Space of Almost-Zero Sequences is not Closed in 2-Sequence Space
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Theorem
Let $\struct {\ell^2, \norm {\, \cdot \,}_2}$ be the normed 2-sequence vector space.
Let $\struct {c_{00}, \norm {\, \cdot \,}_2}$ be the normed vector space of almost-zero sequences.
Then $\struct {c_{00}, \norm {\, \cdot \,}_2}$ is not closed in $\struct {\ell^2, \norm {\, \cdot \,}_2}$.
Proof
Let $\sequence {x_n}_{n \mathop \in \N}$ be a sequence in $c_{00}$ such that:
- $\ds x_n := \tuple {1, \frac 1 2, \ldots \frac 1 n, 0, \ldots}$
Let $\ds x := \tuple {1, \frac 1 2, \ldots, \frac 1 n, \ldots}$ with $n \in \N_{>0}$.
We have that $x \in \ell^2 \setminus c_{00}$ where $\setminus$ denotes set difference.
Then:
\(\ds \norm {x_n - x}_2^2\) | \(=\) | \(\ds \sum_{k \mathop = n \mathop + 1}^\infty \frac 1 {k^2}\) | Definition of $p$-Norm | |||||||||||
\(\ds \) | \(<\) | \(\ds \sum_{k \mathop = n \mathop + 1}^\infty \frac 1 {k \paren {k - 1} }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \sum_{k \mathop = n \mathop + 1}^\infty \paren {\frac 1 {k - 1} - \frac 1 k}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 n\) | Definition of Telescoping Series |
Pass the limit $n \to \infty$
Then:
- $\ds \lim_{n \mathop \to \infty} \norm {x_n - x} = 0$
Hence, $\struct {c_{00}, \norm {\, \cdot \,}}$ does not contain its limit points.
By definition, it is not closed.
$\blacksquare$
Sources
- 2017: Amol Sasane: A Friendly Approach to Functional Analysis ... (previous) ... (next): Chapter $\S 1.4$: Normed and Banach spaces. Sequences in a normed space; Banach spaces