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Math 4A03: Assignment 3: SOLUTIONS
Due Friday, October 22 by 11:00 am
Total of Marks= 50
Problem 1. [15 marks] # 54 on p.121
Solution. The interior of a connected set need not be connected. For example, consider the set A consisting of the union of the closed disks of radius 1
in the plane centered at (−1, 0) and (1, 0), respectively, i.e.
A = {(x, y) ∈ R2 , (x + 1)2 + y 2 ≤ 1} ∪ {(x, y) ∈ R2 , (x − 1)2 + y 2 ≤ 1}.
Then, A is clearly path-connected, since the point (0, 0) belongs to both disks
and if the point (x, y) belongs to one of the disks, the line segment from (x, y)
to (0, 0) is contained in that disk. Thus, any two points in A can be connected
by a path consisting of two line segments passing though (0, 0). hence, A is
◦
connected. On the other hand, the interior of A, A, is the union of two open
disjoint disks,
◦
A = {(x, y) ∈ R2 , (x + 1)2 + y 2 < 1} ∪ {(x, y) ∈ R2 , (x − 1)2 + y 2 < 1},
and is therefore disconnected.
Problem 2. [10 marks] # 55 on p.121
Solution. (a) The closure of a disconnected set need not be disconnected.
For example, the set (0, 1) ∪ (1, 2) is a disconnected subset of the real line (as
the union of two disjoint non-empty open subsets) but its closure is the interval
[0, 2], which is of course connected.
(b) The interior of a disconnected set need not be disconnected. For example,
the set [0, 1] ∪ {2} is disconnected, but its interior is the interval (0, 1), which is
connected.
Problem 3. [12 marks] # 57 on p. 121
Solution. (a) If f : M → R is continuous, then the image of M under f ,
f (M ), is connected if M is connected. Hence, it follows that if f is integervalued, then f must be constant, since otherwise f (M ) would clearly not be
connected.
1
2
(b) Since, between two distinct irrational numbers, we can always find a rational one, it follows that every continuous function with values in the irrational
numbers defined on a connected set must be constant. Indeed, if it were not,
every value between two distinct irrational numbers in its range would have to
be reached, by the intermediate value theorem, which contradicts the fact that
some number between them must be rational.
Problem 4. [13 marks] # 58 on p. 121
Solution. If the point (x0 , y0 , z0 ) belongs to the cone
C = {(x, y, z) ∈ R3 , x2 + y 2 = z 2 },
then so does the point (tx0 , ty0 , tz0 ), if t is any real number since
(tx0 )2 + (ty0 )2 = t2 (x20 + y02 ) = t2 z02 = (tz0 )2 .
Hence, given two points in C the line segment from one of the points to the
origin (0, 0, 0) followed by the line segment from the origin to the other point
is a continuous path between the two point. More explicitely, if (x0 , y0 , z0 ) and
(x1 , y1 , z1 ) belong to C, a continuous path in C joining them is given by
½
(1 − t) (x0 , y0 , z0 ), 0 ≤ t ≤ 1
ϕ(t) =
(t − 1) (x1 , y1 , z1 ), 1 ≤ t ≤ 2.