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Let $M$ be a metric space with the metric $d$, $X$ a subset of $M$, and $r>0$. Also, $B(x;r)$ is the open ball centered at $x$ with radius $r$.

Defining $\begin{equation*}\begin{aligned} B(X;r)=\bigcup_{x \in X}B(x;r), d(a,X)= \underset{x \in X}{\operatorname{inf}}d(a,x)\end{aligned}\end{equation*}$,

prove $\begin{equation*}d(a,X)=\operatorname{inf}\text{{$r>0; a \in B(X;r)$}}.\end{equation*}$

amWhy
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Marcelo
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  • What are your thoughts? What have you tried? Where are you stuck? – Robert Israel Mar 17 '24 at 17:27
  • Given $r>0$ and $a \in B(X;r)$, $a$ must be in some ball, so $\exists,x \in X$ such that $a\in B(x;r)$. By the open ball definition, $d(a,x)<r$. I've concluded so far that each element of ${r>0;a \in B(X;r)}$ is greater than an element of ${d(a,x); x\in X}$. I have no idea how to prove these sets have the same infimum. – Marcelo Mar 17 '24 at 19:46

1 Answers1

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If $t < d(a,X)$, that says $t < d(a,x)$ for every $x \in X$, so $a \notin B(x;t)$ for such $x$, and thus $a \notin B(X;t)$, so if $a \in B(X;r)$ we must have $t < r$. Thus $t \le \inf\{r > 0;\; a \in B(X;r)\}$. This says $d(a,X) \ge \inf \{r > 0; \; a \in B(X;r)\}$ (i.e. if this were not true, you could take $t$ so $d(a,X) < t < \inf \{r > 0; \; a \in B(X;r)\}$).

On the other hand, if $t > d(a,X)$, there must be $r$ with $0 < r < t$ and $d(a,x) = r$, and then $a \in B(x;r)$ so $a \in B(X;r)$. Thus $t \ge \inf\{r > 0;\; a \in B(X;r)\}$. This says $d(a,X) \le \inf \{r > 0; \; a \in B(X;r)\}$.

Combining the two, we must have $ $d(a,X) = \inf {r > 0; ; a \in B(X;r)}$.

Robert Israel
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