14
6. September 15th: Naive set theory: basic operations. Relations.
6.1
.
The useful abbreviation
∀
x
∈
A, p
(
x
) stands for
‘the statement
p
(
x
)
is true
for all
x
in the set
A
’.
In other words,
‘for all
x
, if
x
is an element of
A
, then
p
(
x
)
is true’.
More formally, it is shorthand for
∀
x
(
x
∈
A
)=
⇒
p
(
x
)
.
What is the formal way to interpret the abbreviation
∃
x
∈
A, p
(
x
)?
Answer:
∃
x
∈
A, p
(
x
) should be interpreted as
∃
x
(
x
∈
A
)
∧
p
(
x
)
.
This should make sense without much further discussion:
there exists an
x
in the set
A
such that
p
(
x
)
is true
really means
there exists an
x
such that
x
is in
A
and
p
(
x
)
is true.
If this is not formal enough for you, remember that
∃
x, a
(
x
) is equivalent
to
¬
(
∀
x,
¬
a
(
x
) (this was discussed in
§
3.2). By the same token,
∃
x
∈
A, p
(
x
) is
equivalent to
¬
(
∀
x
∈
A,
¬
p
(
x
)), or, formally,
(*)
¬
(
∀
x,
(
x
∈
A
⇒ ¬
p
(
x
))
.
Now recall that
a
=
⇒
b
is equivalent to (
¬
a
)
∨
b
; therefore, (*) is equivalent to
¬
(
∀
x,
(
x
±∈
A
)
∨ ¬
p
(
x
))
therefore to
∃
x,
¬
((
x
±∈
A
)
p
(
x
))
and Fnally to
∃
x,
(
x
∈
A
)
∧
p
(
x
)
by one of De Morgan’s laws.
±
6.2
.
Prove parts (7) and (8) of Theorem 6.1.
Answer:
These statements are
S
±
(
A
∩
B
) = (
S
±
A
)
∪
(
S
±
B
)
S
±
(
A
∪
B
) = (
S
±
A
)
∩
(
S
±
B
)
.
They are set-theoretic incarnations of De Morgan’s laws. ±or instance, consider
the second formula; let
p, q
be the statements
x
∈
A
,
x
∈
B
respectively, and let
x
∈
S
. Then
x
∈
S
±
(
A
∪
B
)
means
x
±∈
A
∪
B
, that is,
¬
(
x
∈
A
∪
B
), that is,
¬
((
x
∈
A
)
∨
(
x
∈
B
)), by deFnition
of union of sets. This is
¬
(
p
∨
q
)
.