2
b)
C
×
B
×
A
=
{
(0
,x,a
)
,
(0
,x,b
)
,
(0
,x,c
)
,
(0
,y,a
)
,
(0
,y,b
)
,
(0
,y,c
)
,
(1
,x,a
)
,
(1
,x,b
)
,
(1
,x,c
)
,
(1
,y,a
)
,
(1
,y,b
)
,
(1
,y,c
)
}
.
c)
C
×
A
×
B
=
{
(0
,a,x
)
,
(0
,a,y
)
,
(0
,b,y
)
,
(0
,b,y
)
,
(0
,c,x
)
,
(0
,c,y
)
,
(1
,a,x
)
,
(1
,a,y
)
,
(1
,b,x
)
,
(1
,b,y
)
,
(1
,c,y
)
,
(1
,c,y
)
}
.
d)
B
×
B
×
B
=
{
(
x,x,x
)
,
(
x,x,y
)
,
(
x,y,x
)
,
(
x,y,y
)
,
(
y,x,x
)
,
(
y,x,y
)
,
(
y,y,x
)
,
(
y,y,y
)
}
.
38.
Let
S
=
{
sets
x

x
6∈
x
}
.
a)
Show that the assumption that
S
∈
S
leads to a contradiction.
Proof.
Suppose, by way of contradiction, that
S
∈
S
. Then
S
does not satisfy the criterion for membership in
S
, since
S
consists
only of those sets which are not members of themselves. Hence we
conclude
S
6∈
S
, contradicting our assumption.
±
b)
Show that the assumption that
S
6∈
S
leads to a contradiction.
Proof.
Suppose, by way of contradiction, that
S
6∈
S
. Then
S
satisﬁes the criterion for membership in
S
, since it is not a member
of itself. Hence we conclude
S
∈
S
, contradicting our assumption.
±
Note to the students: The deﬁnition of
S
is paradoxical. This
problem should absolutely disturb you. I encourage you to look up
Russell’s Paradox on Wikipedia. In short, the way around this
is to say that
S
cannot be a set. So what follows is an alternate
solution to parts a) and b).
Proof.
Suppose by way of contradiction that
S
is a set.
S
is deﬁned
by the proposition
∀
x
(
x
∈
S
⇔
x
6∈
x
), where we are quantifying
over sets. We apply universal instantiation with
x
=
S
to obtain
S
∈
S
⇔
S
6∈
S
. This proposition is false, a contradiction. We
conclude that
S
is not a set.
±