Appendix to Chapter III
311
tends to 0, which shows that, in X, one has x = lim xp: every x E X is thus
a limit of points of X, as every real number is the limit of rational numbers.
It remains to prove that X is a complete metric space; an exercise consisting of applying the definitions and the triangle inequality, as in every case
where one knows no more.
4 - Continuous functions
The concept of continuity extends immediately to functions defined on a
metric space X (so also to functions defined on any subset E of X) and with
values in another metric space Y. Such a function f is said to be continuous
at a point a of E if it satisfies these clearly equivalent conditions:
(Cl) for every r > 0 there exists an r' > 0 such that, for x E E,
d(a, x) < r' ===> d[J(a), f(x)] < r;
(C2) for every ball BeY with centre f(a) there exists a ball B' c X with
centre a such that f(B' n E) c B;
(C3) f(x) tends to f(a) when x tends to a:
lim f(x) = f(a).
x-a
A function or map defined on X is said to be continuous on X, or continuous
for short, if it is so at every point of X.
There is a simple relation between the concepts of continuous function
and of an open set: a map f of X into Y is continuous if and only if, for
every open V in Y, the inverse image U = f-l(V) of V under f is open
inX.
Necessity of the condition: let a E U and b = f(a) E V. If V is open
it contains a ball B = B(b, r). If f is continuous at a by (2) there exists a
ball B' with centre a such that f(B') C B. One has B' C U by definition of
U = f-l(V) (Chap. I), so that U is open.
Sufficiency of the condition: consider an a EX, put b = f (a), and consider
an open ball B = B(b, r) in Y. Let U be its inverse image under f. By
hypothesis, U is open. Since U contains a, it also contains a ball B' with
centre a. One has f(B') C B, whence continuity by (C2).
The preceding result immediately yields a result as trivial as it is fundamental: Let X, Y and Z be metric spaces, f a map of X into Y and 9 a map
of Y into Z. Suppose that f and 9 are continuous. Then the composite map
h = 9 0 f of X into Z is continuous.
For let W be open in Z. The set V = g-I(W) is open in Y since 9
is continuous. Since f is also continuous, U = f- 1 (V) is open in X. Now
U = h-l(W), qed.
311
tends to 0, which shows that, in X, one has x = lim xp: every x E X is thus
a limit of points of X, as every real number is the limit of rational numbers.
It remains to prove that X is a complete metric space; an exercise consisting of applying the definitions and the triangle inequality, as in every case
where one knows no more.
4 - Continuous functions
The concept of continuity extends immediately to functions defined on a
metric space X (so also to functions defined on any subset E of X) and with
values in another metric space Y. Such a function f is said to be continuous
at a point a of E if it satisfies these clearly equivalent conditions:
(Cl) for every r > 0 there exists an r' > 0 such that, for x E E,
d(a, x) < r' ===> d[J(a), f(x)] < r;
(C2) for every ball BeY with centre f(a) there exists a ball B' c X with
centre a such that f(B' n E) c B;
(C3) f(x) tends to f(a) when x tends to a:
lim f(x) = f(a).
x-a
A function or map defined on X is said to be continuous on X, or continuous
for short, if it is so at every point of X.
There is a simple relation between the concepts of continuous function
and of an open set: a map f of X into Y is continuous if and only if, for
every open V in Y, the inverse image U = f-l(V) of V under f is open
inX.
Necessity of the condition: let a E U and b = f(a) E V. If V is open
it contains a ball B = B(b, r). If f is continuous at a by (2) there exists a
ball B' with centre a such that f(B') C B. One has B' C U by definition of
U = f-l(V) (Chap. I), so that U is open.
Sufficiency of the condition: consider an a EX, put b = f (a), and consider
an open ball B = B(b, r) in Y. Let U be its inverse image under f. By
hypothesis, U is open. Since U contains a, it also contains a ball B' with
centre a. One has f(B') C B, whence continuity by (C2).
The preceding result immediately yields a result as trivial as it is fundamental: Let X, Y and Z be metric spaces, f a map of X into Y and 9 a map
of Y into Z. Suppose that f and 9 are continuous. Then the composite map
h = 9 0 f of X into Z is continuous.
For let W be open in Z. The set V = g-I(W) is open in Y since 9
is continuous. Since f is also continuous, U = f- 1 (V) is open in X. Now
U = h-l(W), qed.