Subset of Normed Vector Space is Everywhere Dense iff Closure is Normed Vector Space/Sufficient 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 $D^- = X$.
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 $X$.
$\blacksquare$
Sources
- 2017: Amol Sasane: A Friendly Approach to Functional Analysis: Chapter $1$: Normed and Banach spaces