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let $M$ a metric space, $a\in M$, $r>0$ and $F = M − B(a, r)$. Show that if $d(x, F ) = 0$, then $x ∈ F $. Proof: let's get a contradiction assuming that $x\notin F$. We have that: $|d(x,F)-d(a,F)|\leqslant d(x,a)$, by hypothesis, $d(x,F)=0$, then, $|d(x,F)-d(a,F)|=|0-d(a,F)|=d(a,F)\leqslant d(x,a)$, and now from that, $d(x,a) and $d(a,F)>r$ it turns that: $r. Contradiction!, therefore $x\in F$.

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It's a bit messy and hard to follow. You state some things, but where did you get them?

  1. $|d(x,F)-d(a,F)|\le d(x,a)$ // it may need some explanation how this is coming from the def. $d(x,F)=\inf_{y\in F} d(x,y)$, especially because this hides the essence of this problem.
  2. $d(x,a) // ok, finally I got it: it is because of the assumption $x\notin F$ i.e. $x\in B(a,r)$.
  3. $d(a,F)>r$ // I think it should be $\ge r$. In $\mathbb R^n$ it equals to $r$.

Else, it's good. But can be simplified, no need for contradictionary logic:

So, by 1. we get $d(x,a)\ge d(a,F)$, by 3., $d(a,F)\ge r$, and these together lead to $x\notin B(a,r)$.