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Math 351 Handout 9: Wednesday February 10, 2023

Deﬁnitions: Let f : D → R be a function with domain D ⊆ R. We say that f is

continuous at x ∈ D if for every sequence {x

n

} in D which converges to x, the sequence

{f(x

n

)} converges to f(x).

We say that a function f : D → R is continuous if it is continuous at every point in D.

Fact: The function f : R → R given by f(x) = 2x is continuous at x = 1.

Proof: Suppose that {x

n

} is a sequence in R which converges to 1. Then, {f(x

n

)} = {2x

n

}.

We proved previously that for any real number k, if x

n

converges to x then kx

n

converges

to kx. So in this case {f(x

n

)} converges to 2 · 1 = f (1). Therefore, if {x

n

} is a sequence in

R which converges to 1, then {f(x

n

)} converges to f(1), so f is continuous at 1.

In-class exercises:

(1) If c ∈ R, prove that the constant function f : R → R given by f(x) = c is continuous

at x = 1.

(2) If c ∈ R, prove that the constant function f : R → R given by f(x) = c is continuous.

(3) Prove that the function f : R → R given by f(x) = x is continuous.

(4) Prove that if f : D → R and g : D → R are continuous at a point y ∈ D, then the

function f + g : D → R given by (f + g)(x) = f (x) + g(x) for all x ∈ D is continuous

at y.

(5) Prove that if f : D → R and g : D → R are continuous at a point y ∈ D, then the

function f g : D → R given by (fg)(x) = f (x)g(x) for all x ∈ D is continuous at y.

(6) Prove that any polynomial function is continuous.

(7) Deﬁnition: A subset U of R is open if for any x ∈ U, there exists ϵ > 0 so that

(x − ϵ, x + ϵ) ⊂ U.

Prove that the open interval (−1, 1) is open. Generalize your argument to show

that if a < b, then (a, b) is open.

(8) Prove that if a ≤ b, then [a, b] is not open. (In this notation, [a, a] = {a}.)

(9) Prove that if U is open, then its complement R \ U = {x ∈ R | x /∈ U} is closed.

(10) Prove that if C is closed, then its complement R \ C is open.

1

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