# inner product space proving $S_0 \subseteq S$ implies that $S^{\bot} \subseteq S_0^{\bot}$

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 $$v\in S_0^\top$$.