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- Thread starter dray
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- Feb 7, 2012

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One way is easy: $S$ is closed in $X$. Therefore if $X$ is complete then so is $S$.Let $X$ be a complex normed space and $S$ be the unit sphere. Prove that $X$ is complete if and only if $S$ is complete.

Can anyone point me in the driection of a solution to this?

For the converse implication, suppose that $(x_n)$ is a Cauchy sequence in $X$. Then $(\|x_n\|)$ is a Cauchy sequence of real numbers and so has a limit $L.$ If $L=0$ then $x_n\to0$ and so $(x_n)$ converges. Thus we may assume that $L\ne0$ and hence that $\{\|x_n\|\}$ is bounded away from 0. With that assumption, show that $\bigl(x_n/\|x_n\|\bigr)$ is a Cauchy sequence in $S$. If $S$ is complete then that sequence has a limit $s$. Prove that $(x_n)$ converges to the limit $Ls$.