Consider a Hilbert-Space $H$ and two closed convex sets $A,B\subseteq H$ with non-empty intersection, i.e. $A\cap B\neq \emptyset$.

I am wondering if the following set inclusion holds true: $$ U_{\varepsilon}(A)\cap B\subseteq U_{\varepsilon}(A\cap B),$$

where for $M\subseteq H$, we define $U_\varepsilon(M):=\{x\in H\mid d(x,H)<\varepsilon\}$.

Let me give you some of my thoughts regarding the assumptions above:

  • The assumption $A\cap B\neq \emptyset $ is obviously necessary
  • Without the additional assumption of convexity, one may also construct a counterexample.

Here is why I think the assertion holds under the assumptions made above:

Let $x\in U_{\varepsilon}(A)\cap B$, we then have $\varepsilon > d(x,A)=d(x,A\cap B)$, where I am unsure about the last equality, but so far I have failed to prove it.

Thus the question is if the statement is true or if there is a counterexample.

  • $\begingroup$ The inclusion does not hold e.g. in $\mathbb R^2$ for $B=[0,1]\times\{0\}$ and $A=\{(t,t): 0\le t\le 1\}$. Draw a picture to see this. $\endgroup$ – Jochen Jun 6 '18 at 14:49
  • $\begingroup$ Thank your very much. If we adjust your example slightly and take $A=\{(t, 0.5*t) : 0\leq t\leq 1\}$. It seems even clearer that the inclusion does not hold.. Thank you very much for the input. $\endgroup$ – Tsuyoi Jun 6 '18 at 15:28

In the comments above Jochen provided a counterexample. I just spell it out here to close the question.

Taking $H=\mathbb{R}^2$ and $B=[0,1]\times\{0\}$ and $A=\{(t,0.5t): 0\leq t\leq 1\}$. Then set $\varepsilon = 3/4$, we get on the one hand $$ U_{3/4}(A\cap B)=U_{3/4}(0)$$ and on the other hand $$U_{3/4}(A)\cap B= [0,1]\times\{0\}=B.$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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