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

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

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


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

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

By definition:

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

By Terms in Convergent Series Converge to Zero:

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

By definition of convergent sequence:

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


$\forall \size {x_n} < 1 : \size {x_n}^2 < \size {x_n}$

By comparison test:

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

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

In other words, $\ell^1 \subseteq \ell^2$.


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

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

From Basel Problem:

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

From Harmonic Series is Divergent:

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


Hence, $x \in \ell^2$, but $x \notin \ell^1$.

In other words, $\ell^2 \not \subseteq \ell^1$.


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

Therefore, $\ell^1 \subsetneqq \ell^2$.