Fall 2016, Math 409, Section 502
Solutions to some problems from the seventh assignment
Exercise 1. Let f, g : R → R be continuous functions so that for all q ∈ Q
we have f (q) = g(q). Prove that f = g, i.e. f (x) = g(x) for all x ∈ R.
Solution. Fix x ∈ R. We will show that for an arbitrary positive real number
ε we have |f (x) − g(x)| < ε, which will yield that f (x) = g(x). Fixing ε > 0
and using the definition of continuity on both f and g at x, we obtain δ1 > 0
and δ2 > 0, such that
ε
(∗)
for all y ∈ R with |x − y| < δ1 we have |f (y) − f (x)| < , and
2
ε
(∗∗)
for all y ∈ R with |x − y| < δ2 we have |g(y) − g(x)| < .
2
Set δ = min{δ1 , δ2 }. By the density of Q in R, there exists q ∈ Q with
x − δ < q < x + δ. By (∗) we obtain |f (q) − f (x)| < ε/2 and by (∗∗)
we obtain |g(q) − g(x)| < ε/2. As q ∈ Q, we have f (q) = g(q) and hence
|f (x)−g(x)| = |(f (x)−f (q))+(g(q)−g(x))| 6 |f (x)−f (q)|+|g(q)−g(x)| <
ε/2 + ε/2 = ε.
Second solution. Fix x ∈ R. Then there exists a sequence of rational numbers (qn )n with limn qn = x. By the sequential characterization of continuity
we obtain f (x) = limq f (qn ) and g(x) = limn g(qn ). As f (qn ) = g(qn ) for all
n ∈ N, we conclude f (x) = limn f (qn ) = limn g(qn ) = g(x).
Exercise 2. Let Q = {qn : n ∈ N} be an enumeration of the set of rational
numbers (i.e. for each q ∈ Q there is exactly one n ∈ N with q = qn ). Let
f : R → R be the function defined by the rule
0 if x ∈ R \ Q
f (x) =
1
n if x ∈ Q and x = qn .
Prove that f is continuous at x if and only if x ∈ R \ Q.
Solution. We will show that for all x ∈ R we have limy→x f (y) = 0. By
the definition of f , this will yield that f (x) = limy→x f (x) if and only if
x ∈ R \ Q and the desired result will follow from the limit characterization
of continuity.
Fix x ∈ R and also fix ε > 0. We will find δ > 0 so that for all y ∈ R \ {x}
with |x − y| < δ we have |f (y) − 0| < ε. By the Archimedian property of
R, there exists n0 ∈ N with n0 > 2 and 0 < 1/n0 < ε. Define F = {qn :
1 6 n 6 n0 } \ {x}. Then F is a non-empty finite set not containing x and
therefore δ = min{|x − z| : z ∈ F } exists and it is strictly positive. We will
show that it is the desired number.
Let y ∈ R \ {x} with |y − x| < δ. We will show that |f (y)| < ε. If
y is irrational, then by definition f (y) = 0 and hence the conclusion holds.
Otherwise, y is rational and hence there exists a unique index m with y = qm .
We will show that m > n0 . If this were not the case, then y ∈ {qn : 1 6 n 6
n0 } ∩ (R \ {x}) = F and therefore |x − y| > min{|x − z| : z ∈ F } = δ, which
is absurd. We finally conclude that |f (y)| = |1/m| 6 1/n0 < ε.
Exercise 3. Let a, b be real numbers with a < b and f : [a, b] → [a, b] be a
continuous function. Prove that there exists x ∈ [a, b] with f (x) = x.
Solution. Define g : [a, b] → R with g(x) = x − f (x). Then g is continuous
as the difference of continuous functions. As the image of [a, b] under f is
contained in [a, b], we deduce f (a) > a and therefore g(a) = a − f (a) 6 0.
Similarly, we obtain g(b) = b − f (b) > 0.
If it so happens that g(a) = 0, then f (a) = a and a is the desired number.
Similarly, if g(b) = 0 then f (b) = b and b is the desired number. If neither
of the above happens to be true, then g(a) < 0 < g(b). Applying the
intermediate value theorem to g, we conclude that there if x ∈ (a, b) so that
g(x) = 0, i.e. f (x) = x.
Exercise 4. Let f : R → R be a continuous function and assume that
f [R] ⊆ R \ Q. Prove that f is constant, i.e. there is some c ∈ R with
f (x) = c for all x ∈ R.
Solution. Assume, on the contrary, that f is not constant, i.e. it obtains at
least two different values. That is, there exist x1 , x2 in R with f (x1 ) = c1
and f (x2 ) = c2 , where c1 6= c2 . As x1 and x2 are clearly not equal, we
may assume that x1 < x2 . Using the density of Q in R, choose q ∈ Q
that is between c1 and c2 . By the intermediate value theorem there exists
y ∈ R with x1 < y < x2 so that f (y) = q. This contradicts the assumption
f [R] ⊆ R \ Q and the proof is complete.
√
Exercise 5. Let f : [0, +∞) → R be the function defined by f (x) = x,
for all x ∈ [0, +∞). Prove that f is uniformly continuous.
Solution. Fix ε > 0. Setting δ = ε2 , we will show√that for all non-negative
√
real numbers x, y satisfying |x − y| < δ we have | x − y| < ε.
Let x, y ∈ [0, +∞) with |x − y| < δ. We may assume that x > y, i.e.
x − y > 0. Then
p
√
√
√
√
√
| x − y| = x − y = y + (x − y) − y
q
√ √
√
6 y + (x − y) + 2 y x − y − y
q
√
√
√
= ( y + x − y)2 − y
√
√
√
√
√
= y + x − y − y = x − y < δ = ε.