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?
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 $$\ell_0 = \{ (x_n) \mid x_n = 0 \text{ for all but finitely many } n \} .$$