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Let $V_{1,2,3}$ be Vectorspaces. We want to show the equation above.

My attempt: \begin{align*} (V_1 \cap V_3)+(V_2 \cap V_3) & =\left \{ v: \exists u_1 \in V_1 , \ u_2 \in V_2 \cap V_3, \ v = u_1 + u_2 \right \} \\ & = \left \{v : \exists u_1 \in V_1, \ (u_2 \in V_2 \wedge u_2 \in V_3), \ v = u_1 + u_2 \right \} \\ & = \left \{ v : \exists u_1 \in V_1, \ u_2 \in V_2, \ v = u_1 + u_2 \right \} \\ & \qquad \qquad \cap \left \{ v: \exists u_1 \in V_1, \ u_2 \in V_3, \ v = u_1 + u_2 \right\} \\ & = (V_1 + V_2)\cap(V_1 + V_3) \\ & \overset{V_1 \subset V_3}{=} (V_1+V_2) \cap V_3 \end{align*}

I have a serious doubt that I'm missing something here. Any help is much appreciated.

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I think your main problem is getting bogged down in symbols, a few words here and there can really make things easier to write and understand.


Let's show both inclusions separately.

First we will show that $(V_1 + V_2)\cap V_3 \subseteq (V_1 \cap V_3) + (V_2 \cap V_3)$.

Take $v \in (V_1 + V_2)\cap V_3$. Then it can be written as $v = v_1 + v_2$ for some $v_1\in V_1$ and $v_2 \in V_2$.

Actually we find that $v_1,v_2\in V_3$ automatically. Why? Well $v_1\in V_1 \subset V_3$ and $v\in V_3$, hence $v_2 = v - v_1 \in V_3$ since $V_3$ is a vector space.

Thus $v \in (V_1 \cap V_3) + (V_2 \cap V_3)$.


Now for the other inclusion, $(V_1 \cap V_3) + (V_2 \cap V_3)\subseteq (V_1 + V_2)\cap V_3 $.

Take $v \in (V_1 \cap V_3) + (V_2 \cap V_3)$. Then $v = v_1 + v_2$ where $v_1\in V_1$, $v_2 \in V_2$ and $v_1,v_2\in V_3$. Now $v_3$ is a vector space hence $v_1+v_2\in V_3$ and clearly $v_1+v_2\in V_1+V_2$. So $v \in (V_1+V_2)\cap V_3$.

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  • $\begingroup$ thanks for the answer. My attempt was to show it directly ( without double inclusion). Would that be still possible? $\endgroup$ – sigmatau Dec 30 '13 at 10:56
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    $\begingroup$ I would doubt it is straight forward but probably possible. See one direction of the inclusion didn't need the fact that $V_1\subset V_3$ yet the other one did. At some point in this proof you really are forced to write "since $V_1\subset V_3$". $\endgroup$ – fretty Dec 30 '13 at 11:07

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