Proof technique: Given vector subspaces $U$ and $W$, prove that $U + W = \{ ...\}$ Let $U,W$ be vector subspaces of $V$ such that $U = \{(x, 0, 0) \in \mathbb{R}^3: x \in \mathbb{R} \}$ and $W = \{(0, y, 0) \in \mathbb{R}^3: y \in \mathbb{R} \} $.
Show that $U + W = \{(x, y, 0):x,y \in \mathbb{R} \}$.
We can see that, if we pick any $u \in U$ and any $w \in W$, $u + w = (x^*,y^*,0)$, where $x^*,y^* \in \mathbb{R}$. But once I've found the vector $u + w$, how do I get the corresponding vector subspace $U + W$? I know that, obviously, since $u$ and $w$ were arbitrarily chosen, $u + w$ can be any vector in $U + W$. But how exactly does the fact that $u + w \in U + W$ imply that $U + W = \{(x,y,0):x,y \in \mathbb{R}\}$?
The steps of the proofs would go:

*

*Let $U,W$ be the above mentioned vector subspaces of a vector space $V$, and let $u, w$ be vectors in $U, W$ respectively.


*Then, $u + w = (x^*, y^*, 0)$, where $x^*, y^* \in \mathbb{R}$.
And then?
Although $u$ and $w$ were indeed arbitrarily chosen, don't we still only have exactly one vector $u+w$? So how could I bridge the gap between this single $u + w$ and the actual set of all such vectors $\{(x,y,0):x,y \in \mathbb{R} \}$?
(the example is from Axler's Linear Algebra Done Right page 14)
 A: By definition, two sets $A,B$ are equal if they have the same elements. That is $A \subseteq B$ and $B \subseteq A$. Denoting
$$T=\{(x, y, 0):x,y \in \mathbb{R} \}$$ you have proven that $U+W \subseteq T$. You now have to prove that $T \subseteq U+W$. For this, taking any element $t = (x,y,z) \in \mathbb R^3$ with $z=0$ (i.e. $t \in T$), you have to prove that it exist $u \in U$ and $ w \in W$ such that $u+w = t$. Which can be done easily with
$$(x,y,0) = (x,0,0)+(0,y,0).$$
Note: indeed for any $t \in T$, there is a unique ordered pair $(u,w) \in U \times W$ such that $t=u+w$. However, the uniqueness has nothing to do with the question. It would if you would have to prove that $T = U \oplus W$ ($\oplus$ meaning direct sum).
A: Two prove S is a subset of T how do I proceed?
I start with an arbitrary element in S and show that it belongs to T.
Now what does this mean? I'll put in better way.
Suppose I'm the one trying to prove S is a subset of T. I'll ask you to give me an element in S and I'll prove it is in T.As you are giving me any element of your choice it is arbitrary for me.
This is what we say as,given any element in S I can prove it is in T.
