MA2223: SOLUTIONS TO ASSIGNMENT 2 1. Determine which of

```MA2223: SOLUTIONS TO ASSIGNMENT 2
1. Determine which of the following subsets of R are open.
(a) {x ∈ R : 1 < x < 3}
Reason: We proved in class that in a metric space every open ball
is an open set. Here we have an open ball in R with centre 2 and
(b) S = {x ∈ R : |x| ≥ 1}
Reason 1: The points −1 and 1 are contained in S but they are not
interior points of S. Any open ball with centre 1 and radius r > 0
will contain the point 1 − where = min{1, 2r }. Since this point
lies outside S, 1 is not an interior point. Similar argument for −1.
Reason 2: The complement of S is the open interval (−1, 1) which
is an open ball in R with centre 0 and radius 1 and hence is an open
set. We proved in class that the complement of an open set is a
closed set. This means S is a closed set. Since R is a connected
space the only subsets of R which are both open and closed are ∅
and R. Hence S cannot be open.
Reason 3: Let x be a limit point of S. Then there exists a sequence
(xn ) in S which converges to x. Since the absolute value |.| : R → R
is a continuous function, the sequence (|xn |) converges to |x|. Now
|xn | ≥ 1 for all n and so from Real Analysis we know that |x| ≥ 1.
Thus x ∈ S and so S is closed. As noted in Reason 2, this implies S
is not open.
(c) R\Z = {x ∈ R : x ∈
/ Z}
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MA2223: SOLUTIONS TO ASSIGNMENT 2
Reason: We proved in class that the union of a collection of open
sets is an open set. Here we have a union of open intervals (n, n + 1)
where n ∈ N.
(d) Q
Reason: None of the points in Q are interior points. Let x ∈ Q
be a rational number. Recall from Real Analysis that between any
two real numbers there exists an irrational number. Thus every open
interval (x−r, x+r) (i.e. every open ball in R with centre x) contains
an irrational number. This shows x is not an interior point of Q.
2. Determine which of the following subsets of R2 are open.
(a) {(x, y) ∈ R2 : (x − 3)2 < 4 − (y − 5)2 }
Reason: This is an open ball in R2 with centre (3, 5) and radius 2.
Again we use the fact that in a metric space every open ball is an
open set.
(b) S = {(x, y) ∈ R2 : x > 0 or y > 0}
Reason 1: Write S = U ∪ V where
U = {(x, y) ∈ R2 : x > 0}
V = {(x, y) ∈ R2 : y > 0}
If (a, b) ∈ U then the open ball B((a, b), a2 ) with centre (a, b) and
a
2
is contained in U . Thus U is an open set. Similarly, if
(a, b) ∈ V then the open ball B((a, b), 2b ) with centre (a, b) and radius
b
2
is contained in V . Thus V is an open set. Now S is an open set
(since it is a union of open sets).
Reason 2: Write S = U ∪ V where
U = {(x, y) ∈ R2 : x > 0}
MA2223: SOLUTIONS TO ASSIGNMENT 2
3
V = {(x, y) ∈ R2 : y > 0}
Let p : R2 → R, (x, y) 7→ x be the projection onto the first coordinate. Then p is a continuous map. Let
W = {x ∈ R : x > 0}
Note that W is an open set in R. Now U = p−1 (W ) and so U is an
open set. Similar argument shows V is an open set. As in Reason 2
this implies S is open.
(c) S = {(x, y) ∈ R2 : 1 < x < 3 and y = 0}
Reason 1: None of the points in S are interior points. Let (x, 0) ∈ S.
If B((x, 0), r) is an open ball in R2 with centre (x, 0) and radius
r > 0 then B((x, 0), r) contains the point (x, 2r ). But (x, 2r ) is not
contained in S and so (x, 0) is not an interior point of S.
Reason 2: The complement of S is not closed. For example (2, 0) is
a limit point of the complement of S since the sequence ((2, n1 ))∞
n=1
lies in the complement of S and converges to (2, 0). But (2, 0) ∈ S.
3. Determine which of the following subsets of R3 are open.
(a) S = {(x, y, z) ∈ R3 : y 2 ≥ 1 − z 2 − x2 }
Reason 1: Points which lie on the unit sphere S((0, 0, 0), 1) are contained in S but are not interior points of S. For example, consider
the point (1, 0, 0) ∈ S. If B((1, 0, 0), r) is an open ball then the point
(1 − , 0, 0) where = min{1, 2r } is contained in B((1, 0, 0), r) but not
in S. This shows (1, 0, 0) is not an interior point of S.
Reason 2: S is the complement of the open ball in R3 with centre
(0, 0, 0) and radius 1. This means S is closed. Since R3 is connected,
the only subsets which are both open and closed are ∅ and R3 . Hence
S cannot be open.
Reason 3: Let x ∈ R3 be a limit point of S. Then there exists a
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MA2223: SOLUTIONS TO ASSIGNMENT 2
sequence (xn ) in S which converges to x. Since the Euclidean norm
k.k : R3 → R is continuous, the sequence (kxn k) converges to kxk
in R. Now kxn k ≥ 1 for all n and so from Real Analysis we know
kxk ≥ 1. Hence x ∈ S and so S is closed. As in Reason 2 this implies
S is not open.
(b) S = {(x, y, z) ∈ R3 : x2 + y 2 < 1 and z = 0}
Reason: None of the points in this set are interior points. Let
(x, y, 0) ∈ S. If B((x, y, 0), r) is an open ball with centre (x, y, 0)
then the point (x, y, 2r ) is contained in B((x, y, 0), r) but not in S.
This shows (x, y, 0) is not an interior point of S.
(c) S = {(x, y, z) ∈ R3 : x = cos t, y = sin t, z = t, t ∈ [0, 4π]}
Reason 1: None of the points in this set are interior points. For
example, consider the point (1, 0, 0) ∈ S. If B((1, 0, 0), r) is an open
ball with centre (1, 0, 0) then the point (1, 0, ) where = min{π, 2r }
is contained in B((1, 0, 0), r) but not in S. This shows (1, 0, 0) is not
an interior point of S.
Reason 2: S is the image of the continuous map
T : [0, 4π] → R3 , t 7→ (cos t, sin t, t)
Since [0, 4π] is compact (by Heine-Borel theorem), S is also compact.
Hence S is closed (and bounded) in R3 . As R3 is a connected space,
S cannot be both open and closed.
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