Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

$\begin{equation} L \cap (M + N) = (L \cap M) + (L \cap N) \end{equation}$.

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.

share|cite|improve this question
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
up vote 3 down vote accepted

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, $L∩(M+N)=L$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.