Method of Proof in Showing Something is Smallest (Subspace) I am reading a proof that shows the sum of subspaces is the smallest subpsace containing all the summands (It is a vector space over $\mathbb{R^n}$).
The author of the book goes to show first it is a subspace.
Then, it goes to show each subspace is contained in the sum, and then, it goes on to show every subspace of the vector space containing each subspace also contains the sum. I am little confused this second part showing it is the smallest. If you could elaborate in general showing something is the smallest in math logic, I would appreciate that, too. Thanks.
 A: This actually just comes from the definition of "smallest". I'm taking this from Munkres' Topology 2nd edition: Let $A_0 \subset A$. We say that $a \in A_0$ is the smallest element of $A_0$ if $a \leq x$ for every $x \in A_0$. (Note $\leq$ is just an arbitrary order relation; in your example it is $\subset$). 
Let $X = U + W$, where $U$ and $W$ are subspaces of a vector space $V$. So in this case $A$ is the set of all subspaces of $V$ and $A_0 \subset A$ is the set of all subspaces of $V$ that contain each of the summands. By showing that $X$ is a subspace that contains both $U$ and $W$ we have shown that $X \in A_0$, completes one part of the definition. The next part of the definition is to show that if $Y$ is any other element of $A_0$, then $X \subset Y$. If this is the case than we can call $X$ the smallest subspace of $V$ that contains both $U$ and $W$. This is what the author was doing when he showed that any subspace containing $U$ and $W$ must also contain $U + W$.
The confusing part was probably the word smallest being applied to subsets since you might only expect to see that word with numbers. But that notion can be generalized to any order relation, which is what the author was using here. If you haven't seen order relations, then I would recommend at least checking out all the basic definitions. You should be able to find them in any textbook having a section about relations. Though I wouldn't worry about it too much since it will probably rarely come up in linear algebra.
A: Let's give them names; call the summands $U$ and $V$, and their sum $W = U+V$. The second part is showing that $W$ is the smallest subspace containing $U$ and $V$. It does this by showing that there are no smaller subspaces containing $U$ and $V$, i.e., if we have a subspace $S$ containing $U$ and $V$, then $S$ must also contain $W$.
In fact, one can equivalently define $U+V$ to be the set intersection of all subspaces which contain $U$ and $V$.
A: Suppose we have a sum, $M = U + W= \{u + w: u \in U \text{ and } w \in W\}$.  Suppose $U$ and $W$ are both contained in $Z$.  We will now show that their sum $M$ is contained in $Z$.  Let $m \in M$.  Then by definition there exists $u \in U \text{ and } w\in W \text{ such that } u+w=m.$  Since $U$ is contained in $Z$, $u \in Z$.  Similarly since $W$ is contained in $Z$, $w \in Z$.  Thus by closure under addition $u + w = m \in Z$.  
