Proof involving subspaces I encountered this question in a document I found on a  google search, it bugged me because my perception keeps telling me I'm wrong no matter what I do.

Let $U$, $W$ and $Z$ be subspaces of a vector space $V$ and the following conditions are satisfied:
  
  
*
  
*$U + Z = W +Z$  
  
*$U \cap Z = W\cap Z$  
  
*$U \subset W$  
  
  
  Prove that $U=W$

So I tried proving by contradiction that $U$ is not equal to $W$,  by saying there exists some vector $w \in W \setminus U$ and  if $w \in W \cap Z$ then $w \in U \cap Z$ which means $w \in  U$. At this momment my mind is shouting  "This isn't it", and I think I should be using a different approach.
 A: $$ \frac{U}{U \cap Z} \cong \frac{U + Z}{ Z} $$ $$\frac{W}{W \cap Z} \cong \frac{W + Z}{ Z}$$  But $U + Z = W + Z $ so $$\frac{U}{U \cap Z} \cong \frac{W}{W \cap Z} = \frac{W}{U \cap Z}$$ This implies that $$\dim(U) = \dim (W) $$ but $U \subseteq W $ and so $$U = W $$
A: You have your $w\in W\setminus U$. The vector $0\in Z$ so 
$$w=w+0\in W+Z=U+Z\;,$$
and there must be $u\in U$ and $z\in Z$ such that $w=u+z$. $U\subseteq W$, so $w-u\in W$,and therefore
$$w-u=z\in W\cap Z=U\cap Z\;.$$
But then $z\in U$ and hence $w=u+z\in U$, contradicting the choice of $w$.
A: No contradiction and no considerations about dimension. Let $w\in W$; then $w\in W+Z$, so $w=u+z$, for some $u\in U$ and $z\in Z$.
But now $z=w-u\in W$, since $U\subseteq W$. Therefore $w-u\in W\cap Z$, so $w-u\in U$, since $U\cap Z=W\cap Z$. Thus
$$
w=(w-u)+u\in U.
$$
Another view at this. The lattice of subspaces of a vector space is modular; this means that, if $U\subseteq W$, then
$$
U+(W\cap Z)=(U+Z)\cap W
$$
for every subspace $Z$ (see Wikipedia). Under our hypotheses the left hand side is $U+(U\cap Z)=U$, while the right hand side is $(W+Z)\cap W=W$. Thus $U=W$.
Proving the modular identity is a good exercise.
A: First, $U+Z=W+Z\implies \dim(U+Z)=\dim(W+Z)\implies$$$\dim U+\dim Z-\dim (U\cap Z)=\dim W +\dim Z-\dim(W\cap Z)\implies$$ $$\dim U-\dim(U\cap Z)=\dim W-\dim(W\cap Z)\ \ \ \ (\star)$$
But $\dim(U\cap Z)=\dim(W\cap Z)$, by assumption. So putting this into $(\star)$, we get $$\dim U=\dim W$$
And then, since $U\subseteq W$, we know that $U=W$.
A: First two conditions says that, dim $U =$ dim $ W. $ So using the fact that $U \subset W,$ we have $U= W.$
WARNING: This process can only be applied if all the subspaces involved (i.e. $U, W, Z$) are finite dimensional. But this is not a part of the assumption. Thanks to egreg for pointing it out.
