Let X = R 2 . Given u ( ⋅ , ⋅ ) such that one of the level curves y = f ( x ) is discontinuous at x = t , construct (or show the existence of) two points p , l ∈ X such that l is a limit point of both L ( p ) and U ( p ) but is contained in exactly one of the two

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Let   X  =            R        2   . Given   u  (  ⋅  ,  ⋅  ) such that one of the level curves   y  =  f  (  x  ) is discontinuous at   x  =  t , construct (or show the existence of) two points   p  ,  l  ∈  X such that l is a limit point of both   L  (  p  ) and   U  (  p  ) but is contained in exactly one of the two

scrapbymarieix · Accepted Answer

Step 1The fact that L(p) and U(p) are closed implies that the function f associated with   c  =  u  (  p  ) is continuous. Indeed, if f is such that   {  (  x  ,  y  )  ∈            R        2    ,  u  (  x  ,  y  )  =  u  (  p  )  }  =  {  (  x  ,  f  (  x  )  )  ,  x  ∈      R    } , you can write : [the next equality is false]  L  (  p  )  =  {  (  x  ,  y  )  ∈            R        2    ,  y  ≤  f  (  x  )  }and L(p) is closed if and only if f is upper-semicontinuous. (This is a well-known property of upper-semicontinuous functions : see for instance here.)Now, the same reasoning with U(p) implies that f is also lower-semicontinuous. Therefore, f is actually continuous. (I think this proves the claim you quote ?)EDIT : Second attempt : If   l  ∈  Xis a limit point of both L(p) and U(p), since both sets are closed, you have   l  ∈  L  (  p  ) and   l  ∈  U  (  p  ) . Therefore (and this is actually independant of the continuity of level functions f) you cannot find an l as in your question.

Let X=mathbb(R)^2. Given u(cdot , cdot) such that one of the level curves y=f(x) is discontinuous at x=t, construct (or show the existence of) two points p, l in X such that l is a limit point of both L(p) and U(p) but is contained in exactly one of the two

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