Complete Lattice, Complemented Lattice, Modular Lattice of Subspaces of a Vector Space 
Let $V$ be a vector space over a division ring $D$ and $S$ the set of all subspaces on $V$, partially ordered by set theoretic inclusion.
(i) $S$ is a complete lattice
(ii) $S$ is a complemented lattice; that is for each $V_1\in S$ there exists $V_2\in S$ such that $V=V_1+V_2$ and $V_1\bigcap V_2=0$, so that $V=V_1\oplus V_2$.
(iii) $S$ is a modular lattice; that is, if $V_1,V_2,V_3\in S$ and $V_3\subset V_1$ then $$V_1\cap (V_2+V_3)=(V_1\cap V_2)+V_3.$$

I need help with this problem. I'm not sure where to start. I think I understand what a complete lattice, a complemented lattice and a modular lattice. I'm just not sure how to start. I have ideas but I'm not sure if they are correct. For example I think the LUB subspace is the intersection of all subspaces while the GLB is the vector space V. Any help/hint would be appreciated. I hope somebody could point me in the right direction. Thanks!
 A: Hints:


*

*The meet of a family is the intersection.  The join is the sum.

*Take a basis of $V/V_1$ and lift it to $V$.

*Take an element of $V_1 \cap (V_2 + V_3)$.  That you can write it as both $v_1 \in V_1$, and $v_2 + v_3$ for $v_2 \in V_2$ and $v_3 \in V_3$.  That means that $v_2 = v_1 - v_3 \in V_1$.
A: This is exercise 5 page 189 in Hungerford's Algebra book.
item (a): If $E$ is any non-empty set of subspaces of $V$, let $U_E= \bigoplus_{W \in E} W$ and $L_E = \bigcap_{W \in E} W$. Just check the definition to see that $U_E$ is l.u.b. of $E$ and $L_E$ is the g.l.b. of $E$.
item (b): Given any $V_1 \in S$, $V_1$ is a vector space over $D$ and, by Theorem IV.2.4, there is a $B_1$ a base of $V_1$. Since $V_1 \subseteq V$, we have that $B_1$ is a linearly independent subset of $V$. So, by Theorem IV.2.4, there is $B$ a basis of $V$ such that $B_1 \subseteq B$. Let $B_2= B \setminus B_1$. We have that $B_2$ is a a linearly independent subset of $V$. Let $V2$ be the subspace of $V$ generated by $B_2$. It is straight forward to check that $V=V_1 +V_2$ and $V_1 \cap V_2= \emptyset$.
item (c): Suppose $V_1,V_2,V_3\in S$ and $V_3\subseteq V_1$.
Given any $v \in V_1\cap (V_2+V_3)$ then there is $v_2 \in V_2$ and $v_3 \in V_3$, such that $v=v_2+v_3$. So, $v_2 = v - v_3 \in V_1$ (because $V_3\subseteq V_1$). It follows that
$v_2 \in V_1 \cap V_2$, so $v= v_2+v_3 \in (V_1 \cap V_2) + V_3$.
So we have proved that $$V_1\cap (V_2+V_3) \subseteq(V_1\cap V_2)+V_3$$
Now, since $ (V_1\cap V_2) \subseteq V_1\cap (V_2+V_3)$ and  $ V_3 \subseteq V_1\cap (V_2+V_3)$ (because $V_3\subseteq V_1$), we have  $(V_1\cap V_2)+V_3 \subseteq V_1\cap (V_2+V_3) $. So, we have
$$V_1\cap (V_2+V_3) = (V_1\cap V_2)+V_3$$
