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Continuity of set-valued map

  1. Sep 30, 2014 #1
    1/ Prove that the set-valued map F defined by
    F : [0, 2π] ⇒ R2 as
    F(α) := {λ(cos α, sin α) : λ ≥ 0}.
    is continuous,
    but not upper semicontinuous at any α ∈ [0, 2π].
    2/ What is the fact that " F is continuous if it is both u.s.c. and l.s.c".
    I would like illustrate that and thank you.
     
  2. jcsd
  3. Oct 1, 2014 #2

    WWGD

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    Do you know the definitions of u.s.c and l.s.c?
     
  4. Oct 1, 2014 #3
    Definition(u.s.c.)
    Let X and Y be two topological spaces and F:X→P(Y)\{φ} be a set-valued map, we say that F is upper semicontinuous at x₀∈X, u.s.c. for short, if for any open V containing F(x₀) there exists a neighborhood N(x₀) of x₀ such that F(x)⊆V for all x∈N(x₀). We say that F is upper semicontinuous if it is so at every x∈X.
    Definition(l.s.c.)
    Let X and Y be two topological spaces and F:X→P(Y)\{φ}. We say that F is lower semicontinuous at x₀, l.s.c. for short, if for every open set V in Y with V∩F(x₀)≠φ, there exists a neighborhood N(x₀) for x₀ such that V∩F(x)≠φ for all x∈N(x₀). F is called lower semicontinuous if it is lower semicontinuous at each x∈X.
     
  5. Oct 4, 2014 #4
    It appears to be lower semicontinuous but not upper semicontinuous.

    To see it's lower semicontinuous, fix an open set ##V\subseteq \mathbb R^2## which intersects ##F(\alpha)## for some given ##\alpha\in[0,2\pi]##. That is, ##\lambda(\cos\alpha,\sin\alpha) \in V## for some ##\lambda\geq0##. You can check that for ##\beta## sufficiently close to ##\alpha##, the openness of ##V## implies ##\lambda(\cos\beta,\sin\beta) \in V##, and in particular, ##V## intersects ##F(\beta)##.

    To see it's not upper semicontinuous at any ##\alpha\in [0,2\pi]##, consider the set ##V = \{\lambda(\cos\beta, \sin\beta): \enspace \beta\in \mathbb R, \enspace \lambda \in (-1, \infty), \enspace \lambda|\alpha-\beta|<1\}##. You can verify that ##V## is an open superset of ##F(\alpha)##, and that ##F(\beta) \nsubseteq V## for any ##\beta\neq\alpha##.
     
  6. Oct 4, 2014 #5

    WWGD

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    I see this as just the equivalent of the definition of continuity at a point; from Wiki:

    337874d01d7807fe9881a5c60fab239a.png , but I am confused at your statement that it is continuous but not u.s.c for ## \alpha \in [0, 2\pi ] ## , since continuity implies u.s.c. Maybe you want continuity for ## \alpha ## outside of ## [0, 2\pi] ## ?

    Sorry to nitpick so much, but in my understanding, your definition of lower semicontinuity implies continuity. What def. of continuity are you using, the inverse image of open/closed is open/closed?
     
    Last edited: Oct 4, 2014
  7. Oct 5, 2014 #6
    thank you very much, , economicsnerd.
    Also like to thank WWGD.
     
  8. Oct 5, 2014 #7
    The file attachment in down, please help me.
     

    Attached Files:

  9. Oct 5, 2014 #8
    The usual definition of a set-valued map being continuous is that it's both upper and lower semicontinuous. So of course, it can't be continuous without being lower semicontinuous.
     
  10. Oct 29, 2014 #9

    Stephen Tashi

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    Browsing the web, it appears that the study of set valued functions is on the frontier of mathematical research. The PDF http://pareto.uab.es/~adaniilidis/DP_2010.pdf gives various definitions related to the continuity of set valued functions and distinguishes between continuity and "strict continuity".
     
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