MA2224 (Lebesgue integral) Tutorial sheet 1
[January 22, 2016]
Name: Solutions
1. If S = {s1 , s2 , . . .} is a countably infinite set and t ∈
/ S, show that S ∪{t} is also countably
infinite.
Solution: We can list the elements of S ∪ {t} by inserting t at the beginning, that is as
S ∪ {t} = {t, s1 , s2 , . . .}.
2. Show that N × N is countable. [Hint: how did we show that Q is countable?]
Solution: We can tabulate the elements of N × N, where the tuples (1, 1), (1, 2), (1, 3), . . .
with first entry 1 form the first row, those with first entry 2 the second row and so on.
S × T = { (1, 1), (1, 2), (1, 3), (1, 4) . . .
(2, 1), (2, 2), (2, 3), (2, 4) . . .
..
.
}
and then list those in a single list by using a zig-zag type pattern
N × N = {(1, 1), (2, 1), (1, 2), (1, 3), (2, 2), (3, 1), (4, 1), (3, 2), . . .}
3. Let f : [0, 2π] → R be given by f (x) = cos(2x). Find f −1 ([2, ∞)) and f −1 ([0, 1/2]).
Solution: f −1 ([2, ∞)) = {x ∈ [0, 2π] : f (x) ∈ [2, ∞)} = {x ∈ [0, 2π] : cos(2x) ∈
[2, ∞)} = ∅ because cos(θ) ∈ [−1, 1] always.
For the other case f −1 ([0, 1/2]), a graph helps
We have cos(2x) = 1/2 at 2x = π/3, 2x = 2π − π/3, 2x = 2π + π/3 and 2x = 4π − π/3,
that is at x = π/6, x = π − π/6, x = π + π/6 and x = 2π − π/6.
We have cos(2x) = 0 at x = π/4, 3π/4, 5π/4, 7π/4.
So
f
−1
h π π i 3π
π
π 5π
7π
π
∪
([0, 1/2]) =
,
,π −
∪ π+ ,
∪
, 2π −
6 4
4
6
6 4
4
6
Richard M. Timoney
2