Let $F:X\to2^Y$ be a set-valued map from a metric space to the subsets of another metric space. We say it is upper semi-continuous if for every $\epsilon$ and every $x_0\in X$ there exists $\delta$ such that $d(x,x_0)<\delta\implies F(x)\subseteq B(F(x_0),\epsilon)=\bigcup_{t\in F(x_0)}B(t,\epsilon)$.
My teacher introduced that, and, assuming by "image of a set via $F$" he meant:
$$F(A)=\bigcup_{t\in A}F(t),$$
from the way he put it, it seems the following theorem should be true.
Theorem
If $F:X\to2^Y$ is upper semi-continuous, then $F(K)$ is compact for all compact sets $K\subseteq X$.
I googled for a proof, and was unable to find one. I cannot think of one myself. How do I prove this theorem?
Note: For the purposes the course uses this for, $F$ can be assumed to be compact and convex-valued, and $X,Y$ can be assumed to be $\mathbb{R}^n,\mathbb{R}^m$, or even $n=m=1$, if needed.
Update
Comment by @user254665:
Counter-example :If $Y$ is not compact and $X$ is not empty and $F(x)=Y$ for every $x\in x$...... BTW your assumption about the meaning of the image of a set is correct.
In effect, readig the notes more carefully, I find:
Allora, sotto quese due condizioni, cioè la semicontinuità superiore e la compattezza, vale quella bella cosa, che l'immagine di un compatto è un compatto. E questaamm è la proprietà fondamentale che hanno queste mappe.
That is:
Then, under these two conditions, that is upper semicontinuity and compactness, that nice thing holds, that the image of a compact set is a compact set. And this, uuh, is the fundamental properties that these maps have.
So it seems he actually stated the following:
Let $f$ be USC and compact-valued. Then $F(K)$ is compact whenever $K$ is.
Can we prove this?
