# 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.

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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 $y$.

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It seems that our examples are very similar, only that I used irrational numbers in place where you have used rationals. (I have intersected them with the interval [0,1] in order to get a compact-valued multifunction.)\\Of course, I have to say that your answer is much more detailed. – Martin Sleziak Oct 8 '11 at 11:30
@Martin Sleziak: Indeed, they are quite similar. But I think $[0,1]\setminus\mathbb{Q}$ is not compact. Maybe you mean the values of your function are bounded? – LostInMath Oct 8 '11 at 12:45
You're right, of course. – Martin Sleziak Oct 8 '11 at 13:02
@LostInMath Is it necessary the case that the only open set covering $\mathbb{Q}$ is $\mathbb{R}$? Let $q_n$ be a enumeration of the rationals, and let $V_n=(q_n-1/4^n,q_n+1/4^n)$. Then $\bigcup V_n$ covers all of $\mathbb{Q}$ but since $\sum |V_n|<1$, $\bigcup V_n$ cannot be $\mathbb{R}$. – Jyotirmoy Bhattacharya Oct 8 '11 at 17:07
@Jyotirmoy Bhattacharya: You are absolutely right. Thanks for pointing out the mistake. Fortunately the example still works, but we have to be a little more careful proving the upper hemicontinuousness. I fixed the answer. – LostInMath Oct 8 '11 at 21:45

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 $x_0=y=0$.

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Nice example. I was just pondering this exact question whether adding the compactness assumption makes the claim true. By the way, if I'm not mistaken, altering your second example just a little bit: $\Gamma(x)=[x−1,x]\cup[−x,−x+1]$ for $x<0$, shows that the claim is false even if we assumed the values are compact and demanded only that $y$ is in the sets $\Gamma(x),x\in U$ (rather than in the interiors of those). – LostInMath Oct 8 '11 at 14:30
Martin, @LostInMath. Thanks! But I'm afraid I'm a bit slow today. Could you please mention the $x_0$ and $y$ used in your examples. – Jyotirmoy Bhattacharya Oct 8 '11 at 16:58
$x_0=0$ and any $y$ which is in the interior of $\Gamma(x_0)$ should work. (I hope I did not miss something.) – Martin Sleziak Oct 8 '11 at 17:12
@MartinSleziak I think taking $y=0$ in the second example will not work since $0$ is in the interior of $\Gamma(x)$ for all $x>-1$. – Jyotirmoy Bhattacharya Oct 8 '11 at 17:24
@JyotirmoyBhattacharya I've changed the second example - this is the function I had in mind. Sorry for my mistake. – Martin Sleziak Oct 8 '11 at 17:32