# Continuity of set-valued map

## Main Question or Discussion Point

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.

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WWGD
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2019 Award
Do you know the definitions of u.s.c and l.s.c?

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.

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##.

WWGD
Gold Member
2019 Award
I see this as just the equivalent of the definition of continuity at a point; from Wiki:

, 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:
thank you very much, , economicsnerd.
Also like to thank WWGD.

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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.

Stephen Tashi