Sets Proof (A ⊆ λB) ⇔ (B ⊆ λA)

[sponsoredwalk]

Just doing some of the set theory questions at the start of a calculus book & I'm kind of
confused about how to prove the following:

(A ⊆ λB) ⇔ (B ⊆ λA)

(Note: λB denotes the complement relative to the universal set, as with & λA)

I'm trying to get used to proving this as if I'm unfurling the definitions & forming a chain of
implications as opposed to the hand waving I used to do It's late & I could be
just making a careless mistake, please let me know what you think.

First the definitions:

(A ⊆ B) := {x ∈ S | (x ∈ A) ⇒ (x ∈ B)}

λB := {x ∈ S | (x ∈ S) ⋀ (x ∉ B)}

So I think that the proof would go as follows:

x ∈ (A ⊆ λB) ⇒ [ (x ∈ A) ⇒ (x ∈ S) ⋀ (x ∈ λB) ] ⇒ [((x ∈ A) ⇒ (x ∈ S)) ⋀ ((x ∈ A) ⇒ (x ∈ λB)) ]

I'm just confused now, at first I read the question without reading the ⇔ (B ⊆ λA)
part of it, just to see could I arrive at it naturally like I had in the other questions but
I just can't see the way to move here

I think I have these proofs down & am following the best method, a quick verification
of what I'm doing is the proof that (C ⊆ A) ⋀ (C ⊆ B) ⇔ C ⊆ (A ∩ B), just to make sure
I'm not assuming things or making careless mistakes:
x ∈ [(C ⊆ A) ⋀ (C ⊆ B)] ⇒ [((x ∈ C) ⇒ (x ∈ A)) ⋀ ((x ∈ C) ⇒ (x ∈ B))] ⇒ [(x ∈ C) ⇒ (x ∈ A) ⋀ (x ∈ B)] ⇒ [(x ∈ C) ⇒ (x ∈ (A ∩ B))]
which shows that [(C ⊆ A) ⋀ (C ⊆ B)] ⇒ [C ⊆ (A ∩ B)] & you just do it backwards to show that ⇔ holds. I think that's right, yeah?

 

[lanedance]

$$(A \subseteq B^c) \Leftrightarrow (B \subseteq A^c)$$

it should help to notice that if A is a subset of B^c, A and B have no elements in common, so
$$A \cap B = \emptyset$$

 

[lanedance]

Shoudl pretty much follow from there

 

[sponsoredwalk]

Yeah I see what you mean by that & I do understand that. I constructed a big truth table
& found the logical equivalence that corresponds to the scenario on the right & came to
the answer. Please let me illustrate it briefly:

$\begin{displaymath}<br /> \begin{array}{|c|c|c|c|c|c|c|c|c|c|c|c|c|c}<br /> A<br /> & \lnot{}A<br /> & B<br /> & \lnot{}B<br /> & A\lor{}B<br /> & \lnot{}(A\lor{}B)<br /> & A\land{}B<br /> & \lnot{}(A\land{}B)<br /> & B\land{}A<br /> & \lnot{}(B\land{}A)<br /> & (A\Rightarrow{}B)<br /> & (\lnot{}B\Rightarrow{}\lnot{}A)<br /> & (B\Rightarrow{}A)<br /> & (A\Leftrightarrow{}B) \\<br /> \hline<br /> 0 & 1 & 0 & 1 & 0 & 1 & 0 & 1 & 0 & 1 & 1 & 1 & 1 & 1 \\<br /> 0 & 1 & 1 & 0 & 1 & 0 & 0 & 1 & 0 & 1 & 1 & 1 & 0 & 0 \\<br /> 1 & 0 & 0 & 1 & 1 & 0 & 0 & 1 & 0 & 1 & 0 & 0 & 1 & 0 \\<br /> 1 & 0 & 1 & 0 & 1 & 0 & 1 & 0 & 1 & 0 & 1 & 1 & 1 & 1 \\<br /> \hline<br /> \end{array}<br /> \end{displaymath}$

Using this we'll see some equivalent situations:

$(A \ \subseteq \ B^c) \ \Rightarrow \ [ \ (x \ \in \ A) \ \rightarrow \ (x \ \in \ B^c) \ ] \Rightarrow \ [ \ (x \ \not\in \ A^c) \ \rightarrow \ (x \ \not\in \ B) \ ]$

Noticing on the truth table that (¬B ⇒ ¬A) ⇔ (A ⇒ B)

$[ \ (x \ \not\in \ A^c) \ \rightarrow \ (x \ \not\in \ B) \ ] \ \Rightarrow \ [ \ (x \ \in \ B) \ \rightarrow \ (x \ \in \ A^c) \ ] \Rightarrow (B \ \subseteq \ A^c)$

I guess it's just a matter of knowing the logical equivalences well enough to recognise
when to substitute them in, thanks a lot