Is F continuous if it is both upper and lower semicontinuous?

[moh salem]

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.

 

[WWGD]

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

 

[moh salem]

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.

 

[economicsnerd]

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]

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?

 

[moh salem]

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

 

[moh salem]

The file attachment in down, please help me.

 

[economicsnerd]

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]

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".