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Notes: $d$ is the distance $(|| \cdot ||)$, $\mathbf{a}$ represents the singleton $\{\mathbf{a}\}$, $\overline{V}$ is the closure of $V \ne \varnothing$.

My attempt:

$\Rightarrow$: $d(\mathbf{a},V) = 0 \iff (\forall \varepsilon>0)(\exists \mathbf{v}\in V)(|| \mathbf{a}-\mathbf{v}|| < \varepsilon).$ We want to prove that $\mathbf{a}$ is an adherent point of $V$. So we have to show that there exists an $R >0$ such that $B(\mathbf{a},R) \cap V \ne \varnothing$. Now we now that $d(\mathbf{a},\mathbf{v}) < \varepsilon$ for a certain $\mathbf{v} \in V$ (1). This implies that $\mathbf{v} \in B(\mathbf{a},\varepsilon)$ (2). From (1) and (2) we get that $B(\mathbf{a},\varepsilon) \cap V \ne \varnothing$, and thus $\mathbf{a}$ is an element of the closure $\overline{V}$ of $V$.

$\Leftarrow$: suppose $\mathbf{a} \in \overline{V}$. By definition there is an $R > 0$ such that $B(\mathbf{a},R) \cap V \ne \varnothing$. Take an element $\mathbf{v} \in V$ from this intersection. This means that $d(\mathbf{a},\mathbf{v}) < R = \varepsilon$. Since $\mathbf{v}$ was chosen randomly, we can conclude that $d(\mathbf{a},V) = 0$.

The proofs seem very similar. Can I unify them in a way so that a sequence of $\iff$'s could be used?

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  • $\begingroup$ A good way of thinking about this to see both implications at once is to say "$d(a,V)=0$ iff there are points in $V$ that come arbitrarily close to $a$ iff there is a sequence in $V$ that converges to $a$." In such statements seeing the underlying simple geometric picture is more important than the technicalities of the proof (which easily follow from the 'qualitative proof' I gave anyway). $\endgroup$ – AlephNull Feb 13 at 20:58
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The intuition behind your attempt feels right. There are, however, just a few mistakes here and there. In your second implication for instance, it is not just that "there is an $R >0$ such that $B(a,R) \cap V \neq \emptyset$;" we actually have $B(a,R) \cap V \neq \emptyset$ for every $R>0$. It is this fact that forces us to conclude with $\text{d}(a,V)=0$; if it is anything else other than $0$, then we will arrive at a contradiction by the fact mentioned.

As to your question:

$$\text{d}(a,V) = 0 \iff (\forall \varepsilon >0)(\exists v\in V)(||a-v||<\varepsilon) \iff (\forall \varepsilon >0)(\exists v\in V\cap B(a,\varepsilon)) \\ \iff a\in V\cup V'=\overline{V}.$$

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