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Math 0420 Homework 4
Exercise 3.2.1 Using the definition of continuity directly prove that f :
R → R defined by f (x) := x2 is continuous.
Scratch Work: we know, |x − c| < δ,
|f (x) − f (c)| = |x2 − c2 |
= |x − c||x + c|
≤ (|x| + |c|)|x − c|
≤ (|c| + 1 + |c|)|x − c|
In the last step, we let δ ≤ 1
Now we might hope that (|c| + 1 + |c|)|x − c| < (2|c| + 1)δ = .
That is,
δ = min{1,
}
2|c| + 1
},
Proof. For any c ∈ R, let > 0, choose δ = min{1. 2|c|+1
Then for any x such that |x − c| < δ, we have |x − c| < 1 then,
|x| − |c| ≤ |x − c| < 1 ⇒ |x| < |c| + 1
Therefore,
|f (x) − f (c)| = |x2 − c2 |
= |x − c||x + c|
≤ (|x| + |c|)|x − c|
≤ (|c| + 1 + |c|)|x − c|
< (2|c| + 1)δ
= (2|c| + 1)
2|c| + 1
=
By the definition of continuity, we can f (x) is continuous.
x if x is rational
using
x2 if x is irrational
the definition of continuity directly prove that f if continuous at 1 and discontinuous at 2.
Exercise 3.2.3 let f : R → Rdefined byf (x) := {
Scratch Work: To show the continuity at 1, we need to find appropriate
δ.
if x is rational, |f (x) − f (1)| = |x − 1| < δ, we only need δ = .
if x is irrational, |f (x) − f (1)| = |x2 − 1| = |x − 1||x + 1| ≤ |x − 1|(|x| + 1) <
3|x − 1| = 3δ, we need δ = min{1, 3 }(The idea here is similar to exercise 3.2.1)
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Thus, we want δ = min{1, , 3 }.
Proof. For the continuity at 1, Let > 0, choose δ = min{1, 3 }, then,
|x − 1| < δ =⇒ |x + 1| ≤ |x − 1| + 2 < 3
Hence,
|f (x) − 1| = {
if x is rational |x − 1|
≤ 3|x − 1| < 3δ = if x is irrational |x2 − 1| = |x + 1||x − 1|
By the definition of continuity, we have f (x) is continuous at 1.
For the discontinuity at 2, First let me write the definition of discontinuity
here.
Definition: f : R → R is discontinuous at c, if there exists > 0, for any
δ > 0, there exists x satifying |x − c| < δ, |f (x) − f (c)| ≥ .
Now take = 1, then for any δ, we can find a irrational number x in (2, 2+δ),
then we have,
|f (x) − f (2)| = |x2 − 2| = x2 − 2 > 22 − 2 = 4 − 2 = 2 > 1
Then, by the definition of discontinuity, we have f is discontinuous at 2.
Exercise 3.2.9 Give an example of functions f : R → R and g : R → R
such that the function defined by h(x) := f (x) + g(x) is continuous but f and
g are not continuous. Can you find f and g that are nowhere continuous, but h
is a continuous function.
0 if x > 0
1 if x > 0
Solution : let f (x) := {
and g(x) := {
1 if x ≤ 0
0 if x ≤ 0
Then both f (x) and g(x) are discontinuous at 0, However, h(x) := f (x) +
g(x) = 1 for all x. which is continuous everywhere.
1 if x is rational
−1 if x is rational
and g(x) := {
0 if x is irrational
0 if x is irrational
Then both f (x) and g(x) are discontinuous everywhere, However, h(x) :=
f (x) + g(x) = 0 for all x. which is continuous everywhere.
let f (x) := {
Exercise 3.2.10 let functions f : R → R and g : R → R be continuous
functions. Suppose for all rational numbers r, f (r) = g(r). Show that f (x) =
g(x) for all x.
Proof. Let c ∈ R, Consider the sequence of intervals [c, c + n1 ] for all n ∈
N. Select a rational number xn from each interval and we will get a rational
sequence {xn } which converges to c.
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Then f (xn ) = g(xn ) for all n ∈ N.
Also by the continuity of f (x) and g(x), we have
lim f (xn ) = lim f (x) = f (c)
n→∞
x→c
and
lim g(xn ) = lim g(x) = g(c).
n→∞
x→c
Therefore, f (c) = g(c), ∀c ∈ R.
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