Prove that $f : mathbbR rightarrow mathbbR$ is lower semi-continuous

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Prove that $f : mathbbR rightarrow mathbbR$ is lower semi-continuous if and only if if the set$ (x,y) in mathbbR^2 : y ge f(x) $ is closed in $mathbbR^2.$



My Proof :



$f$ is lower semicontinuous on $ mathbbR$ if and only if $-f $ is upper semi-continuous on $mathbb R$ . so $f$ is lower semi-continuous on $mathbbR$ if and only the set$(x,y) in mathbbR^2: y ge f(x) $ is closed in $mathbbR^2$



Is it correct..???



Thanks in advance




real-analysis




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    Prove that $f : mathbbR rightarrow mathbbR$ is lower semi-continuous if and only if if the set$ (x,y) in mathbbR^2 : y ge f(x) $ is closed in $mathbbR^2.$



    My Proof :



    $f$ is lower semicontinuous on $ mathbbR$ if and only if $-f $ is upper semi-continuous on $mathbb R$ . so $f$ is lower semi-continuous on $mathbbR$ if and only the set$(x,y) in mathbbR^2: y ge f(x) $ is closed in $mathbbR^2$



    Is it correct..???



    Thanks in advance




    real-analysis




    share|cite|improve this question
























      up vote
      0
      down vote

      favorite









      up vote
      0
      down vote

      favorite











      Prove that $f : mathbbR rightarrow mathbbR$ is lower semi-continuous if and only if if the set$ (x,y) in mathbbR^2 : y ge f(x) $ is closed in $mathbbR^2.$



      My Proof :



      $f$ is lower semicontinuous on $ mathbbR$ if and only if $-f $ is upper semi-continuous on $mathbb R$ . so $f$ is lower semi-continuous on $mathbbR$ if and only the set$(x,y) in mathbbR^2: y ge f(x) $ is closed in $mathbbR^2$



      Is it correct..???



      Thanks in advance




      real-analysis




      share|cite|improve this question














      Prove that $f : mathbbR rightarrow mathbbR$ is lower semi-continuous if and only if if the set$ (x,y) in mathbbR^2 : y ge f(x) $ is closed in $mathbbR^2.$



      My Proof :



      $f$ is lower semicontinuous on $ mathbbR$ if and only if $-f $ is upper semi-continuous on $mathbb R$ . so $f$ is lower semi-continuous on $mathbbR$ if and only the set$(x,y) in mathbbR^2: y ge f(x) $ is closed in $mathbbR^2$



      Is it correct..???



      Thanks in advance





      real-analysis





      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Aug 20 at 13:30









      Bernard

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      111k635103










      asked Aug 20 at 11:52









      stupid

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