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I'm pretty sure it's not closed under vector addition but verification would be appreciated.

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    If by $x1$ you mean the first coordinate of $x$, then the set consists actually of those vectors whose all other coordinates except possibly the first are zero.2011-10-18
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    Look at the case $n=2$ for concreteness. Look at the vector $(a,b)$. To say its norm is $|a|$ is to say that $a^2+b^2=a^2$, so it's a fancy way of saying $b=0$. For sure it is a subspace! Argument for general $n$ is the same.2011-10-18

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