$W_1^\perp + W_2^\perp = (W_1 \cap W_2 )^\perp$: Can a set be a function? Can two such "functions" be composed?

How do I prove the following proposition: $$W_1^\perp + W_2^\perp = (W_1 \cap W_2 )^\perp$$ Note that there have been other questions (here, here) just asking about the $\subseteq$ inclusion, so please don't close this as a duplicate of those.

For Mr. Z:

Let $v\in W_1^\perp + W_2^\perp$ be the sum $v_1+v_2$, where $v_1\in W_1^\perp$ and $v_2\in W_2^\perp$, and let $w\in W_1\cap W_2$. This means $\langle v,w\rangle=\langle v_1+v_2,w\rangle=\langle v_1,w\rangle+\langle v_2,w\rangle=0+0=0$. Hence, $v\in (W_1 \cap W_2 )^\perp$.

I'm just trying to reason this out; that is, so I can understand it and not just "rote it," as it were. It seems worthwhile to see if by the very definitions we can come to show that the set definitions are the same without having to show "both ways." In what follows is my novice attempt at that:

$$---Data---$$

$$X^{\perp}:=\{y\in V~:~\langle x,y \rangle = 0 ~\forall~ x\in X\subset V\}$$ $$A\cap B := \{x~:~x\in A~ \wedge~x\in B\}$$ $$W_1+W_2=\{v\in V~:~v=w_1+w_2,~\text{with $w_i \in W_i$}\}$$

I feel that if you see these things as functions, that is, $X^{\perp}(x,X,y,V)$ and $A\cap B (x,A,B)$, then we can just see $(A\cap B)^{\perp}$ as a composition somehow. Is that possible? I actually don't even know where to start. I'm thinking can I show that

$$(A\cap B)^{\perp}:=\{statements\}$$

is the same as

$$A^{\perp}+B^{\perp}:=\{statements\},$$

where the statements are clearly the same.

$$---References---$$

[1] Cohn's Classic Algebra (page 220)

[2] Shen's Basic Set Theory (page 001)

[3] Artin's Algebra (page 095)

Solution 1:

I assume that you're dealing with finite-dimensional inner product spaces, as I don't think it holds in general, otherwise. I use here the fact that in a finite-dimensional inner product space, we have for any subspace $X$ that $(X^\perp)^\perp=X$. Hence, we may equivalently show that $$(W_1^\perp+W_2^\perp)^\perp=W_1\cap W_2.$$

Suppose $v\in (W_1^\perp+W_2^\perp)^\perp,$ so that $\langle v,w_1+w_2\rangle=0$ for all $w_1\in W_1^\perp$ and $w_2\in W_2^\perp$. From this you should see that $\langle v,w_1\rangle=0$ for all $w_1\in W_1^\perp$ (why?), so $v\in (W_1^\perp)^\perp=W_1$. Likewise, $v\in W_2,$ so $v\in W_1\cap W_2.$

On the other hand, suppose that $v\in W_1\cap W_2$. Note that $W_1\cap W_2\subseteq W_1,$ so what can we say about $\langle v,w_1\rangle$ for $w_1\in W_1^\perp$? Can you take it from there?