Past exam of the algebra, analysis and differential equations 1 course of the University of Oxford / 2012 / 4

A subset $\Omega \subseteq {\mathbb{R}}^n$ is an open set if every $x\in \Omega$ admits a radius $r>0$ such that the open ballis contained in $\Omega$. A subset $F\subseteq {\mathbb{R}}^n$ is a closed set if its complement ${\mathbb{R}}^n\setminus F$ is an open set. Equivalently, $F$ contains the limit of every convergent sequence of its points.

The distance to a set is nonnegative and finite here: any fixed $y_0\in F$ gives $0\le f(x)\le |x-y_0|$. For every $y\in F$, the triangle inequality givesTaking the infimum over $y\in F$ yields $f(x)\le |x-x^{\prime }|+f(x^{\prime })$. Interchanging $x$ and $x^{\prime }$ provesThus $f$ is a Lipschitz function with Lipschitz constant one, so it is continuous. In the epsilon-delta definition of continuity, choose $\delta =\epsilon$.

If $x\in F$, using $y=x$ in the infimum gives $0\le f(x)\le 0$, so $f(x)=0$.

Conversely, if $f(x)=0$, the definition of the infimum allows a choice $y_j\in F$ with $|x-y_j|<1/j$. The sequence $y_j$ is therefore a convergent sequence with limit $x$. Since $F$ is a closed set, its limit $x$ belongs to $F$. HenceClosedness matters: without it, zero distance to a set characterizes its closure, which can contain additional points.

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.