MATH 2200: Quiz #2 – SOLUTIONS

/5 Problem 1: Prove the following: If f is continuous on D and k ∈ R, then kf is continuous on D.

Suppose f : D → R is continuous at c ∈ D, and let k ∈ R. Let ε > 0 be given. It suffices to show∃δ > 0 such that |x− c| < δ =⇒ |(kf)(x)− (kf)(c)| < ε.

If k = 0 then kf = 0 so|(kf)(x)− (kf)(c)| = |0− 0| = 0 < ε

for any δ > 0, so we are done.

So suppose k 6= 0. Then we have ε/|k| > 0, and since f is continuous at c we can choose δ > 0 suchthat

|x− c| < δ =⇒ |f(x)− f(c)| < ε

|k|.

Suppose |x− c| < δ. Then

|(kf)(x)− (kf)(c)| = |kf(x)− kf(c)|= |k||f(x)− f(c)|

< |k| · ε|k|

= ε.

Thus kf is continuous at c. Since c ∈ D was arbitrary we have that kf is continouous on D.

/5 Problem 2: Given an example of a function f : R→ R that is continuous at exactly one point. Sketcha proof.

13/04/2017

f(x) =

{x if x is rational

0 if x is irrational

Claim 1: f is continouous at 0.

Proof. Let ε > 0 be given. Let δ = ε, and suppose |x− 0| < δ. Then

|f(x)− f(0)| = |f(x)− 0| = |f(x)|≤ |x| (by defn. of f)

< δ (by assumption)

= ε (by choice of δ).

Thus, ∀ε > 0, ∃δ > 0 such that |x− 0| < δ =⇒ |f(x)− f(0)| < ε. So f is continuous at 0.

Claim 2: ∀c 6= 0, f is discontinuous at c.

Proof. Suppose c 6= 0 is irrational, and let (xn) be a sequence of rationals with xn → c (such asequence exists, by the density of rationals in R). Then

lim f(xn) = limxn (since each xn is rational)

= c (by choice of (xn))

6= 0 (by assumption).

= f(c) (since c is irrational).

Thus lim f(xn) 6= f(c) so f is not continuous at c.

Suppose, on the other hand, that c 6= 0 is rational, and let (xn) be a sequence of irrationals withxn → c (such a sequence exists, by the density of irrationals in R). Then

lim f(xn) = lim 0 = 0 (since each xn is irrational)

6= c (by assumption)

= f(c) (since c is rational).

Thus lim f(xn) 6= f(c) so f is not continuous at c.

13/04/2017