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I want to proof that

$\mbox p \in \overline A \iff d(p, A)= \inf\{d(p,a):a\in A\}=0$

My attempt for the first implication is as follows. We know that $\overline A= A \cup A'$, it would suffice to prove that for $a \in A$ or $a \in A'$ ($a$ is a limit point) the proposition is fulfilled; let's see for the accumulation point, for the other case I suppose it is analogous.

let $a \in A'$, note that $d(p, a)\geq 0$ let's see that strict inequality is not held.

Let $d(p, a) >0$, take $0<r<d(p, A)$ note that $a \notin B(p; r)$ because $d(p, a)\geq r$ for any $a \in A$, so $B(p; r)-\left\lbrace p \right\rbrace \cap A =\emptyset$, i.e. p is not an accumulation point, but this is a contradiction so that $d(p, a) =0$.

Now, I don't really know how to make the other implication. Any suggestions? any help is appreciated.

Martin Sleziak
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Wrloord
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    https://math.stackexchange.com/questions/385127/in-a-metric-x-d-prove-that-for-each-subset-a-x-in-bara-if-and-only-i – Caio Mar 29 '23 at 05:45
  • Note that $\overline{A}$ can be characterized as the set of points $a\in X$ of the ambient metric space $X$ such that there exists a sequence $\left(a_n\right){n\in\mathbb{N}}\subset A$ such that $\lim{n\to\infty}a_n = a \Longleftrightarrow \lim_{n\to\infty}d(a_n,a) = 0$. – Cartesian Bear Mar 29 '23 at 05:49

2 Answers2

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To prove the other implication, we want to show that if $d(p, A) = \inf {d(p, a): a \in A} = 0$, then $p \in \overline{A}$.

Let $d(p, A) = 0$. This means that for any $\epsilon > 0$, there exists an $a \in A$ such that $d(p, a) < \epsilon$. Since $\epsilon$ is arbitrary, we can say that for any $\epsilon > 0$, there exists an $a \in A$ such that $a$ is in the open ball $B(p; \epsilon)$. This implies that for any $\epsilon > 0$, we have $B(p; \epsilon) \cap A \neq \emptyset$. As a result, $p$ is a limit point of $A$, which means $p \in A'$.

Now we know that $p \in A'$, and since $\overline{A} = A \cup A'$, it follows that $p \in \overline{A}$.

So, we've shown that if $d(p, A) = 0$, then $p \in \overline{A}$. Combining this with your proof for the first implication, we have demonstrated that $p \in \overline{A} \Leftrightarrow d(p, A) = \inf {d(p, a): a \in A} = 0$.

Alon Yariv
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Assume that $d(p, A)= \inf\{d(p,a):a\in A\}=0$ but that $p \notin \overline{A}$. Then $p$ is in the open set which is the complement of $\overline{A}$, so some open ball of radius $r>0$ surrounds it and has empty intersection with $\overline A$, so $\inf\{d(p,a):a\in A\} \ge r$, which is a contradiction.

mcd
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