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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.

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

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