Is $\ell_0$ space a Banach space?

Question

$\ell_0$ space contains all infinite sequence which only has finite nonzero terms. Could anyone tell me whether it is a Banach space? Is it possible for us to find a Cauchy convergent series which is not convergent in this space?

Answer 1

Notice that $\dim \ell_0 = \aleph_0$.

It is known that the algebraic dimension of any infinite-dimensional Banach space is uncountable.

Therefore $\ell_0$ cannot be a Banach space with respect to any norm.

Answer 2

Your vector space has Hamel dimension $\aleph_0$. Every infinite-dimensional Banach space has Hamel dimension ${}\ge 2^{\aleph_0}$. So there is no norm on your vector space making it a Banach space. See Let $X$ be an infinite dimensional Banach space. Prove that every Hamel basis of X is uncountable.

Answer 3

It depends from the norm that you are fixing on it. For example you have that $l_0\subset l^\infty$ and so you can consider on $l^0$ the norm induced by $l^\infty$. In this norm the space cannot be a Banach space because $l^\infty$ is a Banach Space and so every his subset is a Banach Space (with respect to the norm induced by it) if and only if is closed in $l^\infty$ but in our case the subset $l_0$ is not closed in $l^\infty$ because for example if you define $x_n(i):=0$ if $i>n$ and $\frac{1}{i}$ otherwise than the succession $\{x_n\}_n\subset l_0$ it is convergent to $x(i)=\frac{1}{i}$ in $l^\infty$ that is not in $l_0$.

Now you can think to find a $p \geq 1$ such that $l_0$ is closed in $l^p$ (oviously you have always that $l_0\subset l^p$ ) because $l^p$ is a Banach Space and so $l_0$ is a Banach Space (with respect to the norm induced by $l^p$) but it is always false. Infact for any $p\geq 1$ you can define $x_n(i)=0$ if $i>n$ and $\frac{1}{i^2}$ otherwise. In this case you have that the sequence $\{x_n\}_n\subset l_0$ is convergent in $l^p$ norm to $x(i)=\frac{1}{i^2}$ that is not in $l_0$

You can prove that is not possible fix a norm on $l_0$ such that it is a Banach Space with respect to that norm because every Banach Space has Hamel dimension at least $2^{\aleph_0}$