Problem: Let $V$ be an inner product space, $S$ and $S_0$ be subsets of $V$, and $W$ be subsets of $V$, and $W$ be a finite-dimensional subspace of $V$. Prove
1). $S_0 \subseteq S$ implies that $S^{\bot} \subseteq S_0^{\bot}$
Brief sketch: Let $\beta={v_1,…v_k}$ be a basis for $S_0$, $\beta'{v_1',..,v_k'}$ bais for $S_0^{\bot}$, $\gamma={w_1,…,w_k}$ basi for $S$ and $\gamma'={w_1',…,w_k'}$ basis for $S^{\bot}$.
by the theorem, we can conclude the statement.
Theorem: suppose $S={v_1,v_2,..,v_k}$ is an orthonormal set in an $n$-dimensional inner product space $V$, then $S$ can be extended to an orthonormal basis ${v_1,v_2,…,v_k,v_{k+1},…,v_n}$ for $V$.
Will that be enough?
Best Answer
No, it will not be enough. What does “Let $\beta={v_1,...v_k}$ for $S_0$, $\beta'{v_1',..,v_k'}$ for $S_0^{\bot}$” mean? I have no idea.
Besides, you don't need bases here. If $v\in S^\top$, then, for each $w\in S$, $\langle v,w\rangle=0$. In particular (since $S_0\subset S$), for each $w\in S$, $\langle v,w\rangle=0$, and so $v\in S_0^\top$.