Q:

Showing LUB of subspaces equals the sum of the subspaces

Let $V$ be a vector space and let $S(V)$ denote the set of all subspaces of the vector space $V$. The set $S(V)$ of all subspaces of a vector space $V$ is partially ordered by set inclusion. The {$0$} is is the smallest element in $S(V)$ and the entire space $V$ is the largest element. And lets define the sum of two vector spaces as:- $$S + T = \{u + v|u \in S, v \in T\}$$ where $S$ and $T$ are vector spaces.
My question is why is the least upper bound (lub) of $S$ and $T$ equal to the $S + T$. 
$$S+T = \operatorname{lub} \{ S,T \}.$$ Intuitively, I would guess the union of S and T to be the lub. i.e. $$  S \cup T = \operatorname{lub} \{ S,T \}. $$ Why is my intuition wrong? (What did I miss?)

A:

By definition, the least upper bound of $\{S,T\}$ is the smallest subspace of $V$ that contains both $S$ and $T$.
In contrast, $S \cup T$ is not even a subspace (except in the cases where $S\subseteq T$ or $T \subseteq S$).

