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I arrived at the following problem during the day. For $\iota\in I$, let $A_\iota\subseteq\mathbb R^d$ be non-empty and closed. For $\varepsilon>0$ let $B_\varepsilon(A_\iota)=\bigcup_{x\in A_\iota}B_\varepsilon(x).$ Do we have $$\bigcap_{\varepsilon>0}\bigcup_{\iota\in I}\overline{B_\varepsilon(A_\iota)}=\bigcup_{\iota\in I}A_\iota$$ and if not, is there a chance that we have $$\bigcap_{\varepsilon>0}\overline{conv}\Bigl(\bigcup_{\iota\in I}\overline{B_\varepsilon(A_\iota)}\Bigr)=\overline{conv}\bigl(\bigcup_{\iota\in I}A_\iota\bigr)?$$

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3 Answers 3

up vote 1 down vote accepted

The first statement doesn't hold when $d=1$, $I=\Bbb N^*$, $A_{i}=\{i^{-1}\}$: $x$ is in the LHS but not in the RHS.

For the second statement, fix $\varepsilon>0$ and $x$ in the LHS. Then we can find an integer $N$ and $c_j\in [0,1],x_j\in\Bbb R^d,j\in [N]$ such that $\lVert x-\sum_{j=1}^Nc_jx_j\rVert<\varepsilon$, where $x_j\in\bigcup_{\iota\in I}\overline{B_{\varepsilon}(A_{\iota})}$ and $\sum_jc_j=1$. Now, let $i_j\in I$ and $y_j\in A_{i_j}$ such that $\lVert x_j-y_j\rVert<\varepsilon$. We have $$\lVert x-\sum_{j=1}^Nc_jy_j\rVert\leqslant 2\varepsilon,$$ and $\sum_{j=1}^Nc_jy_j\in \operatorname{conv}\bigcup_{\iota\in I}A_{\iota}$.

Note that we didn't use the fact that we work in $\Bbb R^d$ (we can replace it by any normed space), and that $A_i$ are closed.

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$A_\iota = \{\frac{1}{\iota}\}$ is a counterexample to the first statement ($I = \mathbb{N}$). For all $\epsilon > 0$, $0 \in \bigcup_{i\in I} \overline{B_\epsilon(A_\iota)}$, hence $0$ is a member of the left-hand side. Yet $0$ lies in none of the $A_\iota$, hence $0$ isn't a member of the right-hand side.

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Here is a more geometric argument for the second part:

Let $L = \cap_{\epsilon>0} \overline{\mathbb{co}}( \cup_{i \in I} \overline{B_\epsilon(A_i)})$, $R = \overline{\mathbb{co}} (\cup_{i \in I} A_i)$.

We want to show $L=R$.

For all $\epsilon >0$, we have $\cup_{i \in I} A_i \subset \cup_{i \in I} \overline{B_\epsilon(A_i)}$, hence $\overline{\mathbb{co}} (\cup_{i \in I} A_i)\subset \overline{\mathbb{co}}( \cup_{i \in I} \overline{B_\epsilon(A_i)})$, and since this is true for all $\epsilon>0$, we have $R \subset L$.

Now suppose $x_0\notin R$. Then $R$ and $\{x_0\}$ are closed, convex sets, so by the Hahn Banach theorem there is a linear functional $\phi$ that separates the sets, that is, there exists $\gamma_1, \gamma_2$ such that $\phi(x_0) < \gamma_1 < \gamma_2 < \phi(y)$ for all $y \in R$.

Now suppose $y \in B_\epsilon(A_i)$, then for some $a \in A_i$, we have $\|y-a\| < \epsilon$. Since $|\phi(y)-\phi(a)| \leq \|\phi\| \|y-a\| \leq \epsilon \|\phi\|$, we have $\phi(y) \geq \phi(a) - \epsilon \|\phi\|$. We can choose $\epsilon$ small enough so that $\phi(x_0) < \gamma_1 < \phi(y)$, for all $y \in B_\epsilon(A_i)$.

Note that the $\epsilon$ only depends on $\|\phi\|, \gamma_1$ and not on $i \in I$. Consequently, $B_\epsilon(A_i) \subset C = \phi^{-1}[\gamma_1,\infty)$ for all $i\in I$. Note that $C$ is a closed convex set, hence we have $\overline{B_\epsilon(A_i)} \subset C$ for all $i\in I$, and hence $\cup_{i \in I} \overline{B_\epsilon(A_i)} \subset C$, and again, since $C$ is closed and convex, we have $\overline{\mathbb{co}}( \cup_{i \in I} \overline{B_\epsilon(A_i)}) \subset C$, from which it follows that $L \subset C$. Since $\phi(x_0) < \gamma_1$, it follows that $x_0 \notin L$. Hence $L \subset R$.

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