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The distance between a point $a \in \mathbb{R}$ and a set $X \subset \mathbb{R}$ is defined as $$d(a,X) := \inf\{|x-a|: x \in X\}.$$ How to prove if $X$ is closed, then there is a $b \in X$ such that $d(a,X) = |b-a|$?

I've constructed a decreasing sequence converging to $d$ as follows: Given $r > d(a,X)$, there is a $x \in X$ such that $|x-a| < r$. Repeating the process with $r_{n+1} := \frac{d+r_n}{2}$ we get the inequality:

$$d \leq |x_n-a| < r_n$$

It's easy to prove that $r_n \mapsto d$, and therefore $|x_n-a| \mapsto d$. If i could show the set $A := \{|x-a|: x\in X\}$ is closed, the result would be immediate. This is somehow my second question, is true that for every closed set $X$, the set $|X| := \{|x|: x\in X\}$ is closed?

Be free to contribute alternative proofs, i would appreciate.

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    Anyone know if this is true in any metric space? That is, for metric space $X$ with metric $d$ and $E \subseteq X$ for $x \in X$ define the distance from $x$ to $E$ by $\Delta(x, E) = \inf_{z \in E} d(x,z)$. It seems like it should be but the proofs below depend on the Heine–Borel Theorem as it applies specifically to $\mathbb{R}$. EDIT: A simple search just answered my question in the negative, see [here](http://math.stackexchange.com/questions/938367/distance-between-a-point-and-a-closed-set-in-metric-space). I'll just leave this here though in case anyone else coming across this is curious2015-12-11

2 Answers 2