Subset of Normed Vector Space is Everywhere Dense iff Closure is Normed Vector Space
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
- 2017: Amol Sasane: A Friendly Approach to Functional Analysis ... (previous) ... (next): $\S 1.4$: Normed and Banach spaces. Sequences in a normed space; Banach spaces