Is "False" in logic analogous to "Null set" in set theory? I have been doing proofs in elementary set theory, and so far, just using definitions (like below) and applying propositional logic has sufficed. 
A ⋃ B = e ∈ A ∨ e ∈ B
A ⊂ B = e ∈ A ⟹ e ∈ B
A' = e ∉ A = ¬(e ∈ A)

So the proofs are as follows:


*

*Convert set theory operations to their "logical" definitions

*Shuffle the symbols using logical identities

*Convert back from logic land to set theory land


Here is my question:


*

*Is the logical analogous to the null set - Ø - the boolean false?

*Is the logical analogous to the universal set - U - the boolean true?


More formally, are these definitions correct?
Ø = {e | false}
U = {e | true}

Here's my proof for: A ⊂ B ⟹ A ⋂ B’ = Ø, for example, where I use false for Ø:
A ⊂ B ⟹ A ⋂ B’ = Ø
≡ {Definition of Set Intersection and Subset, Definition of Ø}
[e ∈ A ⟹ e ∈ B] ⟹ [e ∈ A ∧ e ∈ B’ = false]
≡ {Exportation: A ⟹ [B ⟹ C] ≡ [A ∧ B] ⟹ C}
[e ∈ A ∧ e ∈ B] ⟹ [e ∈ A ∧ e ∈ B’ = false]

Context 1. e ∈ A
Context 2. e ∈ B

e ∈ A ∧ e ∈ B’
≡ {Context 1}
e ∈ B’ 
≡ {Definition of ‘}
¬(e ∈ B)
≡ {Context 2, Contradiction}
false
≡ {Definition of Ø}
Ø

Is the use of false for Ø valid in the proof above?
 A: You're basically right, but I'll flesh out a pedantic point. Identifying a set $S$ with the unary predicate $\varphi$ for which $\forall e(e\in S\iff\varphi(e))$, a set is $\emptyset$ is identified with the unary $\varphi$ that always returns "false", not with "false" itself. (This is like confusing a constant function with the value it returns; it's a subtle distinction, but an easy one to make if e.g. you define a function as a certain kind of set of ordered pairs.) The usual choice for an explicit statement of this $\varphi$ is that $\varphi(e)$ iff $e\neq e$.
A: Yes, they are analogous.  You can show the analogy a bit more directly like this:
For boolean logic, we have:
$P \land \bot \Leftrightarrow \bot$  (or: $P\cdot 0 = 0$)
and
$P \lor \bot \Leftrightarrow P$ (or: $P+0=P$)
while for sets we have:
$P \cap \emptyset = \emptyset$
and
$P \cup \emptyset = P$
And yes, if you know that all the sets that you are working with are subsets of some 'universal' set $U$, you have:
$P \cap U = P$
and
$P \cup U = U$
just as in boolean logic you have:
$P \land \top \Leftrightarrow P$  (or: $P\cdot 1 = P$)
and
$P \lor \top \Leftrightarrow \top$ (or: $P+1=1$)
