If $U$ and $W$ are subsets of $V$ and $U \subset W$ then $W^{\perp}
\subset U^{\perp}$
I am trying to understand the given proof: 
I absolutely do not understand how we arrive at the conclusion.
If we let $v \in V$ then for all $w \in W$ we have for some $v \in V$ the set $W^{\perp} = \{v \in V \mid \; \langle v, w \rangle = 0, \; \forall w \in W \}$. Since $U \subset W$, then for all $u \in U$ we have: $U^\perp = \{ v \in V \mid \; \langle v , u \rangle = 0 , \; \forall u \in U \}$ for the same $v \in V$.
Now my issue is with how this implies that $W^\perp \subset U^\perp$. If I talk about examples in $\mathbb{R}^n$ with the standard basis then I can see the conclusion, but I do not see how to do this generally.
Best Answer
The proof starts with taking an arbitrary $v\in W^\bot$, and proves that $v\in U^\bot$.
Therefore, it proves the statement:
$$\forall v: v\in W^\bot\implies v\in U^\bot,$$ and this statement is, by definition, equivalent to the statement $W^\bot\subset U^\bot$.