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I am struggling with this question from Halmos's text, please ignore the imperative language.

"Suppose that $L, M$ and $N$ are subspaces of a vector space. Show that the equation

$L \cap (M + N) = (L \cap M) + (L \cap N)$

is not necessarily true.

Since each of these subspaces has origin in them, clearly there intersections could not be empty. I wasn't able to formulate an example where this result did n't hold. Any help would be highly appreciated.

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    @Leandro Thanks for that i suspected that was wrong. I think the correct result is L ∩ (M + (L ∩ N)) = (L ∩ M) + (L ∩ N). Although being new to proofs i am unsure how to prove that one too. Any tips on how to approach such a proof would be much appreciated.2012-02-08

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Work for example in $\mathbb{R}^2$.

Let $M$ be the set of multiples of the vector $(1,0)$, let $N$ be the set of multiples of $(0,1)$. It's your turn, you can choose $L$.

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    @Hardy: Definitely $L\cap M$ just contains the $0$ vector. Think of it geometrically. You can think of $L$ as all points on the line $y=x$, and $M$ as the points on the $x$-axis. The only point they have in common is the origin. And I checked the standard usage of $U+V$, when both are subspaces $W$. It is the set of all points of $W$ of the shape $u+v$, where $u\in U$ and $v\in V$.2012-02-08