prove $(V1⊕...⊕Vk)∩(W1⊕...⊕Wk)=(V1∩W1)⊕...⊕(Vk∩Wk)$

Question

I was thinking an hour over this question and I really didn't make any progress. Though to ask here and maybe get a hint ;) .

Question : if $V=U_1⊕...⊕U_k$ and for each $i=1,...,k$ given $V_i,W_i ⊆ U_i$.

Prove : $(V_1⊕...⊕V_k)∩(W_1⊕...⊕W_k)=(V_1∩W_1)⊕...⊕(V_k∩W_k)$

I though to make this proof as in Discreme math when you assume $v∈$ one side and prove $v∈$ the second side so you prove $⊆$ from both sides and it yelds $\rightarrow$ $=$ symbol but I failed to do so.

I got stuck in the middle of the way in the both side of the proof.

any hints ideas please ? Thank you and have a nice day!

Answer 1

I think I got the answer but I'm not sure. First let's notice that because $Vi,Wi⊆Ui$.

Intersection between this two can only occur it their index is the same (because $V_1$ can't intersect with $W_2$ because $U1⊕...⊕Uk$, $\rightarrow$ $U_1∩...∩U_i={0}$.

Lets assume $v∈(V_1⊕...⊕V_k)∩(W_1⊕...⊕W_k)$

then $v∈(V_1⊕...⊕V_k)$ And $v∈(W_1⊕...⊕W_k)$ and because intersection can only occur between the same index $\Rightarrow$ $v∈(V_1∩W1)⊕...⊕(V_k∩W_k)$.

Second side is pretty much the same , is it ok to prove like that ?