1
Solution to Problem 2.
Deﬁne
X
0
=
X

g
(
Y
) and
Y
0
=
Y

f
(
X
).
For any
x
∈
X
0
, consider the sequence
x, g
(
f
(
x
))
,g
(
f
(
g
(
x
))))
,....
Deﬁne
A
⊂
X
to be all elements of such sequences.
y
∈
Y
0
, consider the sequence
g
(
y
)
(
f
(
g
(
y
))
(
f
(
g
(
f
(
g
(
y
))))
Deﬁne
B
⊂
X
to be all elements of such sequences.
Finally let
C
=
X

(
A
∪
B
). Then
A
∩
B
=
/
±
,
A
∩
C
=
/
±
,
B
∩
C
=
/
±
, and
of course
X
=
A
∪
B
∪
C
. Deﬁne
h
(
x
)=
f
(
x
)i
f
x
∈
A
,
g

1
(
x
f
x
∈
B
,
f
(
x
f
x
∈
C
.
Then
h
:
X
→
Y
is bijective.
It is kind of fun to apply this to the case
X
=[

1
,
1],
Y
=(

1
,
1) and
f
(
x
x/
2
(
y
y
.
In that case,
(1)
X
0
=
{
1
,
1
}
, and
A
=
{±
1
,
±
1
/
2
,
±
1
/
4
,...
}
.On
A
, we set
h
(
x
x/
2;
(2)
Y
0

1
,
1)

[

1
/
2
,
12], and
B

1
,
1)

(
A
∪{
0
}
). On
B
, we set
h
(
x
x
;
(3)
C
=
{
0
}
, and
h
(0)=0.
.
There are many entertaining cases, like setting up a bijective correspondance
between
R
and
R
2
.
Note that this construction is completely constructive. Suppose that instead
of supposing
f
and
g
injective, we had supposed them surjective. I have been
asking around, more particularly our logician Justin Moore, and it turns out