# Is $\ell^1$ a Banach space which is uncountably infinite dimensional?

We know that $$ℓ^1=\left\{(x_n)_n :\sum\limits_{n=0}^\infty |x_n|<\infty \right\}$$ with the usual norm $$||(x_n)_n||=\sum\limits_{n=0}^\infty |x_n|$$ is a $$\bf Banach$$ space which is infinite dimensional.

We can easily check that the set $$\{e_n\}_0^\infty$$ such that: $$e_n=(0,0,\cdots,0,\underbrace{1}_{n^{th}},0,\cdots)$$ is a set of linearly independent vectors in $$ℓ^1$$.

But there is a theorem saying that if a $$V$$ is a Banach space then either $$V$$ is finite dimensional or $$\bf uncountably$$ infinite dimensional.

But we see that the set $$\{e_n\}_0^\infty$$ is countable.

What I did not understand?

• $e_n$ is not an "algebraic" or Hamel basis of the space, meaning not every vector is a finite linear combination of the vectors $e_n$. Any Hamel basis of an infinite dimensional Banach space has to be uncountable. Commented Mar 10, 2020 at 22:16
• $\{e_1\}$ is also a set of linearly independent vectors in $\ell^1$. Does that mean $\ell^1$ is $1$-dimensional? Commented Mar 10, 2020 at 22:17

The set $$B = \{e_n\}$$ is linearly independent, but not a basis of the vector space $$\ell^1$$. In fact, the set of all linear combinations of elements of $$B$$ is $$\ell_0 = \{ (x_n) \mid x_n = 0 \text{ for all but finitely many } n \} .$$
• Take $x_n = 1/n^2$. Then $x_n \ne 0$ for all $n$ and $\sum \lvert x_n \rvert = \pi/6$. Commented Mar 10, 2020 at 23:46