Homework 7. Solutions
1. Show that the unit circle C = {(x, y) ∈ R2 : x2 + y 2 = 1} is
connected. Hint: Use polar coordinates and Theorem 2.15.
In terms of polar coordinates, the points on the unit circle are the
points that have the form (cos θ, sin θ). Consider the function
f : [0, 2π) → R2 ,
f (θ) = (cos θ, sin θ).
Each component of f is continuous, so f itself is continuous. Since
the domain of f is an interval, it is connected, so the image of f is
connected as well. In other words, the unit circle C is connected.
Homework 7. Solutions
2. Let U = {(x, y) ∈ R2 : xy 6= 0}. How many connected components
does this set have? Hint: One of the components is (0, ∞) × (0, ∞).
To say that xy 6= 0 is to say that each of x, y is either positive or
negative. Thus, U can be expressed as the union of the sets
U1 = (0, ∞) × (0, ∞),
U2 = (0, ∞) × (−∞, 0),
U3 = (−∞, 0) × (0, ∞),
U4 = (−∞, 0) × (−∞, 0).
Each of these sets is a product of intervals and thus connected. It
remains to show that they are actually connected components.
Let A be a connected subset of U and let pi : A → R denote
the projection on the ith variable. By continuity, each pi (A) is an
interval that does not contain 0, so it is contained in either (−∞, 0)
or (0, ∞). Thus, A itself is contained in one of the sets Uj .
Homework 7. Solutions
3. Show that there is no surjective continuous function f : A → B, if
A = (0, 3) ∪ (3, 6),
B = (0, 1) ∪ (1, 2) ∪ (2, 3).
Hint: Look at the restriction of f on each subinterval. You will need
to use Theorem 2.11 and also some parts of Theorem 2.15.
Note that A = A1 ∪ A2 is the union of two disjoint open intervals
and B = B1 ∪ B2 ∪ B3 is the union of three disjoint open intervals.
Since f : A → B is continuous, each restriction f : Ai → B must be
continuous, so each f (Ai ) must be connected.
Since f (Ai ) is a connected subset of B1 ∪ B2 ∪ B3 , it lies within
either B1 or B2 ∪ B3 . If it actually lies in B2 ∪ B3 , then it lies within
either B2 or B3 . Thus, each f (Ai ) is contained in a single Bj and
the function f is not surjective, contrary to assumption.
Homework 7. Solutions
4. Let (X, T ) be a topological space and suppose A1 , A2 , . . . , An are
connected subsets of X such that Ak ∩ Ak+1 is nonempty for each k.
Show that the union of these sets is connected. Hint: Use induction.
When n = 1, the union is equal to A1 and this set is connected by
assumption. Suppose that the result holds for n sets and consider
the union of n + 1 sets. This union has the form
U = A1 ∪ · · · ∪ An ∪ An+1 = B ∪ An+1 ,
where B is connected by the induction hypothesis. Since An+1 has
a point in common with An , it has a point in common with B. In
particular, the union B ∪ An+1 is connected and the result follows.