proof-verification




      real-analysis proof-verification






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 30 at 20:26









      Pedro Tamaroff♦

      97.5k10153299




      97.5k10153299










      asked Jan 30 at 20:12









      New2MathNew2Math

      16715




      16715






















          2 Answers
          2






          active

          oldest

          votes


















          2












          $begingroup$

          Assume that there is no fixed point, $f(x)ne x$ for all $xinmathbb{R}$. Then, by continuity, either $f(x)>x$ for all $xinmathbb{R}$, or $f(x)<x$ for all $xinmathbb{R}$. Assume the former. Then, for each $xin[a,b]$, one has $f(x)>xge a$, that is, $f(x)>a$. Hence $anotin f([a,b])$, and we are done. The second case can be considered in a similar way.






          share|cite|improve this answer









          $endgroup$





















            0












            $begingroup$

            There is no reason to split your argument into counting finitely/infinitely many things. The moment you have a single fixed point, you're done. Or if you don't have any fixed points, then you will have at least one(infinite or not doesn't matter) $z$ and one $z'$ (guaranteed by the assumption that $[a,b]$ is contained in the image) such that




            $f(z)>z$
            and $f(z′)<z′$




            Once you have this, you can just apply intermediate value theorem once as in the standard proof (see below).



            PS I cannot understand what you wrote starting from the line




            Without loss of generality assume there are only finitely many Points for which $z<f(z)$.





            The standard proof. Without loss $[a,b]= [0,1]$. $f$ is continuous, so by extreme value theorem, it attains its min $mle 0$ and max $Mge 1$ at the points $x_0$ and $x_1$ respectively. Define for $tin[0,1]$, $x_t = tx_0 + (1-t)x_1$ and consider $F(t) = f(x_t)-x_t$. We have $F(0) = m - x_0 ≤ -x_0le 0$, and $F(1) = M - x_1 ≥ 1-x_1 ge 0.$ By intermediate value theorem, there is a point $s$ in $[0,1]$ such that $F(s) = 0$. This corresponds to a point $x_s in [0,1]$ such that $f(x_s) = x_s$, as needed.






            share|cite|improve this answer









            $endgroup$














              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',
              autoActivateHeartbeat: false,
              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%2f3094045%2fshow-that-if-fa-b-rightarrow-mathbbr-is-continuous-and-a-b-subseteq-f%23new-answer', 'question_page');
              }
              );

              Post as a guest















              Required, but never shown

























              2 Answers
              2






              active

              oldest

              votes








              2 Answers
              2






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes









              2












              $begingroup$

              Assume that there is no fixed point, $f(x)ne x$ for all $xinmathbb{R}$. Then, by continuity, either $f(x)>x$ for all $xinmathbb{R}$, or $f(x)<x$ for all $xinmathbb{R}$. Assume the former. Then, for each $xin[a,b]$, one has $f(x)>xge a$, that is, $f(x)>a$. Hence $anotin f([a,b])$, and we are done. The second case can be considered in a similar way.






              share|cite|improve this answer









              $endgroup$


















                2












                $begingroup$

                Assume that there is no fixed point, $f(x)ne x$ for all $xinmathbb{R}$. Then, by continuity, either $f(x)>x$ for all $xinmathbb{R}$, or $f(x)<x$ for all $xinmathbb{R}$. Assume the former. Then, for each $xin[a,b]$, one has $f(x)>xge a$, that is, $f(x)>a$. Hence $anotin f([a,b])$, and we are done. The second case can be considered in a similar way.






                share|cite|improve this answer









                $endgroup$
















                  2












                  2








                  2





                  $begingroup$

                  Assume that there is no fixed point, $f(x)ne x$ for all $xinmathbb{R}$. Then, by continuity, either $f(x)>x$ for all $xinmathbb{R}$, or $f(x)<x$ for all $xinmathbb{R}$. Assume the former. Then, for each $xin[a,b]$, one has $f(x)>xge a$, that is, $f(x)>a$. Hence $anotin f([a,b])$, and we are done. The second case can be considered in a similar way.






                  share|cite|improve this answer









                  $endgroup$



                  Assume that there is no fixed point, $f(x)ne x$ for all $xinmathbb{R}$. Then, by continuity, either $f(x)>x$ for all $xinmathbb{R}$, or $f(x)<x$ for all $xinmathbb{R}$. Assume the former. Then, for each $xin[a,b]$, one has $f(x)>xge a$, that is, $f(x)>a$. Hence $anotin f([a,b])$, and we are done. The second case can be considered in a similar way.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 30 at 20:29









                  VladimirVladimir

                  5,413618




                  5,413618























                      0












                      $begingroup$

                      There is no reason to split your argument into counting finitely/infinitely many things. The moment you have a single fixed point, you're done. Or if you don't have any fixed points, then you will have at least one(infinite or not doesn't matter) $z$ and one $z'$ (guaranteed by the assumption that $[a,b]$ is contained in the image) such that




                      $f(z)>z$
                      and $f(z′)<z′$




                      Once you have this, you can just apply intermediate value theorem once as in the standard proof (see below).



                      PS I cannot understand what you wrote starting from the line




                      Without loss of generality assume there are only finitely many Points for which $z<f(z)$.





                      The standard proof. Without loss $[a,b]= [0,1]$. $f$ is continuous, so by extreme value theorem, it attains its min $mle 0$ and max $Mge 1$ at the points $x_0$ and $x_1$ respectively. Define for $tin[0,1]$, $x_t = tx_0 + (1-t)x_1$ and consider $F(t) = f(x_t)-x_t$. We have $F(0) = m - x_0 ≤ -x_0le 0$, and $F(1) = M - x_1 ≥ 1-x_1 ge 0.$ By intermediate value theorem, there is a point $s$ in $[0,1]$ such that $F(s) = 0$. This corresponds to a point $x_s in [0,1]$ such that $f(x_s) = x_s$, as needed.






                      share|cite|improve this answer









                      $endgroup$


















                        0












                        $begingroup$

                        There is no reason to split your argument into counting finitely/infinitely many things. The moment you have a single fixed point, you're done. Or if you don't have any fixed points, then you will have at least one(infinite or not doesn't matter) $z$ and one $z'$ (guaranteed by the assumption that $[a,b]$ is contained in the image) such that




                        $f(z)>z$
                        and $f(z′)<z′$




                        Once you have this, you can just apply intermediate value theorem once as in the standard proof (see below).



                        PS I cannot understand what you wrote starting from the line




                        Without loss of generality assume there are only finitely many Points for which $z<f(z)$.





                        The standard proof. Without loss $[a,b]= [0,1]$. $f$ is continuous, so by extreme value theorem, it attains its min $mle 0$ and max $Mge 1$ at the points $x_0$ and $x_1$ respectively. Define for $tin[0,1]$, $x_t = tx_0 + (1-t)x_1$ and consider $F(t) = f(x_t)-x_t$. We have $F(0) = m - x_0 ≤ -x_0le 0$, and $F(1) = M - x_1 ≥ 1-x_1 ge 0.$ By intermediate value theorem, there is a point $s$ in $[0,1]$ such that $F(s) = 0$. This corresponds to a point $x_s in [0,1]$ such that $f(x_s) = x_s$, as needed.






                        share|cite|improve this answer









                        $endgroup$
















                          0












                          0








                          0





                          $begingroup$

                          There is no reason to split your argument into counting finitely/infinitely many things. The moment you have a single fixed point, you're done. Or if you don't have any fixed points, then you will have at least one(infinite or not doesn't matter) $z$ and one $z'$ (guaranteed by the assumption that $[a,b]$ is contained in the image) such that




                          $f(z)>z$
                          and $f(z′)<z′$




                          Once you have this, you can just apply intermediate value theorem once as in the standard proof (see below).



                          PS I cannot understand what you wrote starting from the line




                          Without loss of generality assume there are only finitely many Points for which $z<f(z)$.





                          The standard proof. Without loss $[a,b]= [0,1]$. $f$ is continuous, so by extreme value theorem, it attains its min $mle 0$ and max $Mge 1$ at the points $x_0$ and $x_1$ respectively. Define for $tin[0,1]$, $x_t = tx_0 + (1-t)x_1$ and consider $F(t) = f(x_t)-x_t$. We have $F(0) = m - x_0 ≤ -x_0le 0$, and $F(1) = M - x_1 ≥ 1-x_1 ge 0.$ By intermediate value theorem, there is a point $s$ in $[0,1]$ such that $F(s) = 0$. This corresponds to a point $x_s in [0,1]$ such that $f(x_s) = x_s$, as needed.






                          share|cite|improve this answer









                          $endgroup$



                          There is no reason to split your argument into counting finitely/infinitely many things. The moment you have a single fixed point, you're done. Or if you don't have any fixed points, then you will have at least one(infinite or not doesn't matter) $z$ and one $z'$ (guaranteed by the assumption that $[a,b]$ is contained in the image) such that




                          $f(z)>z$
                          and $f(z′)<z′$




                          Once you have this, you can just apply intermediate value theorem once as in the standard proof (see below).



                          PS I cannot understand what you wrote starting from the line




                          Without loss of generality assume there are only finitely many Points for which $z<f(z)$.





                          The standard proof. Without loss $[a,b]= [0,1]$. $f$ is continuous, so by extreme value theorem, it attains its min $mle 0$ and max $Mge 1$ at the points $x_0$ and $x_1$ respectively. Define for $tin[0,1]$, $x_t = tx_0 + (1-t)x_1$ and consider $F(t) = f(x_t)-x_t$. We have $F(0) = m - x_0 ≤ -x_0le 0$, and $F(1) = M - x_1 ≥ 1-x_1 ge 0.$ By intermediate value theorem, there is a point $s$ in $[0,1]$ such that $F(s) = 0$. This corresponds to a point $x_s in [0,1]$ such that $f(x_s) = x_s$, as needed.







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Jan 30 at 20:37









                          Calvin KhorCalvin Khor

                          12.5k21439




                          12.5k21439






























                              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.




                              draft saved


                              draft discarded














                              StackExchange.ready(
                              function () {
                              StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3094045%2fshow-that-if-fa-b-rightarrow-mathbbr-is-continuous-and-a-b-subseteq-f%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

                              MongoDB - Not Authorized To Execute Command

                              Image
                              .everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ height:90px;width:728px;box-sizing:border-box; } 0 I have successfully enabled authorization on MongoDB and I have created an account on the admin database and then I created an account for my database called test. The following connection string to connect to my test database works successfully: mongo --host 192.168.17.52 --port 27017 -u user1 -p password --authenticationDatabase test Only problem I have now is, I cannot execute commands such as: show dbs. I get the following error when I try to do so: "errmsg" : "not authorized on admin to execute command { listDatabases: 1.0, lsid: { id: UUID("a1d5bc0d-bc58-485e-b232-270758a89455") }, $db: "admin...
                              Read more

                              in spring boot 2.1 many test slices are not allowed anymore due to multiple @BootstrapWith

                              Image
                              1 I tried to upgrade a yummy sandwich made of two test slices (@JsonTest and @JdbcTest in my case, crunchy test code in between) adding spring boot 2.1 flavour to it. But it seems it was not much of a success. I cannot annotate my tests with many @...Test since they are now each bringing their own XxxTestContextBootstrapper. It used to work when they all used same SpringBootTestContextBootstrapper. @RunWith(SpringRunner.class) @JdbcTest @JsonTest public class Test { @Test public void test() { System.out.printn("Hello, World !"); } } The error I get from BootstrapUtils is illegalStateException : Configuration error: found multiple declarations of @BootstrapWith for test class I understand I might be doing something wrong here but is there an easy way I could load both Json and Jdbc contex...
                              Read more

                              How to fix TextFormField cause rebuild widget in Flutter

                              Image
                              .everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ height:90px;width:728px;box-sizing:border-box; } 0 I've got the same issue at #9280. I tried all solutions from that topic with/without GlobalKey for Scaffold and Form but the problem wasn't fixed. I tried to build a authentication form with login and signup. But the keyboard doesn't show when I tap the TextFormField at the first time. Then I try to input some letter from physical keyboard, the widget is rebuilt. Like this: login screen Nothing to be log when I run flutter run --verbose, so I logged manually. Every tapping on field, the widget called dispose -> init -> build. [√] Flutter (Channel beta, v1.0.0, on Microsoft Windows [Version 10.0.17763.195], ...
                              Read more