105
Fixed-Point Theory for Generalized Metric Spaces
Now let f be strictly contracting on X, and assume that x and y are two
distinct fixed points of f . Then we get �(x, y) = �(f (x), f (y)) < �(x, y), which
is impossible. So the fixed point of f is unique in this case.
•
We next give an iterative proof of a special case of Theorem 4.5.1.
4.5.2 Theorem Let (X, �, Γ) be a spherically complete, dislocated generalized ultrametric space with Γ = {2
−α | α ≤ γ} for some ordinal γ. We order
Γ by 2
−α < 2
−β if and only if β < α, and denote 2
−γ by 0. Thus, Γ is the set
Γ γ+1 of Remark 4.3.2. If f : X → X is any strictly contracting function on
X, then f has a unique fixed point.
Proof: Let x ∈ X. Then we have f (x) ∈ f (X) and �(f (x), x) ≤ 2
−0 ,
since 2
−0 is the maximum possible distance between any two points in X.
Now, �(f (f (x)), f (x)) ≤ 2
−1 ≤ 2
−0 since f is strictly contracting, and by
(U4), it follows that �(f
2 (x), x) ≤ 2
−0 . By the same argument, we obtain
�(f
3 (x), f
2 (x)) ≤ 2
−2 ≤ 2
−1 , and therefore �(f
3 (x), f (x)) ≤ 2
−1 . In fact, an
easy induction argument along these lines shows that �(f
n+1 (x), f
m (x)) ≤
2
−m for m ≤ n. Again by (U4), we obtain that the sequence of balls of the
form B 2 −n (f
n (x)) is a descending chain (with respect to set-inclusion) if n is
increasing and, therefore, has non-zero intersection B ω since X is assumed to
be spherically complete. We therefore conclude that there is x ω ∈ B ω with
�(x ω , f
n (x)) ≤ 2
−n for each n ∈ N.
Next, for each n ∈ N, we now argue as follows. Since �(f (x ω ), f
n+1 (x)) <
�(x ω , f
n (x)) ≤ 2
−n and also �(x ω , f
n+1 (x)) ≤ 2
−(n+1) ≤ 2
−n , we therefore
obtain �(f (x ω ), x ω ) ≤ 2
−n . Since this is the case for all n ∈ N, it follows that
�(f (x ω ), x ω ) ≤ 2
−ω .
It is straightforward to cast the above observations into a transfinite induction argument, and we obtain the following construction. Choose x ∈ X
arbitrarily. For each ordinal α ≤ γ, we define f
α (x) as follows. If α is a successor ordinal, then f
α (x) = f (f
α−1 (x)), as usual. If α is a limit ordinal, then we
choose f
α (x) as some x α which has the property that �(x α , f
β (x)) ≤ 2
−β , noting that the existence of such an x α is guaranteed by spherical completeness
of X.
The resulting transfinite sequence f
α (x) has the property that, for all
α ≤ γ, �(f
α+1 (x), f
α (x)) ≤ 2
−α . Consequently, �(f
γ+1 (x), f
γ (x)) = 2
−γ = 0,
and therefore f
γ (x) must be a fixed point of f .
Finally, x γ = f
γ (x) can be the only fixed point of f . To see this, suppose
y = x γ is another fixed point of f . Then we obtain �(y, x γ ) = �(f (y), f (x γ )) <
�(y, x γ ), from the fact that f is strictly contracting, and this is impossible. •
Fixed-Point Theory for Generalized Metric Spaces
Now let f be strictly contracting on X, and assume that x and y are two
distinct fixed points of f . Then we get �(x, y) = �(f (x), f (y)) < �(x, y), which
is impossible. So the fixed point of f is unique in this case.
•
We next give an iterative proof of a special case of Theorem 4.5.1.
4.5.2 Theorem Let (X, �, Γ) be a spherically complete, dislocated generalized ultrametric space with Γ = {2
−α | α ≤ γ} for some ordinal γ. We order
Γ by 2
−α < 2
−β if and only if β < α, and denote 2
−γ by 0. Thus, Γ is the set
Γ γ+1 of Remark 4.3.2. If f : X → X is any strictly contracting function on
X, then f has a unique fixed point.
Proof: Let x ∈ X. Then we have f (x) ∈ f (X) and �(f (x), x) ≤ 2
−0 ,
since 2
−0 is the maximum possible distance between any two points in X.
Now, �(f (f (x)), f (x)) ≤ 2
−1 ≤ 2
−0 since f is strictly contracting, and by
(U4), it follows that �(f
2 (x), x) ≤ 2
−0 . By the same argument, we obtain
�(f
3 (x), f
2 (x)) ≤ 2
−2 ≤ 2
−1 , and therefore �(f
3 (x), f (x)) ≤ 2
−1 . In fact, an
easy induction argument along these lines shows that �(f
n+1 (x), f
m (x)) ≤
2
−m for m ≤ n. Again by (U4), we obtain that the sequence of balls of the
form B 2 −n (f
n (x)) is a descending chain (with respect to set-inclusion) if n is
increasing and, therefore, has non-zero intersection B ω since X is assumed to
be spherically complete. We therefore conclude that there is x ω ∈ B ω with
�(x ω , f
n (x)) ≤ 2
−n for each n ∈ N.
Next, for each n ∈ N, we now argue as follows. Since �(f (x ω ), f
n+1 (x)) <
�(x ω , f
n (x)) ≤ 2
−n and also �(x ω , f
n+1 (x)) ≤ 2
−(n+1) ≤ 2
−n , we therefore
obtain �(f (x ω ), x ω ) ≤ 2
−n . Since this is the case for all n ∈ N, it follows that
�(f (x ω ), x ω ) ≤ 2
−ω .
It is straightforward to cast the above observations into a transfinite induction argument, and we obtain the following construction. Choose x ∈ X
arbitrarily. For each ordinal α ≤ γ, we define f
α (x) as follows. If α is a successor ordinal, then f
α (x) = f (f
α−1 (x)), as usual. If α is a limit ordinal, then we
choose f
α (x) as some x α which has the property that �(x α , f
β (x)) ≤ 2
−β , noting that the existence of such an x α is guaranteed by spherical completeness
of X.
The resulting transfinite sequence f
α (x) has the property that, for all
α ≤ γ, �(f
α+1 (x), f
α (x)) ≤ 2
−α . Consequently, �(f
γ+1 (x), f
γ (x)) = 2
−γ = 0,
and therefore f
γ (x) must be a fixed point of f .
Finally, x γ = f
γ (x) can be the only fixed point of f . To see this, suppose
y = x γ is another fixed point of f . Then we obtain �(y, x γ ) = �(f (y), f (x γ )) <
�(y, x γ ), from the fact that f is strictly contracting, and this is impossible. •