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)
Is this proof about metric space correct? (help)
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analysis