The Order of Orthogonality

Question

I would like to show that $B\subset A$ implies $A^{\bot}\subset B^{\bot}$.

Note the meaning behind this: The bigger a subset, the smaller its orthogonal should be.
Let $x$ be in the complement of A. Then $\langle x,y\rangle =0$ for all $y\in A$. But since $B\subset A$, $\langle x,y\rangle=0$ for all $y\in B$. Is it true?

Answer 1

$$x\in A^\bot\iff x\notin A\overbrace{\Longrightarrow}^{B\subset A} x\notin B\iff x\in B^\bot$$

Answer 2

Yes, your argument is correct and straightforward.

Although true on the level of set inclusions, one should be aware that the implication $\:A^\perp\subset B^\perp\:$ merely depends on the vector subspace $\,\operatorname{span}(B)\,$ being a subspace of $\,\operatorname{span}(A)\,$ (as implied by $B\subset A$), and not on the "bigness" of $A$ and $B$ in the sense of sets.

A note regarding terminology: $V^\perp$ is usually called "orthogonal complement" only if $V$ is a subspace, else $V\oplus V^\perp$ won't yield the entire ambient vector space.