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.
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 $$U_{3/4}(A)\cap B= [0,1]\times\{0\}=B.$$