# 1-Sequence Space is Proper Subset of 2-Sequence Space

## Theorem

Let $\ell^1$ and $\ell^2$ be the $1$-sequence space and $2$-sequence space respectively.

Then $\ell^1$ is a proper subset of $\ell^2$.

## Proof

### $\ell^1$ is a subset of $\ell^2$

Let $\sequence {x_n}_{n \mathop \in \N}$ be a sequence in $\ell^1$.

By definition of $1$-sequence space:

$\ds \sum_{n \mathop = 0}^\infty \size {x_n} < \infty$
$\ds \lim_{n \mathop \to \infty} \size {x_n} = 0$

By definition of convergent sequence:

$\exists N \in \N : \forall n > N : \size {x_n} < 1$

Furthermore:

$\forall \size {x_n} < 1 : \size {x_n}^2 < \size {x_n}$
$\ds \sum_{n \mathop = 0}^\infty \size {x_n}^2 < \infty$

Hence:

$\sequence {x_n}_{n \mathop \in \N} \in \ell^2$

That is:

$\ell^1 \subseteq \ell^2$

$\Box$

### $\ell^2$ is not a subset of $\ell^1$

Let $x := \tuple {1, \dfrac 1 2, \ldots, \dfrac 1 n, \ldots}$ with $n \in \N_{>0}$.

From Basel Problem:

$\ds \sum_{n \mathop = 1}^\infty \frac 1 {n^2} < \infty$
$\ds \sum_{n \mathop = 1}^\infty \frac 1 n$

Hence:

$x \in \ell^2$

but:

$x \notin \ell^1$

That is:

$\ell^2 \nsubseteq \ell^1$

$\Box$

By definition, $\ell^1$ and $\ell^2$ are not equal.

Therefore:

$\ell^1 \subsetneqq \ell^2$

$\blacksquare$