Let $R$ be a completed normed ring, eg Banach algebra. Suppose that $F$ is a free $R$module of infinite rank with a norm defined by the square root of sum of all norms of its components. If $F'$ is a dense submodule of $F$, does $F'$ contain a basis of $F$?
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This is not so. Let $R=\mathbb C$ and consider the module $F$ to be the finite sequences in $\ell^2({\mathbb C})$. Let $F'$ be the submodule of all sequences $(z_n)$ with $\sum_nz_n=0$. Then $F'$ is dense in $F$. To see this, let $(w_n)$ be in $F$ and let $a=\sum_nw_n$. Let $N$ be a natural number such that $w_n=0$ for $n\ge N$. For $j\in\mathbb N$ let $z(j)$ be the element of $F'$ given by $z(j)_n=w_n$ if $n < N$ and $z(j)_n=\frac 1j$ if $N\le j\le N+j1$. Then $\ z(j)w\=\sqrt{j a^2/j^2}=\sqrt{a^2/j}$ tends to zero, so $F'$ is dense in $F$. 

