Ludzie pragną czasami się rozstawać, żeby móc tęsknić, czekać i cieszyć się z powrotem.

16 Let f : D → C and g: E → C be functions defined on D
and E respectively, where D and E are subsets of C satisfying f (D) ⊂ E.
Let w be a limit point of D, and let l be an element of E. Suppose that
lim f (z) = l and that the function g is continuous at l. Then lim g(f (z)) =
z→w
z→w
g(l).
Proof Let ε > 0 be given. Then there exists some η > 0 such that |g(z) −
g(l)| < ε for all z ∈ E satisfying |z − l| < η. But then there exists δ > 0
such that |f (z) − l| < η for all z ∈ D satisfying 0 < |z − w| < δ. Thus if
0 < |z − w| < δ then |f (z) − l| < η, and therefore |g(f (z)) − g(l)| < ε. Hence lim g(f (z)) = g(l).
z→w
13
1.10
The Intermediate Value Theorem
Proposition 1.17 Let f : [a, b] → Z continuous integer-valued function de-
fined on a closed interval [a, b]. Then the function f is constant.
Proof Let
S = {x ∈ [a, b] : f is constant on the interval [a, x]},
and let s = sup S. Now s ∈ [a, b], and therefore the function f is continuous
at s. Therefore there exists some real number δ satisfying δ > 0 such that
|f (x) − f (s)| < 1 for all x ∈ [a, b] satisfying |x − s| < δ. But the function f
2
is integer-valued. It follows that f (x) = f (s) for all x ∈ [a, b] satisfying
|x − s| < δ. Now s − δ is not an upper bound for the set S. Therefore
there exists some element x0 of S satisfying s − δ < x0 ≤ s. But then
f (x) = f (s) = f (x0) = f (a) for all x ∈ [a, b] satsifying s ≤ x < s + δ, and
therefore the function f is constant on the interval [a, x] for all x ∈ [a, b]
satisfying s ≤ x < s + δ. Thus x ∈ [a, b] ∩ [s, s + δ) ⊂ S. In particular s ∈ S.
Now S cannot contain any elements x of [a, b] satisfying x > s. Therefore
[a, b] ∩ [s, s + δ) = {s}, and therefore s = b. This shows that b ∈ S, and thus
the function f is constant on the interval [a, b], as required.
Theorem 1.18 (The Intermediate Value Theorem) Let a and b be real num-
bers satisfying a < b, and let f : [a, b] → R be a continuous function defined
on the interval [a, b]. Let c be a real number which lies between f (a) and f (b)
(so that either f (a) ≤ c ≤ f (b) or else f (a) ≥ c ≥ f (b).) Then there exists
some s ∈ [a, b] for which f (s) = c.
Proof Let c be a real number which lies between f (a) and f (b), and let
gc: R \ {c} → Z be the continuous integer-valued function on R \ {c} de-
fined such that gc(x) = 0 whenever x < c and gc(x) = 1 if x > c. Suppose
that c were not in the range of the function f . Then the composition func-
tion gc ◦ f : [a, b] → R would be a continuous integer-valued function defined
throughout the interval [a, b]. This function would not be constant, since
gc(f (a)) 6= gc(f (b)). But every continuous integer-valued function on the in-
terval [a, b] is constant (Proposition 1.17). It follows that every real number c
lying between f (a) and f (b) must belong to the range of the function f , as
required.
Corollary 1.19 Let f : [a, b] → [c, d] be a strictly increasing continuous func-
tion mapping an interval [a, b] into an interval [c, d], where a, b, c and d are
real numbers satisfying a < b and c < d. Suppose that f (a) = c and f (b) = d.
Then the function f has a continuous inverse f −1: [c, d] → [a, b].
14
Proof Let x1 and x2 be distinct real numbers belonging to the interval [a, b]
then either x1 < x2, in which case f (x1) < f (x2) or x1 > x2, in which case
f (x1) > f (x2). Thus f (x1) 6= f (x2) whenever x1 6= x2. It follows that the
function f is injective. The Intermediate Value Theorem (Theorem 1.18)
ensures that f is surjective. It follows that the function f has a well-defined
inverse f −1: [c, d] → [a, b]. It only remains to show that this inverse function
is continuous.
Let y be a real number satisfying c < y < d, and let x be the unique real
number such that a < x < b and f (x) = y. Let ε > 0 be given. We can then
choose x1, x2 ∈ [a, b] such that x − ε < x1 < x < x2 < x + ε. Let y1 = f (x1)