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Let $E$ be a vector space, and $F, G, H$ subspaces of $E$ such that $G\subset F$.

The exercise is to show that:

$F\cap (G+H) = G + (H\cap F)$

I understand this identity as sets, but as spaces and subspaces I don't really know how to prove it with the formalism required, so I am looking for the way to work with this concepts of space and subspace.

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  • $\begingroup$ You don't have a weel defined $+$ for sets. For $\{a,b\} + \mathbb Z$ to be defined you must give a definition of $a+1$ and so on. But takingtwo subsets of a vector space you can define such a $+$ $\endgroup$ – InfiniteLooper Mar 3 '19 at 19:58
  • $\begingroup$ This is the modular law for the lattice of subspaces of $E$. $\endgroup$ – darij grinberg Mar 3 '19 at 20:00
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When you use the operation $+$, I will assume you mean that $G+H = \{g +h \in E: g \in G, h \in H\}$.

Let $x \in F\cap(G+H)$. Since $F\cap(G+H) \subset G+H$, $x$ may be expressed as $x = g + h$ for some $g \in G$ and $h \in H$. However, since $F\cap(G+H) \subset F$, we have $x \in F$. Observe that $h = x-g$. Clearly, $x-g \in H$. Now, since $G \subset F$, $g \in F$. Thus, since $ x \in F$ and $g \in F$, $x-g \in F \implies x-g \in H\cap F$. But observe that $x = g + (x-g) \implies x \in G + (H\cap F) \implies F \cap (G+H) \subset G+(H \cap F)$.

Now, let $y \in G + (H\cap F)$. So, $y =g + r$ for some $g \in G$ and $r \in H\cap F$. We note that since $H \cap F \subset H$, $r \in H$. Thus, $y = g + r \in G+H$. However, since $H \cap F \subset F$, we also have $r \in F$. Again, since $g \in G$ and $G \subset F$, we have $g \in F$. Thus, since $F$ is a subspace and $r, g \in F$, $r + g \in F$. Thus, $y \in F$ and $y \in G+H$. It follows that $y \in F \cap (G+H) \implies G+(H \cap F) \subset F \cap (G+H)$. The result follows.

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For the first inclusion, take $x \in F \cap (G+H) \Rightarrow x\in F \wedge x\in (G+H)$. Then exist $g \in G$ and $h \in H$ such that $x=g+h \in F$. As $G \subseteq F$ we have to $g \in F$ and how $F$ is a subspace $\Rightarrow$ $h \in F$. So $h \in H \cap F$ and $x=g+h \in G+(H\cap F)$.

For the second inclusion, take $x \in G+(H\cap F)$ so exits $g \in G \subseteq F$ and $y \in H \cap F$ such that $x=g+y$. As $y \in F$ and $F$ is subspace $\Rightarrow x \in F$. Also, $y \in H$ so $x=g+y \in G+H$, finally $x \in F \cap(G+H)$

$\textbf{Observation :}$ If you have $x+y \in F$ and $x \in F$ where $F$ is a subspace. Necessarily you have to $y \in F$ because $x+y=f \in F \Rightarrow y=f-x$ and how $F$ is a subspace : $y \in F$

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