Definition of orthogonal subspace In the book I'm using to study there is this definition: given a vectorial space $E$ with an internal product:


*

*given $u \in E$, the set of all vectors in $E$ that is orthogonal to $u$ is a subspace of $E$;

*given a subset $M$ of $E$, the set of all v in E such that v is orthogonal to $u$, for any $u$ in $M$ is a subspace of $E$.


Question: can $M$ be any set? Doesn't it need to be, itself, a subspace of $E$? And them, the newly created subspace would called $M$-orthogonal?
 A: As my analysis prof used to say "you can define whatever you want as long as you are consistent" ;) note that the orthogonal set does not change, when you replace $M$ by its span. This means, you can take the vector space generated by $M$ and everything reduces to the case you are familiar with.
Added: We want to prove the following equality of sets
$$ \{ x\in E \ \vert \ \forall m\in M : \langle x, m\rangle=0 \} = \{ x\in E \ \vert \ \forall m\in span(M) : \langle x, m\rangle=0 \}$$
The inclusion $\supseteq$ is clear. Let us prove the other inclusion. Let $x$ be orthogonal to $M$, we want to prove that it is orthogonal $span(M)$. Let $m\in span(M)$, then we can write $m=\alpha_1 m_1 + \dots + \alpha_n m_n$ where $\alpha_i \in \mathbb{C}$ and $m_i \in M$ and we get
$$ \langle x , m \rangle = \alpha_1 \langle x, m_1 \rangle + \dots + \alpha_n \langle x, m_n \rangle = \alpha_1 \cdot 0 + \dots + \alpha_n \cdot 0 = 0. $$
The first equality is the bilinearity of the inner product and the second equality is the assumption that $x$ is orthogonal to $M$. Thus, $x$ is orthogonal to $span(M)$.
