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Let $(X, d)$ be a metric space and $K \subset X$ be closed. Prove or disprove (providing a counterexample) that for every $x \in X$, there exists an $y \in K$ such that $\operatorname{dist}(x,K) = d(x,y)$.

Notation: $\operatorname{dist}(x,K)=\inf(y \in K: d(x,y))$.

I am not sure whether I understand the question correctly as by definition of the distance between a point and a set, it requires existence of some point $y \in K$. So, I am not sure what is the point of the question and why such point would not exist (as if would not it be contradictory to the definition of $\operatorname{dist}$)?

Andrés E. Caicedo
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Sanctuary
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  • Ok, assume $K$ not empty. Consider $\mathbb R$ and $K = (a,b)$ with $a < b$. Consider for simplicity $x < a$ then $d(x,K)= |x-a|$ but $a \notin K$ (because $K$ is open) and thus there is no $y \in K$ such that $d(x,y)=d(x,K)$. Thus, the fact that $K$ is closed is relevant. – Mauro ALLEGRANZA Jan 19 '20 at 14:52
  • In general this is not true even if $K$ is closed. Consider the space $(0,1)\cup {2}$ (as a subspace of the real line). Then $K=(0,1)$ is closed, $d(K,2)=1$ and for any $y\in K$ you have that $d(2,y)>1$. The missing ingredient is that for this to hold you need $X$ to be a closed metric space. – Shai Deshe Jan 19 '20 at 15:02
  • @ShaiDeshe thanks, I see where my misunderstanding came from now – Sanctuary Jan 20 '20 at 14:13

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A counterexample: Let $K$ be the set $\{ x \in \mathbb{Q}: x > \sqrt{2}\} \subset \mathbb{Q}$, with $\mathbb{Q}$ given the Euclidean metric: $d(x,y) = |x-y|$. K is closed, but there is no $y\in K$ such that $d(0,y) = dist(0, K)$.

wilkersmon
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