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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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The square root of the sum is not a norm. You need the square root of the sum of squares of norms. Next if $F'$ contains a basis of $F$, it is equal to $F$. –  doug Jun 21 '12 at 11:00
Yes. Thanks for your comments. –  yeshengkui Jun 23 '12 at 8:14

1 Answer 1

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 $N\le j\le N+j-1$. Then $\| z(j)-w\|=\sqrt{j |a|^2/j^2}=\sqrt{|a|^2/j}$ tends to zero, so $F'$ is dense in $F$.

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Thanks for your answer. That's nice. I just realizes for any proper dense submodule $F'$ is a counterexample. –  yeshengkui Jun 23 '12 at 8:15

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