Showing that closed set is the preimage of $0$ of the distance function

Question

Let $(X,d)$ be a metric space. For a nonempty subset $A$ of $X$ we define the real valued distance function $\rho_A$ by $\rho_A(x) := \inf\{d(x,a) : a \in A\}$ for any $x \in X$.

Now it is quite intuitive, that $$A = \rho_A^{-1}(\{ 0 \})$$ whenever $A$ is closed. The inclusion $A \subseteq \rho_A^{-1}(\{0\})$ is trivial, but somehow I do not exactly know what to use to show the other inclusion. Any hint would be nice.

Have you tried to show that $\overline{A}=\rho_A^{-1}(0)$? – Maik Pickl Jan 13 '17 at 21:57 — Maik Pickl, Jan 13 '17 at 21:57
@MaikPickl I thought about it. – TheGeekGreek Jan 13 '17 at 21:58 — TheGeekGreek, Jan 13 '17 at 21:58

score 3 · Accepted Answer · answered Jan 13 '17 at 21:57

Suppose $\rho_A(x) = 0$, hence $$ \inf_{a \in A} d(x,a) = 0 $$ By definition of the infimum, given $n \in \mathbf N$, there is $x_n \in A$ with $$ d(x_n,a) < \frac 1n $$ Then $x_n \to a$. As $x_n \in A$, and $A$ is closed, we have $a \in A$.

Therefore $\rho_A^{-1}(\{0\}) \subseteq A$.

Showing that closed set is the preimage of $0$ of the distance function

1 Answers1