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Decomposing $f + g > t$



The 2019 Stack Overflow Developer Survey Results Are In
Unicorn Meta Zoo #1: Why another podcast?
Announcing the arrival of Valued Associate #679: Cesar ManaraMinimal dense subset of $mathbbQ cap [0,1]$A classic example in measure theory, hold the measure theoryStep-by-step help using the distributive law in set theoryProving set function is measureCheck the identity $Acap (B - C)=(Acap B) - (Acap C)$Classification of an open set in realSuppose $A^text' $ is the set of limit point of $A$. Prove $A^text' $ is closed set.Show finite unions of rectangles form an algebraSymmetric difference as a equivalent relation-equivalence classesReformulation of intersection of two unions










0












$begingroup$


For this purpose of this question, define



$$
leftf > tright = left f(x) > tright
$$



I was told that



$$
leftf + g > t right = cup_r in mathbbQ leftf > r right cap leftg > t - r right
$$



I want to prove it, but I got stuck. The following equation seems to be a nice starting point, but I cannot prove it either.



$$
leftf + g > t right = cup_r in mathbbR leftf > r right cap leftg > t - r right
$$



Any hint?










share|cite|improve this question









$endgroup$
















    0












    $begingroup$


    For this purpose of this question, define



    $$
    leftf > tright = left f(x) > tright
    $$



    I was told that



    $$
    leftf + g > t right = cup_r in mathbbQ leftf > r right cap leftg > t - r right
    $$



    I want to prove it, but I got stuck. The following equation seems to be a nice starting point, but I cannot prove it either.



    $$
    leftf + g > t right = cup_r in mathbbR leftf > r right cap leftg > t - r right
    $$



    Any hint?










    share|cite|improve this question









    $endgroup$














      0












      0








      0





      $begingroup$


      For this purpose of this question, define



      $$
      leftf > tright = left f(x) > tright
      $$



      I was told that



      $$
      leftf + g > t right = cup_r in mathbbQ leftf > r right cap leftg > t - r right
      $$



      I want to prove it, but I got stuck. The following equation seems to be a nice starting point, but I cannot prove it either.



      $$
      leftf + g > t right = cup_r in mathbbR leftf > r right cap leftg > t - r right
      $$



      Any hint?










      share|cite|improve this question









      $endgroup$




      For this purpose of this question, define



      $$
      leftf > tright = left f(x) > tright
      $$



      I was told that



      $$
      leftf + g > t right = cup_r in mathbbQ leftf > r right cap leftg > t - r right
      $$



      I want to prove it, but I got stuck. The following equation seems to be a nice starting point, but I cannot prove it either.



      $$
      leftf + g > t right = cup_r in mathbbR leftf > r right cap leftg > t - r right
      $$



      Any hint?







      real-analysis general-topology elementary-set-theory






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Apr 8 at 6:34









      nalzoknalzok

      291217




      291217




















          1 Answer
          1






          active

          oldest

          votes


















          2












          $begingroup$

          Write $f(x)+g(x) >t$ as $f(x) >t-g(x)$. Since there is a rational number between any two real numbers we can find $r in mathbb Q$ such that $f(x) >r >t-g(x)$. Can you finish the proof now?






          share|cite|improve this answer









          $endgroup$












          • $begingroup$
            Gotcha. Thanks! Basically $leftf > t - g right = cup_r in mathbbQ leftf > r > t - g right = cup_r in mathbbQ leftf > rright cap leftr > t - g right$
            $endgroup$
            – nalzok
            Apr 8 at 6:39












          Your Answer








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          1 Answer
          1






          active

          oldest

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          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          2












          $begingroup$

          Write $f(x)+g(x) >t$ as $f(x) >t-g(x)$. Since there is a rational number between any two real numbers we can find $r in mathbb Q$ such that $f(x) >r >t-g(x)$. Can you finish the proof now?






          share|cite|improve this answer









          $endgroup$












          • $begingroup$
            Gotcha. Thanks! Basically $leftf > t - g right = cup_r in mathbbQ leftf > r > t - g right = cup_r in mathbbQ leftf > rright cap leftr > t - g right$
            $endgroup$
            – nalzok
            Apr 8 at 6:39
















          2












          $begingroup$

          Write $f(x)+g(x) >t$ as $f(x) >t-g(x)$. Since there is a rational number between any two real numbers we can find $r in mathbb Q$ such that $f(x) >r >t-g(x)$. Can you finish the proof now?






          share|cite|improve this answer









          $endgroup$












          • $begingroup$
            Gotcha. Thanks! Basically $leftf > t - g right = cup_r in mathbbQ leftf > r > t - g right = cup_r in mathbbQ leftf > rright cap leftr > t - g right$
            $endgroup$
            – nalzok
            Apr 8 at 6:39














          2












          2








          2





          $begingroup$

          Write $f(x)+g(x) >t$ as $f(x) >t-g(x)$. Since there is a rational number between any two real numbers we can find $r in mathbb Q$ such that $f(x) >r >t-g(x)$. Can you finish the proof now?






          share|cite|improve this answer









          $endgroup$



          Write $f(x)+g(x) >t$ as $f(x) >t-g(x)$. Since there is a rational number between any two real numbers we can find $r in mathbb Q$ such that $f(x) >r >t-g(x)$. Can you finish the proof now?







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Apr 8 at 6:38









          Kavi Rama MurthyKavi Rama Murthy

          74.4k53270




          74.4k53270











          • $begingroup$
            Gotcha. Thanks! Basically $leftf > t - g right = cup_r in mathbbQ leftf > r > t - g right = cup_r in mathbbQ leftf > rright cap leftr > t - g right$
            $endgroup$
            – nalzok
            Apr 8 at 6:39

















          • $begingroup$
            Gotcha. Thanks! Basically $leftf > t - g right = cup_r in mathbbQ leftf > r > t - g right = cup_r in mathbbQ leftf > rright cap leftr > t - g right$
            $endgroup$
            – nalzok
            Apr 8 at 6:39
















          $begingroup$
          Gotcha. Thanks! Basically $leftf > t - g right = cup_r in mathbbQ leftf > r > t - g right = cup_r in mathbbQ leftf > rright cap leftr > t - g right$
          $endgroup$
          – nalzok
          Apr 8 at 6:39





          $begingroup$
          Gotcha. Thanks! Basically $leftf > t - g right = cup_r in mathbbQ leftf > r > t - g right = cup_r in mathbbQ leftf > rright cap leftr > t - g right$
          $endgroup$
          – nalzok
          Apr 8 at 6:39


















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