How to prove that subspace of Banach space is also Banach space?
I believe that's only true if subspace is closed. However, I don't understand why we can't take non-closed subspace. Is there any counterexample of non-closed subspaces that will not be Banach although space was Banach?
Best Answer
It is true that if $X$ is a Banach space and $F\subseteq X$ is a subspace. Then $F$ is a Banach space with the restricted norm of $X$ into $F$ if and only if $F$ is closed in $X$.
A counterexample is the following.
Consider $X=\ell_{\infty}$ with the supremum norm given by $$||x||_{\infty}=\sup\bigl\{|x(n)|:\,n\in \mathbb{N}\bigr\},$$ for every $x=(x(n))_{n}$ in $\ell_{\infty}$. Now, consider as $F$ the space $c_{00}$ of eventually zero sequences, i.e. $$c_{00}=\bigl\{x\in \ell_{\infty}:\, \exists\, m \in \mathbb{N}\, \text{such that}\, x(n)=0\, \text{for every}\, n\geq m\bigr\}.$$ Then, $c_{00}$ is not a Banach space, for the sequence $(x_k)_k$ given by $$x_k=\biggl(1,\, \frac{1}{2},\,...,\,\frac{1}{k},\,0,\,0,\,...\biggr)$$ is a Cauchy sequence since $||x_k-x_m||_{\infty}=1/k$, for every $k>m$ but not convergent in $c_{00}$. Indeed, for the last claim observe that $(x_k)_k$ converges to the element $$x=\biggl(1,\,\frac{1}{2},\,...,\,\frac{1}{k},\,\frac{1}{k+1},\,...\biggr)$$ in $\ell_{\infty}$ which is not in $c_{00}$.
Edit: Why $(x_k)_k$ is not convergent on $c_{00}$?
Let $x$ be the point in $\ell_{00}$ with $x=(1,1/2,...,1/n,...)$. Then, observe that for every $k$, $$||x-x_k||_{\infty}=||(0,0,...,\frac{1}{k+1},\frac{1}{k+2},...)||_{\infty}=\frac{1}{k+1}.$$ Hence, $||x-x_k||_{\infty}\to 0$ as $k\to \infty$. If ($x_k)_k$ where convergent on $c_{00}$ to some point $y$ we would have $$||x-y||_{\infty}\leq ||x-x_k||_{\infty}+||x_k-y||_{\infty}\to 0,$$ as $k\to \infty$. In other words, $x=y$. But $x$ does not belong to $c_{00}$.