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University of Toronto at Scarborough
Department of Computer and Mathematical Sciences
MAT C27H
2015/16
Problem Set #1
Due date: Thursday, January 21, 2016 at the beginning of class
(1) Text chap. 2.1 question 1
Prove that if a, b ∈ R with a < b then neither [a, b) nor (a, b] is an open subset of R.
Also show that neither is a closed subset of R.
Soln:
There is no open interval (u, v) containing a which is contained in [a, b). (This would
require u < a.) Similarly there is no open interval (u, v) containing b which is contained
in (a, b]. (This would require v > b.)
(a, b] is closed iff the disjoint union of (−∞, a] and (b, ∞) is open. This would require
that there is an open interval (u, v) contained in the disjoint union of (−∞, a] and (b, ∞)
for which a ∈ (u, v). This would require v > a, so (u, v) must intersect the interval [a, b],
which is a contradiction.
(2) Text chap. 2.1 question 2
Prove that [a, ∞) and (−∞, a] are closed subsets of R.
Soln: The complement of [a, ∞) is (−∞, a). This is the union of open intervals (N, a)
where N ranges over all integers less than a. Hence it is open.
Similarly the complement of (−∞, a] is (a, ∞). This is the union of open intervals
(a, N ) where N ranges over the integers greater than a. Hence it is open.
(3) Text chap. 2.1 question 3
Show by example that the union of an infinite number of closed subsets of R is not
necessarily a closed subset of R.
Soln:
The union of [1/n, 1] (for n positive integers) is (0, 1] which is not closed (see question
1 of Chap 2.1)
(4) Text chap. 2.1 question 4
Prove
(a) Z is not open Proof: If Z were open, for each integer n there would be an interval
(a, b) with a < n < b contained in the set of integers. There is no such interval.
(b) Show that the set of prime numbers is a closed subset of the real numbers but not an
open subset.
Proof: The complement of the integers is the union of the set of real numbers which
are not integers. We need to show this set is open. The set of real numbers which are
not integers is a union of open intervals (n, n + 1). The set of integers is thus closed.
The complement of the prime numbers is the union of the non-integers with those
integers which are not prime. If n is an integer which is not prime, then the disjoint
union of interval (n − 1, n) ∪ (n, n + 1) is replaced by the open interval (n − 1, n + 1)
1
. So the complement of the prime numbers is the union of all intervals (n, n + 1) (n a
positive integer) with the intervals (n − 1, n + 1) when n is composite. This is open.
We need to show that the set of prime numbers is not open: If it were open, for each
prime number p there would be an open interval (a, b) with a < p < b contained in
the set of prime numbers. This is not the case, since the prime numbers are integers.
(c) The set of irrational numbers is neither closed nor open in R.
Proof: If the set of irrational numbers were open, for each irrational number there
would be an open interval (a, b) contained in the set of irrational numbers. This is not
possible, since any interval (a, b) contains some rational numbers.
If the set of irrational nubmers were closed, its complement (the set of rational numbers) would be open. This would mean there would be an open interval (a, b) contained
in the set of rational numbers. This is not possible, since any interval (a, b) contains
some irrational numbers.
(5) Text chap. 2.1 question 7(i)
Let S = {0, 1, 21 , 13 , . . . }. Prove S is closed in R.
Proof: We need to show this set contains all its adherent points. The only adherent point
is {0}, which is contained in S.
(Note that (ii) is not closed, because that set does not contain 0.
(6) Text chap. 2.2 question 1
(a) Show that Ra,b ≤ D.
Soln: Draw a square with side length 2 and centre (a, b) containing each point (a, b)
in the disk of radius 1. Here = (1 − r)/8, r defined as in the question
Then
√ the maximum distance from the centre in a square of side length 2epsilon. It is
2 (this is achieved at the four corners of the square).
Then use the Cauchy-Schwarz inequality to get an upper bound on the distance from
0 of a point √
in the square (where the square is centered at (a,b) in D).
This is r + 2.
Using the formula for (= (1
√− r)/8) we see that this number is always less than 1: it
is r(1 − d) + d (where d = 2/8). This is less than (1 − d) + d (since r < 1) in other
words it is s less than 1.
(b) Conclude that D is the union of the Ra,b over all a and b.
Soln: Each point (a, b) ∈ D is contained in one of the Ra,b . and we showed in the
previous exercise that Ra,b ⊂ D.
(c) Conclude from the previous exercise that D is an open set in R2 .
Soln: D is the union of the open sets Ra,b .
(d) Show that every disk
E := (x, y) ∈ R2 : (x − a)2 + (y − b)2 < c2
is open in R2 .
Soln: The map
f : (x, y) 7→ ((x − a)/c, c(y − b)/c) is a bijective affine-linear map which takes E to
D. E is open if and only if D is open, since f is continuous and invertible.
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