Subset of Normed Vector Space is Everywhere Dense iff Closure is Normed Vector Space/Sufficient 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 $D$.

Let $D^- = X$.

Then $D$ is dense.

Proof
Let $X = D^-$.

We have to show, that for every $x \in X$ there is an open ball with an element from $D^-$.

We have that $X = D \cup \paren {X \setminus D}$.

Suppose $x \in X \setminus D$.

Then $x \in D^- \setminus D$.

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

Therefore, there is a sequence $\sequence {d_n}_{n \mathop \in \N}$ in $D$ which converges to $x$.

Thus:


 * $\forall \epsilon \in \R_{> 0} : \exists N \in \N : \norm {x - d_N} < \epsilon$

In other words:


 * $\ds d_N \in D \implies d_N \in \map {B_\epsilon} x$

Therefore:


 * $d_N \in D \cap \map {B_\epsilon} x$.

Suppose $x \in D$.

Let $\epsilon > 0$.

Then $x \in \map {B_\epsilon} x \cap D$.

From both parts and definition we conclude that $D^-$ is dense in $X$.