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

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.