Subset of Normed Vector Space is Everywhere Dense iff Closure is Normed Vector Space/Necessary Condition

Theorem
Let $\struct {X, \norm {\, \cdot \,} }$ is a normed vector space.

Let $D \subseteq X$ be a subset of $X$.

Let $D^-$ be the closure of $D$.

Then $D$ is dense :
 * $D^- = X$

Proof
Let $x \in X \setminus D$.

Suppose $D$ is dense in $X$.

Then:


 * $\forall n \in N : \exists d_n \in D : d_n \in \map {B_{\frac 1 n}} x$

where $\ds \map {B_{\frac 1 n}} x$ is an open ball.

Let $\sequence {d_n}_{n \mathop \in \N}$ be a sequence in $D$.

Then:


 * $\forall n \in \N : \norm {x - d_n} < \frac 1 n$

Hence, $x$ is a limit point of $D$.

In other words, $x \in D^-$.

We have just shown that:


 * $x \in X \setminus D \implies x \in D^-$

Hence:


 * $X \setminus D \subseteq D^-$.

By definition of closure:


 * $D \subseteq D^-$

Therefore:

Thus:


 * $X = D^-$.