95
Fixed-Point Theory for Generalized Metric Spaces
(1.2) If y ∈ T (x) and f (y) = a, then f (y) ∈ T (x).
(1.3) If y ∈ T (x) and y = a, then f
−1 (y) ⊆ T (x).
It is clear that the intersection of the family of all sets closed under these rules
is itself closed under these rules, and hence T (x) exists. Moreover, it is also
clear that each of the sets T (x) is non-empty. Now let T = {T (x) | x ∈ X},
and observe the following facts.
(i) T (a) = {a}. To see this, we note that (1.1), (1.2), and (1.3) are all true
relative to the set {a}. Therefore, by minimality, we have T (a) = {a}.
(ii) If x = a, then a ∈ T (x), and so T (a)∩T (x) = ∅. Hence, either T (a) and
T (x) are equal or they are disjoint. To see this, suppose x = a, and consider
rule (1.3). Clearly, we cannot have a ∈ f
−1 (x); otherwise, f (a) = x, and hence
a = x, which is a contradiction. Thus, rules (1.2) and (1.3) applied repeatedly
and starting with x never place a in T (x), and, by minimality, the process
just described generates T (x).
(iii) If T (x) = T (a) and T (y) = T (a), then either T (x) and T (y) are equal
or they are disjoint. To see this, suppose z ∈ T (x)∩T (y). Then the rules (1.1),
(1.2), and (1.3) under repeated application starting with z force T (x) = T (y).
Thus, the collection T is a partition of X.
(2) We next inductively define a mapping l : T → Z ∪ {∞} on each T ∈ T .
(2.1) We set l(a) = ∞, and this defines l on T = T (a). If T = T (a), we
choose an arbitrary x ∈ T and set l(x) = 0 (of course, x = a) and proceed as
follows.
(2.2) For each y ∈ T with f (y) = a and l(y) = k, let l(f (y)) = k + 1.
(2.3) For each y ∈ T with l(y) = k, let l(z) = k − 1 for all z ∈ f
−1 (y).
We will henceforth assume that all this is done for every T ∈ T so that l is a
function defined on all of X. It is clear that the mapping l is well-defined since
(X, τ ) is a T 1 space.
9 For, if there is a cycle in the sequence f
n (x) of iterates
for some x ∈ X, then we can arrange for some element y in this sequence to
be frequently not in some neighbourhood of a, using the fact that X is T 1 ,
which contradicts the convergence of the sequence f
n (x) to a.
(3) Define a mapping ι : Z ∪ {∞} → R by
0
if k = ∞,
ι(k) =
2
−k
otherwise.
Furthermore, define a mapping δ : X × X → R by
δ(x, y) = max{ι(l(x)), ι(l(y))}
and a mapping d : X × X → R by
δ(x, y) if x = y,
d(x, y) =
0
if x = y.
9 We can weaken the requirement of τ being T 1 by replacing it with the following condition: for every y ∈ X there exists an open neighbourhood U of a with y � ∈ U .
Fixed-Point Theory for Generalized Metric Spaces
(1.2) If y ∈ T (x) and f (y) = a, then f (y) ∈ T (x).
(1.3) If y ∈ T (x) and y = a, then f
−1 (y) ⊆ T (x).
It is clear that the intersection of the family of all sets closed under these rules
is itself closed under these rules, and hence T (x) exists. Moreover, it is also
clear that each of the sets T (x) is non-empty. Now let T = {T (x) | x ∈ X},
and observe the following facts.
(i) T (a) = {a}. To see this, we note that (1.1), (1.2), and (1.3) are all true
relative to the set {a}. Therefore, by minimality, we have T (a) = {a}.
(ii) If x = a, then a ∈ T (x), and so T (a)∩T (x) = ∅. Hence, either T (a) and
T (x) are equal or they are disjoint. To see this, suppose x = a, and consider
rule (1.3). Clearly, we cannot have a ∈ f
−1 (x); otherwise, f (a) = x, and hence
a = x, which is a contradiction. Thus, rules (1.2) and (1.3) applied repeatedly
and starting with x never place a in T (x), and, by minimality, the process
just described generates T (x).
(iii) If T (x) = T (a) and T (y) = T (a), then either T (x) and T (y) are equal
or they are disjoint. To see this, suppose z ∈ T (x)∩T (y). Then the rules (1.1),
(1.2), and (1.3) under repeated application starting with z force T (x) = T (y).
Thus, the collection T is a partition of X.
(2) We next inductively define a mapping l : T → Z ∪ {∞} on each T ∈ T .
(2.1) We set l(a) = ∞, and this defines l on T = T (a). If T = T (a), we
choose an arbitrary x ∈ T and set l(x) = 0 (of course, x = a) and proceed as
follows.
(2.2) For each y ∈ T with f (y) = a and l(y) = k, let l(f (y)) = k + 1.
(2.3) For each y ∈ T with l(y) = k, let l(z) = k − 1 for all z ∈ f
−1 (y).
We will henceforth assume that all this is done for every T ∈ T so that l is a
function defined on all of X. It is clear that the mapping l is well-defined since
(X, τ ) is a T 1 space.
9 For, if there is a cycle in the sequence f
n (x) of iterates
for some x ∈ X, then we can arrange for some element y in this sequence to
be frequently not in some neighbourhood of a, using the fact that X is T 1 ,
which contradicts the convergence of the sequence f
n (x) to a.
(3) Define a mapping ι : Z ∪ {∞} → R by
0
if k = ∞,
ι(k) =
2
−k
otherwise.
Furthermore, define a mapping δ : X × X → R by
δ(x, y) = max{ι(l(x)), ι(l(y))}
and a mapping d : X × X → R by
δ(x, y) if x = y,
d(x, y) =
0
if x = y.
9 We can weaken the requirement of τ being T 1 by replacing it with the following condition: for every y ∈ X there exists an open neighbourhood U of a with y � ∈ U .