Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question

2 Answers 2

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 $y$.

share|improve this answer
    
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
1  
@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 $x_0=y=0$.

share|improve this answer
1  
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
1  
@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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.