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Let $n\in\mathbb{R}^n$ and $K\subseteq\mathbb{R}^n$ a compact subset. Let $\{U_i\}_{i\in I}$ be a locally finite collection (every point has a neighborhood, which intersects only finitly many $U_i$) of open subsets $U_i\subseteq\mathbb{R}^n$, which are disjoint from $K$, i.e. $K\cap U_i=\emptyset$. Let furthermore $K_i\subseteq U_i$ be a compact subset for each $i\in I$. Since $K$ is disjoint from $U_i$, it is disjoint from $K_i$ for each $i$.

Does their exist a neighborhood of $K$, which is disjoint from all the $K_i$?

My thoughts so far:

Since $K$ and $K_i$ are compact and disjoint, they have a proper distance $0<\varepsilon_i=dist(K,K_i)$, but as $I$ can be inifite, $\epsilon_i$ might get arbitrary small. (?)

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Your intuition that this cannot be done is correct.

Take for instance $K=\{0\}$, and $U_i=\left(\frac{1}{i+1},\frac{1}{i}\right)$. Since $K_i\subset U_i$, we have $x<\frac{1}{i}$ for all $x\in K_i$ and all $i\in\mathbb{N}$.

If $U$ is an open neighborhood of $K$, then there certainly exists some integer $i$, such that $\left(0,\frac{1}{i}\right)\subset U$, but then $U$ also contains all elements of $K_i$.

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