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Sep 1, 2013 at 16:25 comment added user35375 Sorry, I meant that $x\in M$, $x\not\in S$ but $3x\in M'$. Is it possible that $3x$ can be written as the linear combination of some vectors in $M\cap S$ but $x$ can't?
Sep 1, 2013 at 15:57 comment added user35375 I'm not following... To show that $M'$ is saturated in $M$, we need to show that if $ax\in M'$ then $x\in M'$. Why is this true? Why can't there be some $x\in M$, $x\not\in S$ but $3x\in M$?
Sep 1, 2013 at 12:23 history answered Lev Borisov CC BY-SA 3.0