MA2223: SOLUTIONS TO ASSIGNMENT 5
1. State which of the following collections is a topology on the set X =
{a, b, c}.
(a) τ1 = {∅, {a}, {a, b, c}}
Answer: Topology.
(b) τ2 = {∅, {a, b}, {c}, {a, b, c}}
Answer: Topology.
(c) τ3 = {∅, {a, b}, {b, c}, {a, b, c}}
Answer: Not a topology.
Reason: The intersection of the sets {a, b} and {b, c} is {b}. But {b}
is not an element of τ3 .
Are any of the topologies Hausdorff?
Answer: No.
Reason: τ1 and τ2 are not Hausdorff since in both cases there do not exist
disjoint open sets which separate a and b.
2. Determine which of the following sets is a connected / compact subspace
of R2 .
(a) S1 = {(x, y) ∈ R2 : x2 + y 2 = 1}
Answer: S1 is connected and compact.
Reason 1: S1 is the image of the continuous map
f : [0, 2π] → R2 ,
t 7→ (cos t, sin t)
We proved in class that every closed interval [a, b] in R is connected
and compact. We also proved that the image of a connected space
under a continuous map is connected and that the image of a compact space under a continuous map is compact.
Reason 2: S1 is bounded since for example it is contained in the
1
2
MA2223: SOLUTIONS TO ASSIGNMENT 5
open ball B((0, 0), 2). Let x ∈ R2 be a limit point for S1 . Then
there exists a sequence (xn ) in S1 which converges to x. We know
the Euclidean norm k.k : R2 → R is continuous and so the sequence
(kxn k) converges to kxk in R. But kxn k = 1 for all n and so kxk = 1.
Thus x ∈ S and so S is closed. We proved in class that a subset of
Rn which is closed and bounded is compact.
Resaon 3: Another way to show S1 is closed is to note that the
preimage of a closed set under a continuous map is closed. The singleton set {1} is closed in R and the Euclidean norm k.k : R2 → R
is continuous. Now S1 is the preimage of {1} under the Euclidean
norm on R2 .
Reason 4: S1 is path connected since if x, y ∈ S1 then we can construct the path,
α : [0, 1] → R2 ,
t 7→
(1 − t) x + t y
k(1 − t) x + t yk
We divide by the norm here to ensure the path is contained in S1 .
Note that α is continuous provided x and y are not antipodal since
in this case the denominator passes through 0. To remedy this we
can patch together two paths via a third point in S1 .
(b) S2 = {(x, y) ∈ R2 : x2 + y 2 = 1 and y 6= 0}
Answer: S2 is not connected and not compact.
Reason 1: S2 can be expressed as a union S2 = U ∪ V where
U = S2 ∩ {(x, y) ∈ R2 : y > 0}
V = S2 ∩ {(x, y) ∈ R2 : y < 0}
U and V are open in the subspace topology on S2 , they are nonempty and disjoint. This shows S2 is not connected.
Reason 2: S2 is not closed since the points (−1, 0) and (1, 0) are
both limit points for S2 but they are not contained in S2 . (1, 0) is a
MA2223: SOLUTIONS TO ASSIGNMENT 5
3
π
π
limit point since the sequence ((cos n+1
, sin n+1
))∞
n=1 lies in S2 and
converges to (1, 0). A similar argument shows (−1, 0) is a limit point
for S2 . We proved in class that a compact subspace of Rn is closed.
Thus S2 is not compact.
3. The subspace T = S1 × S1 of R4 is called a torus.
Is T a connected / compact space?
Answer: T is connected and compact.
Reason 1: We proved in class that the product of two connected spaces
is connected and the product of two compact spaces is compact.
Reason 2: As noted above, S1 is path connected. Suppose (x1 , y1 ) and
(x2 , y2 ) are two points in S1 × S1 . There exists a path
α1 : [0, 1] → S1
joining x1 to x2 and a path
α2 : [0, 1] → S1
joining y1 to y2 . Now we can construct the path
α : [0, 1] → S1 × S1 ,
t 7→ (α1 (t), α2 (t))
joining (x1 , y1 ) and (x2 , y2 ). This shows T is path connected and hence
connected.