# A correspondence is upper lower hemicontinuous if it

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A correspondence is upper (lower) hemicontinuous if it is upper (lower) hemicontinuous at all x X . It is continuous if it is both upper and lower hemicontinuous It is often easier to check sufficient conditions for a correspondence’s continuity that to verify the basic definition directly. Here are two sometimes convenient sets of conditions. Proposition 5.11. Suppose X R m and Y R n . A compact-valued correspondence f : X Y is upper hemicontinuous if, and only if, for any domain sequence x j x and any range sequence y j such that y j f ( x j ) , there exists a convergent subsequence { y j k } such that lim y j k f ( x ) . Proposition 5.12. Suppose A R m and B R n is closed. If f : A B has closed graph and f ( K ) is bounded for any compact K A , then f is upper hemicontinous. 10 To prove that the graph of Γ : A B is closed, show that the graph contains all of its limit points: if ( x n , y n ) ( x, y ) and y n Γ( x n ) for all n (i.e. each ( x n , y n ) is in the graph of Γ), then y Γ( x ) (i.e. the limit ( x, y ) is in the graph of Γ). Proposition 5.13. Suppose A R m , B R n , and f : A B . Then f is lower hemicontinuous if, and only if, for all { x m } ∈ A such that x m x A and y f ( x ) , there exist y m f ( x m ) such that y m y . Exercise 5.14. Suppose Γ : R R is defined by: Γ( x ) = ( { 1 } if x < 1 [0 , 2] if x 1 . 10 The graph of f is the subset G A × B defined by ( a, b ) G if b f ( a ). The image f ( K ) of a set K is defined as f ( K ) = S x K f ( x ). 25
Prove that Γ is upper hemicontinuous, but not lower hemicontinuous. Suppose Γ 0 : R R is defined by: Γ 0 ( x ) = ( { 1 } if x 1 [0 , 2] if x > 1 . Prove that Γ 0 is lower hemicontinuous, but not upper hemicontinuous.
Exercise 5.15. Suppose g : A B is a function. Define Γ : A B by Γ( x ) = { g ( x ) } . Prove that g is a continuous function if and only if Γ is a continuous correspondence.
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