Subset of Normed Vector Space is Everywhere Dense iff Closure is Normed Vector Space/Necessary Condition
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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 $D$ be dense.
Then:
- $D^- = X$
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:
\(\ds X\) | \(=\) | \(\ds D \cup \paren {X \setminus D}\) | ||||||||||||
\(\ds \) | \(\subseteq\) | \(\ds D^-\) | ||||||||||||
\(\ds \) | \(\subseteq\) | \(\ds X\) |
Thus:
- $X = D^-$.
$\blacksquare$
Sources
- 2017: Amol Sasane: A Friendly Approach to Functional Analysis: Chapter $1$: Normed and Banach spaces