Let $A$ and $B$ be subspaces of a Hilbert space $H$.
Consider the following proposition:
$$A \subset B \implies A^\perp \subset B^\perp\qquad (1)$$
I am aware that the following is true:
$$A \subset B \implies B^\perp \subset A^\perp\qquad (2)$$
However, I attempted to "prove" $(1)$ as follows:
To prove $A^\perp \subset B^\perp$, let us show that $$\forall y \in
A^\perp, y \in B^\perp$$
If $y \in A^\perp$, then: $$\forall x \in A, \langle x,y \rangle = 0$$
Because $A \subset B$ is true, meaning, $$\forall x \in A, x \in B$$ it follows that $$\forall x \in B, \langle x, y \rangle = 0$$i.e. $$y \in B^\perp$$
I am convinced that this proof is flawed since I haven't come across $(1)$ in any textbook, instead, proposition $(2)$ always showed up. Any indications as to what went wrong in the above proof is greatly appreciated.
Best Answer
The mistake in this proof is the step that starts "it follows that...". The reason is that there could be $z \in B$ such that $z \not \in A$ and then you can't say anything about $\langle z, y \rangle$.
What you have written is actually almost a proof of $(2)$ [with $A$ and $B$ swapped]. You just had to write instead "because $B \subset A$ is true, meaning $\forall x \in B, x \in A$, it follows that $\forall x \in B, \langle x,y \rangle = 0$..."