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?


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 $v\in S_0^\top$.


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