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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
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).
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 $a \in X$?
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 $a\in X$?
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.
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 $v\in X+(Y\cap Z)$
