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

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