Discontinuities of upper semicontinuous function

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Suppose $f:mathbbR rightarrow mathbbR$ is an upper semicontinuous function (USC), i.e. for all $alpha in mathbbR$, the set $ x in mathbbR : f(x) < alpha $ is open.



What are the set of discontinuities of $f$? For $f = chi_[- infty, 0)$, its discontinuity is $x=0$. Hence, the set of discontinuities of an USC is finite. Do we have example such that set of discontinuities of an USC is countable or uncountable?



Here states that characteristic function of Cantor set has uncountable discontinuities. But there is no justification on why such function is upper semicontinuous.






continuity semicontinuous-functions







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    Suppose $f:mathbbR rightarrow mathbbR$ is an upper semicontinuous function (USC), i.e. for all $alpha in mathbbR$, the set $ x in mathbbR : f(x) < alpha $ is open.



    What are the set of discontinuities of $f$? For $f = chi_[- infty, 0)$, its discontinuity is $x=0$. Hence, the set of discontinuities of an USC is finite. Do we have example such that set of discontinuities of an USC is countable or uncountable?



    Here states that characteristic function of Cantor set has uncountable discontinuities. But there is no justification on why such function is upper semicontinuous.






    continuity semicontinuous-functions







    share|cite|improve this question

























      up vote
      0
      down vote

      favorite









      up vote
      0
      down vote

      favorite











      Suppose $f:mathbbR rightarrow mathbbR$ is an upper semicontinuous function (USC), i.e. for all $alpha in mathbbR$, the set $ x in mathbbR : f(x) < alpha $ is open.



      What are the set of discontinuities of $f$? For $f = chi_[- infty, 0)$, its discontinuity is $x=0$. Hence, the set of discontinuities of an USC is finite. Do we have example such that set of discontinuities of an USC is countable or uncountable?



      Here states that characteristic function of Cantor set has uncountable discontinuities. But there is no justification on why such function is upper semicontinuous.






      continuity semicontinuous-functions







      share|cite|improve this question















      Suppose $f:mathbbR rightarrow mathbbR$ is an upper semicontinuous function (USC), i.e. for all $alpha in mathbbR$, the set $ x in mathbbR : f(x) < alpha $ is open.



      What are the set of discontinuities of $f$? For $f = chi_[- infty, 0)$, its discontinuity is $x=0$. Hence, the set of discontinuities of an USC is finite. Do we have example such that set of discontinuities of an USC is countable or uncountable?



      Here states that characteristic function of Cantor set has uncountable discontinuities. But there is no justification on why such function is upper semicontinuous.






      continuity semicontinuous-functions




      continuity semicontinuous-functions






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      share|cite|improve this question













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      edited Sep 1 at 0:05









      Martin Sleziak

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      43.7k6113260










      asked Oct 17 '16 at 3:31









      Idonknow

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          Since, for a set $Ssubseteq Bbb R$, $$xinBbb R,:,chi_S(x)<alpha=begincasesBbb R&textif alpha>1\Bbb Rsetminus S&textif 0<alphale1\emptyset&textif alpha<0endcases$$



          an indicator function is USC if and only if $S$ is closed, and LSC if and only if $S$ is open.






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            accepted










            Since, for a set $Ssubseteq Bbb R$, $$xinBbb R,:,chi_S(x)<alpha=begincasesBbb R&textif alpha>1\Bbb Rsetminus S&textif 0<alphale1\emptyset&textif alpha<0endcases$$



            an indicator function is USC if and only if $S$ is closed, and LSC if and only if $S$ is open.






            share|cite|improve this answer
























              up vote
              1
              down vote



              accepted










              Since, for a set $Ssubseteq Bbb R$, $$xinBbb R,:,chi_S(x)<alpha=begincasesBbb R&textif alpha>1\Bbb Rsetminus S&textif 0<alphale1\emptyset&textif alpha<0endcases$$



              an indicator function is USC if and only if $S$ is closed, and LSC if and only if $S$ is open.






              share|cite|improve this answer






















                up vote
                1
                down vote



                accepted







                up vote
                1
                down vote



                accepted






                Since, for a set $Ssubseteq Bbb R$, $$xinBbb R,:,chi_S(x)<alpha=begincasesBbb R&textif alpha>1\Bbb Rsetminus S&textif 0<alphale1\emptyset&textif alpha<0endcases$$



                an indicator function is USC if and only if $S$ is closed, and LSC if and only if $S$ is open.






                share|cite|improve this answer












                Since, for a set $Ssubseteq Bbb R$, $$xinBbb R,:,chi_S(x)<alpha=begincasesBbb R&textif alpha>1\Bbb Rsetminus S&textif 0<alphale1\emptyset&textif alpha<0endcases$$



                an indicator function is USC if and only if $S$ is closed, and LSC if and only if $S$ is open.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Oct 17 '16 at 5:20







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