Linear Algebra: show that $F\cap (G+H) = G + (H\cap F)$

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.

• 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 $+$ – InfiniteLooper Mar 3 '19 at 19:58
• This is the modular law for the lattice of subspaces of $E$. – darij grinberg Mar 3 '19 at 20:00

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.
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 : $$y \in F$$