Interior points of correspondence Let $\Gamma(x)$ be a correspondence (i.e. a set-valued function) between two Euclidean spaces which is continuous (i.e. both lower- and upper-hemicontinuous). If $y$ is a point in the interior of $\Gamma(x_0)$ it seems plausible from drawing graphs that there should be an open set $U$ containing $x_0$ such that $y$ is in the interior of $\Gamma(x)$ for all $x \in U$. 
Is this a correct theorem? I would appreciate some references or other pointers.  
 A: It seems that the claim does not hold. Define $\Gamma:\mathbb{R}\to\mathcal{P}(\mathbb{R})$ by $\Gamma(x)=\mathbb{Q}$, if $x\neq 0$, and $\Gamma(0)=\mathbb{R}$.
$\Gamma$ is upper hemicontinuous everywhere. Pick $x\in\mathbb{R}$ and an open neighbourhood $V$ of $\Gamma(x)$. If $x=0$, choose $U=\mathbb{R}$, and if $x\neq 0$, choose $U=\mathbb{R}\setminus\{0\}$. Then $\Gamma(y)$ is a subset of $V$ for each $y\in U$.
$\Gamma$ is lower hemicontinuous everywhere. Pick $x\in\mathbb{R}$ and an open set $V$ intersecting $\Gamma(x)$. Note that $V$ contains a rational number, so we can choose $U=\mathbb{R}$. Certainly $\Gamma(y)$ intersects $V$ for each $y\in U$.
The claim does not hold in this case. The point $y=0$ is in the interior of $\Gamma(0)=\mathbb{R}$, but any open set $U$ containing $0$ contains another point $x$ too. Since the interior of $\Gamma(x)=\mathbb{Q}$ is empty, it certainly does not contain $y$.
A: If I haven't overlooked something, this should be a counterexample:
Define $\Gamma$ on $\mathbb R$ as follows:
$$\Gamma(x)=\begin{cases}
[0,1]\setminus\mathbb Q; &x<0 \\
{[0,1]}; & x\ge 0
\end{cases}$$
The above property fails for $x_0=0$ and any $y$ from the interior of $\Gamma(x_0)$.
EDIT:
The following example shows that the claim is false for compact-valued multifunctions, too:
$$\Gamma(x)=\begin{cases}
[x-1,x]\cup[-x,-x+1]; &x<0 \\
{[-1,1]}; & x\ge 0
\end{cases}$$
Now $x_0=y=0$.
