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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.

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    Have you tried to show that $\overline{A}=\rho_A^{-1}(0)$? – 2017-01-13
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    @MaikPickl I thought about it. – 2017-01-13

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