# Show that $\cap$ may not distribute over $+$

Let $L$, $M$, and $N$ are subspace of some vector space. Show that following equation need not be true.

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

I am trying to prove it by using 'if belongs to LSH then show that it belongs to RHS' argument. I start with the assumption that $v_1 \in M$ and $v_2 \in N$; therefore $v_1 + v_2 \in M+N$ and $v_1 + v_2 \in L$ i.e. $v_1 + v_2 \in LHS$. Now if $v_1 + v_2$ is to belong to RHS, it must happen that $v_1 (v_2) \in (L \cap M)$ and $v_2 (v_1) \in (L \cap N)$ (right?); therefore $v_1$ and $v_2$ both must belong to $L$. This may not be the case.

Is this proof correct? I am confused because the equation looks true and I can not cook-up an example which shows that this may not be the case.

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The only method to show that something "may not be the case" or "is not generally true" is to come up with a counterexample. – Julian Kuelshammer Nov 11 '12 at 10:18
Well, actually, trying to write a proof is not a bad method to identify a counter-example. – Phira Nov 11 '12 at 10:57
@Phira: Yes, but I think Julian's point is that "trying to write a proof" only works when it actually produces a counterexample. If it doesn't then it doesn't prove anything either way. – Henning Makholm Nov 11 '12 at 14:47

Let $M=${$(x,0)|x∈R$} and $M=${$(0,y)|y∈R$} and $L=${$(a,a)|a∈R$} Clearly $L∩M=${(0,0)}$=L∩N$. However, $L∩(M+N)=L$