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In a metric space $X$, let the nonempty compact set $K$ lie in the open set $\Omega$. If $F=X\setminus \Omega$ is nonempty, the distance to a set $d(x,F)$ is a continuous function of $x$: the triangle inequality gives $|d(x,F)-d(y,F)|\le d(x,y)$. Since $F$ is a closed set, this distance is strictly positive on $K$. The extreme value theorem attains its minimum on $K$, soIf $F$ is empty, the extended-real convention gives $d(K,F)=+\infty$. This result provides a single positive neighbourhood radius that stays inside $\Omega$ at every point of $K$. Compactness is essential: in the real line $K=\{1/n:n\ge 1\}\subset (0,2)$ has zero distance between two sets from $(-\infty ,0]\cup [2,\infty )$, although each point of $K$ has positive distance to a set from that complement.

Put $F={\mathbb{R}}^n\setminus \Omega$. This is a closed set. First assume $F$ is nonempty. The Heine-Borel theorem makes the nonempty bounded closed set $K$ a compact set. By the preceding parts, $x\mapsto d(x,F)$ is a continuous function and is strictly positive at every $x\in K$, because $K\subset \Omega$.

The extreme value theorem gives $x_{\ast }\in K$ attaining its minimum. ThereforeThe equality with the distance between two sets follows by taking the two infima successively. Crucially, compactness supplies a point attaining the minimum; positivity at each point alone would not make an arbitrary infimum positive. This is the principle that a compact subset of an open set has positive distance from its complement.

If $\Omega ={\mathbb{R}}^n$, its complement is empty. With the usual extended-real convention $\inf \varnothing =+\infty$, the stated inequality holds in that case too.