Proof $d(\mathbf{a},V) = 0 \iff \mathbf{a} \in \overline{V}$

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?

• 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). – AlephNull Feb 13 at 20:58

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