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\noindent{\bf Problem 1. } Let $A$ be a subset of a metric space $(X,d)$. Show that the distance function to $A$, defined by $d(x,A):=\inf_{y\in A}d(x,y)$, is a continuous function of $x$ and that $d(x,A)=0$ iff $x\in\bar{A}$.

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{\bf Marking key.}

Continuity: 13/25.

Suppose $d(x,z)<\epsilon$. As $d(x,A)=\inf_{y\in A}d(x,y)$, there is some $y\in A$ such that $d(x,y)<d(x,A)+\epsilon$, and then $d(z,y)\leq d(z,x)+d(x,y)<\epsilon+d(x,A)+\epsilon=d(x,A)+2\epsilon$, and so $d(z,A)<d(x,A)+2\epsilon$. By symmetry also $d(x,A)<d(z,A)+2\epsilon$, and so $|d(x,A)-d(z,A)|<2\epsilon$. Hence when $z\to x$ we have that $d(z,A)\to d(x,A)$, so $d(-,A)$ is continuous.

Even better, given $x,z$, $d(x,A)=\inf_{y\in A}d(x,y)\leq \inf_{y\in A}d(x,z)+d{z,y}=d(x,z)+\inf_{y\in A}d(z,y)=d(x,z)+d(z,A)$, and by symmetry $d(z,A)\leq d(x,z)+d(x,A)$. So $|d(x,A)-d(z,A)\leq d(x,z)$ and $d(-,A)$ is continuous.

``Iff'': 12/25.

Deductions:

$(-2)$ used v.s. notation in a metric space.

$(-4)$ unexplained deduction of the $\inf$ inequality from the inequality for a specific point.

$(-4)$ assumed the existence of a minimizer.

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