By Ewald C.-O.
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Therefore y ∈ µ(x) and µ is closed at x. 6 ( second part ) implies that µ = γ ∩ µ is uhc at x. Finally we show that µ is compact valued. Since µ(˜ x) ⊂ γ(˜ x) for all x˜ and γ is compact valued it is enough to show that µ(˜ x) is closed in γ(˜ x) ( in the relative topology ). Equivalently we can show that γ(˜ x)\µ(˜ x) is open in γ(˜ x). Assume this would not be the case. e. t. limn y = y˜ and U (y n , x˜) ∩ γ(˜ x) = ∅ , ∀ n. Clearly limn (y n , x˜, z) = (˜ y , x˜, z) ∈ Gr(U ). t. (y , x˜, z) ∈ Gr(U ) which implies that z ∈ U (y n0 , x˜) ∩ γ(˜ x).
I closed at x ∀ i ⇒ i i i γi uhc at x γi lhc at x γi is closed at x 4. γi has open graph ∀ i ⇒ i γi has open graph Proof. 2 ( sequential characterization of hemi continuity ) and the fact that a sequence in a product space converges if and only if all its component sequences converge. ) are clear. 13. , k and γi : X →→ Yi be correspondences. Then γi : X →→ i Yi := { i x → yi : yi ∈ Y i } i γi (x). 9. Let γi : X →→ Yi be as above. 1. γi uhc and γi compact valued ∀i ⇒ 2. γi lhc ∀i ⇒ i i γi uhc and compact valued.
Let γ, µ be as above. then 46 1. γ, µ uhc ⇒ γ ◦ µ uhc 2. γ, µ lhc ⇒ γ ◦ µ lhc Proof. Exercise ! 12. , k be correspondences. Then the correspondence γi : X →→ i Yi i → x γi (x) i is called the product of the γi . 8. Assume γi are correspondences as above. 1. γi uhc at x and γi (z) compact ∀z ∈ X and ∀ i ⇒ 2. γi lhc at x and γi (z) compact ∀z ∈ X and ∀ i ⇒ 3. γi closed at x ∀ i ⇒ i i i γi uhc at x γi lhc at x γi is closed at x 4. γi has open graph ∀ i ⇒ i γi has open graph Proof. 2 ( sequential characterization of hemi continuity ) and the fact that a sequence in a product space converges if and only if all its component sequences converge.