Subset of Normed Vector Space is Everywhere Dense iff Closure is Normed Vector Space

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Theorem

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

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

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


Then $D$ is dense if and only if $D^- = X$.


Proof

Necessary Condition

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:

\(\ds X\) \(=\) \(\ds D \cup \paren {X \setminus D}\)
\(\ds \) \(\subseteq\) \(\ds D^-\)
\(\ds \) \(\subseteq\) \(\ds X\)

Thus:

$X = D^-$.

$\Box$


Sufficient Condition

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 $X$.

$\blacksquare$


Sources

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