Let $X,Y,Z$ be subspaces of $V$ so that $X$ is a subspace of $Y$. Prove that $Y\cap (X+Z)=X+(Y\cap Z)$ Let $X,Y,Z$ be subspaces of $V$ so that $X$ is a subspace of $Y$. Prove that $Y\cap (X+Z)=X+(Y\cap Z)$
I know that I need to prove that $Y\cap (X+Z)\subseteq X+(Y\cap Z)$ and $X+(Y\cap Z)\subseteq Y\cap (X+Z)$ but I´m having a hard time.
Can you lend me a hand please? I would really appreciate it
 A: This is called a modular law.
It holds for subgroups of any abelian (additive) group, 
thus it holds also for submodules of a module,
and in particular for subspaces of a vector space.
I prefer to write it in the form that is easy to remember:
$\newcommand{\inters}{\cap}$

Let $A$, $X$, $B$ be subgroups of an abelian group.
  If $A\subseteq B$, then $A+(X\inters B)=(A+X)\inters B$.

You see -- you just move the pair of parentheses.
Proof.
$~$Since $A\subseteq A+X$ and $A\subseteq B$, we have $A\subseteq (A+X)\inters B\,$;
since $X\subseteq A+X$, we have also $X\inters B\subseteq(A+X)\inters B\,$;
it follows that $A+(X\inters B)\subseteq (A+X)\inters B\,$.
In order to prove the reverse inclusion,
we consider any $y\in(A+X)\inters B\,$,
that is, $y=a+x=b$ for some $a\in A$, some $x\in X$, and some $b\in B$.
Then $x=b-a\in B$ because $b\in B$ and $a\in A\subseteq B$,
thus $x\in X\inters B$,
whence $y=a+x\in A+(X\inters B)\,$.$~$ Done.
The proof is essentially the same as the one given by Deuteu, just cleaner and clearer.
$\newcommand{\coll}{\mathcal}$
$\newcommand{\Inters}{\bigcap}$
Let $\coll{L}$ be the set of all subgroups of an abelian group
(or of all submodules of a module, or of all subspaces of a vector space).
When $\coll{L}$ is ordered by inclusion, it becomes a lattice:
any $X,Y\!\in\coll{L}$
have the greatest lower bound $X\inters Y$ and the least upper bound $X+Y\!$.
(Actually every indexed family of members $X_i$ of $\coll{L}$
has the greatest lower bound $\Inters_iX_i$ and the least upper bound $\sum_iX_i$
-- the lattice $\coll{L}$ is complete).
The lattice $\coll{L}$ is a modular lattice
because (you guessed it)
it obeys the modular law.
See the article "Modular lattice" in Wikipedia (for starters).
A: As noted, to prove an equality of sets one must prove each side is a subset of the other one.
First, we prove the $\subseteq$ side, i.e. $Y \cap (Z+X) \subseteq X + (Y \cap Z)$:
Take any $a \in Y \cap (Z+X)$. That means $a$ is in both $Y$ and $(Z+X)$. The question is: is $a \in X$?


*

*if $a \in X$, that means that it certainly is in $X+(Y\cap Z)$.

*if $a\notin X$, then it must be in $Z$ (since it is in $Z+X$). Thus, $a$ is in both $Y$ and $Z$, which means it is in $Y\cap Z$ and therefore it is also in $X+(Y\cap Z)$.


Second, prove that $X+(Y\cap Z) \subseteq Y\cap (Z+X)$:
Take any $a \in X+(Y\cap Z)$. Then, $a$ is either in $X$ or in the intersection $Y\cap Z$. Since $X$ is a subset of $Y$, we have that $a$ must be in $Y$. Now, the same question: is $a\in X$?


*

*if $a\in X$, it is also in $Z+X$ and we are done.

*if $a \notin X$, it must be in $Y\cap Z$, which means it certainly is in  $Y\cap (Z+X)$. $\Box$

A: A collection of hints:
Hint: $X + Y = Y$
General hint: if you try to show $A \subset B$ then take a generic element $a \in A$ and show $a \in B$.
Hint about elements: a generic element of the LHS is of the form: $Y \ni y=x+z \in X+Z$ (it lies in the intersection). When does this happen to be in the RHS? In which subspace(s) lives $z = y-x$? (hint: it is not only $Z$)
Another hint about elements: if $x + r$ is a generic element on the RHS, then in particular it is an element in $X+Z$. So it suffices to show that it is also in $Y$. Why is that?
If you use those hints you should be able to solve the exercise yourself - hope it helps.
A: As you said we can do it by double inclusion there is :
If $w \in X+(Y\cap Z)$, we have $w=x+\mu$ with $x\in X$ and $\mu \in (Y\cap Z)$
- $X \subseteq Y$ so $x \in Y$ thus $x,\mu \in Y$ so $w \in Y$- $\mu \in Z$ so $w \in (X+Z)$In conclusion $w \in Y\cap (X+Z)$
If $v \in Y\cap (X+Z)$, $v=\alpha \in Y$ and $v=\beta + \gamma$ with $\beta \in X$ and $\gamma \in Z$$\gamma = \alpha - \beta$ and $\beta \in X \subseteq Y$ so $\gamma \in Y$ thus $\gamma \in Y \cap Z$
In conclusion $v=\beta_{\in X} + \gamma_{\in Y\cap Z}$ so $v\in X+(Y\cap Z)$
