“real: chapter_02” — 2011/5/22 — 23:05 — page 10 — #10
2-10 Real Analysis
We now claim that this f is one-to-one.
Let f (x) = f (y)(x, y ∈ S ∪ B).
Case 1: x, y ∈ P or x, y ∈ S\P or x, y ∈ B.
In this case, it is clear that x = y.
Case 2: x ∈ S\P, y ∈ P or x ∈ P, y ∈ S\P.
This cannot happen because in this case x = f (x) = f (y) = x
2i
for
some i and x ∈ S\P and x
i
∈ P or vice versa.
Case 3: x ∈ P, y ∈ B,orx ∈ B, y ∈ P.
We merely assume that x ∈ P, y ∈ B. The other case can be handled
similarly. In this case, x
2n
= f (x
n
) = f (x) = f (y) = f (y
m
) = x
2m−1
for some n and m. In this case also x
2n
= x
2m−1
, a contradiction.
Case 4: x ∈ S\P, y ∈ B (or x ∈ B, y ∈ S\P). In this case also, we
have x = f (x) = f (y) = f (y
n
) = x
2n−1
∈ P, a contradiction to the
fact that