Solution #1: If f : D → R is continuous at x0 , then for all > 0 there exists a δ > 0 such that
whenever x ∈ D and |x − x0 | < δ, then |f (x) − f (x0 )| < . Since A, B are subsets of D, this
holds in particular whenever x ∈ A or x ∈ B (and |x − x0 | < δ), proving that f : A → R and
f : B → R are continuous at x0 . Vice versa, if f : A → R and f : B → R are continuous at x0 ,
then for all > 0 there exist numbers δA , δB > 0 such that if either x ∈ A, |x − x0 | < δA , or
x ∈ B, |x − x0 | < δB , then |f (x) − f (x0 )| < . Now define δ := min{δA , δB } > 0. If x ∈ D and
|x − x0 | < δ, then it follows that either x ∈ A, |x − x0 | < δ 6 δA , or x ∈ B, |x − x0 | < δ 6 δB ,
and hence |f (x) − f (x0 )| < in both cases, as required for the continuity of f at x0 .
Solution #2: Recall that a function f : D → R is called α-Hölder if there exists a constant C
such that |f (x) − f (y)| 6 C|x − y|α for all x, y ∈ D. Using the -δ-characterization, it is very
easy to see that f is uniformly continuous: given > 0, we need to find some δ > 0 depending
only on such that whenever x, y ∈ D and |x−y| < δ, it follows that |f (x)−f (y)| < ; using the
1
α
) . Alternatively, we can work with the definition
above, it then clearly suffices to take δ = ( 2C
∞
of uniform continuity based on sequences: given any two sequences {xn }∞
n=1 , {yn }n=1 ⊂ D with
|xn − yn | → 0, we need to show that |f (xn ) − f (yn )| → 0; so it suffices to show that if {tn }∞
n=1
is a sequence with tn > 0 for all n ∈ N and tn → 0, then tαn → 0; but this is almost immediate
from the definition of a limit: given > 0, we need to find N ∈ N such that tαn < for n > N ;
1
for this, simply use the assumption that tn → 0 to find N ∈ N such that tn < α for n > N .
Note: Both solutions crucially rely on the fact that the function x 7→ xα is increasing. To prove
this fact, notice that, strictly speaking, we have defined
xα only for α ∈ Q; in this case, write
√
m
n
α = n with m, n ∈ N and use that the functions y = x and z = y m are increasing (which we
already know), so that x1 < x2 implies y1 < y2 and hence z1 < z2 , i.e. xα1 < xα2 as desired.
Solution #3: If f was not constant, then there would exist a, b ∈ [0, 1] with f (a) 6= f (b). Then
a 6= b, so we can assume without loss of generality that a < b. Since the irrationals are dense
(Corollary 1.10), there exists an irrational c between f (a) and f (b). It follows from the IVT
that there exists an x0 ∈ (a, b) such that f (x0 ) = c, contradicting the fact that f (x0 ) ∈ Q.
√
Solution #4: Given x, , define S := {δ > 0 : f (x + δ) < f (x) + } and define δ0 := 2 + 2 x.
Then we need to prove two things: (1) If δ < δ0 then δ ∈ S (this tells us that no number less
than δ0 can be an upper bound of S). (2) If δ > δ0 then δ 6∈ S (this shows that δ0 is an upper
bound of S). Given (1) and (2), it then follows that δ0 = sup S, the least upper bound of S, as
desired. The proofs of (1) and (2) are very
psimilar,√using the√fact that f is strictly increasing:
(1) If δ < δ0 then f (x + δ) < f (x + δ0 ) = x + 2 x + 2 = x + = f (x) + , so that δ ∈ S.
(2) If δ > δ0 then f (x + δ) > f (x + δ0 ) = f (x) + , so that δ 6∈ S.
Bonus Solution: Let x, y ∈ D and define m := x+y
. Notice that |x − m| = |y − m| = |x−y|
. By
2
2
the triangle inequality and the definition of α-Hölder,
α
α
|x − y|
|x − y|
|f (x) − f (y)| 6 |f (x) − f (m)| + |f (m) − f (y)| 6 C
+C
= qC|x − y|α ,
2
2
where q := 21−α . Repeating this argument n times, we find that |f (x) − f (y)| 6 q n C|x − y|α .
Now notice that q < 1 because α > 1. Letting n → ∞, we deduce that |f (x) − f (y)| = 0.