$W_1^\perp + W_2^\perp = (W_1 \cap W_2 )^\perp$: 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)
 A: 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?
A: I'm so confused now :)
Assuming everything is finite-dimensional, here's a hint: Prove $X^\perp\cap Y^\perp\subset (X+Y)^\perp$.
A: I would not say that you are "viewing a set as a function" but that you are "modifying a set using an operator". Indeed, fix a finite dimensional vector space $V$ with a fixed bilinear form $B$ and let $Sub(V)$ be the set of vector subspaces of $V$. Then you can consider the following three operations
$$Sub(V)\to Sub(V)\ \ \ W\mapsto W^{\perp}\ \ \ (\text{where the orthogonal is with respect to $B$})$$
$$Sub(V)\times Sub(V)\to Sub(V)\ \ \ (W_1,W_2)\mapsto W_1+W_2$$
$$Sub(V)\times Sub(V)\to Sub(V)\ \ \ (W_1,W_2)\mapsto W_1\cap W_2$$
The two binary operations really make $Sub(V)$ into a lattice which has many good properties. Now, you want to prove that $W_1^{\perp}+W^{\perp}_2=(W_1∩W_2)^{\perp}$ (for two elements $W_1$ and $W_2\in Sub(V)$). This can be done following the following steps:
Step1. Notice that your thesis is equivalent to prove that $(W_1^{\perp}+W^{\perp}_2)^{\perp}=W_1\cap W_2$. For this you have to use some properties of the function ${}^\perp$, that is, $X^{\perp\perp}=X$ and ($X=Y\ \Leftrightarrow\ X^\perp=Y^\perp$) for all $X,Y\in Sub(V)$ (these properties need $V$ to be finite dimensional);
Step2. $(W_1^{\perp}+W^{\perp}_2)^{\perp}=\{x\in V:B(x,y)=0 \text{ for all $y\in W_1^{\perp}+W^{\perp}_2$}\}=\{x\in V:B(x,y_1)=0 \text{ and }B(x,y_2)=0 \text{ for all $y_1\in W_1^{\perp}$ and $y_2\in W^{\perp}_2$}\}=\{x\in V:x\in W_1^{\perp\perp}\text{ and }x\in W_2^{\perp\perp}\}=W_1^{\perp\perp}\cap W_2^{\perp\perp}=W_1\cap W_2$, where the first equality comes from the definitions, the second uses bilinearity of $B$ (and that $0$ belongs to any subspace of $V$), the third equality follows again by the definition of orthogonal, the fourth one just translates an "and" into an intersection and the last one uses a property you already used in Step1, that is, $X^{\perp\perp}=X$ for all $X\in Sub(V)$.
