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1 Introduction to Topology Professor M. Zuker
Introduction to Topology: Fall 2014
Final Examination – December 10, 2014.
Notation: (X, T ) always refers to a topological space, where X is a non-empty set and T is
the set of “open sets”. The latter is never empty since it always contains X and ∅.
For sets A and B, A ⊂ B always means that A is a proper subset of B. Otherwise, A ⊆ B
is used. If A ⊆ X, Ā and Cl(A) both refer to the “closure of A”. If A ⊆ S, the complement
of A (in S) is denoted by S \ A.
In a metric space, (X, d), B(x, a) = {y ∈ X | d(x, y) < a, a ∈ R+ }. B(x, a) is called “the
open ball of radius a about x”. Similarly, B̄(x, a) = {y ∈ X | d(x, y) ≤ a, a ∈ R+ } is called
“the closed ball of radius a about x”. The closure of B(x, a) is denoted by B(x, a).
Unless otherwise specified, subspaces and quotient spaces of Rn have the “usual” topology
where open sets are unions of open balls.
Specific subspaces: I = [0, 1]. D = {(x, y) ∈ R2 | x2 + y 2 ≤ 1}. (Note that D =P
B̄((x, y), 1) in
2
2
n
n+1
R .) D is called the “unit disk in the plane”. S = {(x1 , x2 , . . . , xn+1 ) ∈ R | n+1
i=1 xi = 1}.
1
2
2
S is the circle of radius one about the origin in R and S is the sphere of radius one about
the origin in R3 . S 1 × S 1 is called “the torus”. If r ∈ R+ , rD refers to the disk D scaled by r.
That is, rD = {(x, y) ∈ R2 | x2 + y 2 ≤ r2 }. Similarly, if (a, b) ∈ R2 , (a, b) + rD refers to the
scaled disk rD translated by (a, b). That is, (a, b)+rD = {(x, y) ∈ R2 |(x−a)2 +(y−b)2 ≤ r2 }
1. For each of the B subsets of R defined below, you must answer “1” if B is a basis for
the usual topology on R, “2” if B is a basis for another topology on R and “3” if B is
not a basis. If you answer “2”, you must state what the resulting topology is. If you
answer “3”, you must prove (provide a counter-example) that B is not a basis for any
topology.
(a) B = {(a, b) | a, b ∈ Q, a < b}.
Solution: 1. This problem was discussed in class.
(b) B = {(a, a + 1/2) | a ∈ R}.
Solution: 3. This is not a basis for any topology. (0, 1/2) ∩ (1/4, 3/4) =
(1/4, 1/2), which is not empty but contains no element of B.
(c) B = {(a, a + ) | a ∈ R, ∈ (0, 1)}.
Solution: 1. (Any open interval (a, b) is the union of overlapping basis elements
of length < 1.)
(d) B = {[a, b] | a, b ∈ R, a < b}.
Solution: 3. This is not a basis because [−1, 0] ∩ [0, 1] = {0}, which contains
no element of B.
(e) B = {[a, b] | a, b ∈ R, a ≤ b}.
Solution: 2. This is a basis for the discrete topology on R because [a − 1, a] ∩
[a, a + 1] = [a, a] = {a}, so all single point sets are open.
2 Introduction to Topology Professor M. Zuker
(f) B = {[a, a + 1/n) | a ∈ Q, n ∈ Z+ }.
Solution: 2. This is a basis for the “lower limit topology” on R. An equivalent
basis is B = {[a, b) | a, b ∈ R, a < b}.
(g) B = ( 2mk , 2nk ) | m, n ∈ Z, m < n, k ∈ Z+ .
Solution: 1. Rational numbers of the form
counts.
m
2k
are dense in R. That’s what
(h) B = {(a − 1/n, a + 1/n) | a ∈ Q, n ∈ Z+ }.
Solution: 1.
2. If (X1 , T1 ) and (X2 , T2 ) are two topological spaces, X1 ' X2 means that X1 is homeomorphic to X2 . Prove that ' is an equivalence relation on topological spaces. This
means you must show that ' is Reflexive, Symmetric and Transitive.
Solution:
R: Let ι be the identity function on X. Then ι is a homeomorphism from X to X.
S: Suppose X1 ' X2 . Then there exists a homeomorphism f : X1 → X2 . Let g = f −1 .
By the definition of homeomorphism, g must be continuous. Thus g is a homeomorphism from X2 to X1 , so X2 ' X1 .
T: Suppose that X1 ' X2 and X2 ' X3 . Then there are homeomorphisms f1 : X1 →
X2 and f2 : X2 → X3 . Let gi = fi−1 for i = 1, 2. Then g1 and g2 are continuous. Let
f = f2 ◦ f1 and let g = g1 ◦ g2 . Both f and g are continuous and g = f −1 , so X1 ' X3 .
3. In R, R2 and R3 , consider the following topological spaces:
(a) R
(b) R2
(c) {(x, y) ∈ R2 | x − y = 2}
(d) {(x, y, z) ∈ R3 | x + y + z = 1}
(e) D
(f) int(D)
(g) 2D \ D
(h) D \ {(0, 0)}
(i) S 2 \ {(1, 0, 0)}
(j) R2 \ (1, 1)
The above ten topological spaces can be partitioned into homeomorphic equivalence
classes. How many equivalence classes are there? For each equivalence class, state
which spaces belong to it.
Solution: Lets start from the beginning and add new equivalencs classes as needed.
(a) R. Place R in equivalence class I.
3 Introduction to Topology Professor M. Zuker
(b) Since R2 is not homeomorphic to R, place R2 in quivalence class II.
(c) This is a line, which is homeomorphic to R, so this space is in equivalence class I.
(d) This is a plane, which is homeomorphic to R2 , so this space is in equivalence class
II.
(e) The closed disk defines a new equivalence class, III.
(f) The interior of D is the same as B((0, 0), 1) and is homeomorphic to R2 , so it
belongs in equivalence class II. A homeomorphism can be defined by f : int(D) → R2 ,
where f (x, y) = ( 1−xx2 −y2 , 1−xy2 −y2 ). This works because x2 + y 2 < 1.
(g) This space defines a new equivalence class, IV.
(h) This space is in fact homeomorphic to the previous one, 2D \ D. Removing a single
point from the interior of a disk is equivalent to removing a “subdisk” contained in
the interior of a disk. It suffices to provide an explicit homeomorphism, which is easy.
Define f : D \ {(0, 0)} → 2D \ D by f (x, y) = (ax, ay), where
1 + x2 + y 2
.
a= p
x2 + y 2
(i) The sphere with a single point removed is homeomorphic to the plane. This space
is in equivalence class II.
(j) This should have been written as R2 \ {(1, 1)}. That is, remove a single point from
R2 . This space is homeomorphic to the complement of the disk (D) and defines a new
equivalence class, V.
Summary: I: a, c ; II: b, d, f, i ; III: e ; IV: g, h ; V: j
4. Define F : R2 → R2 by F (x, y) = (cos(2πx), sin(2πy)). Then ∼F is an equivalence
relation on R2 defined by (x, y) ∼F (x0 , y 0 ) ↔ F (x, y) = F (x0 , y 0 ). Let X = R2 / ∼F .
(a) To what well known topological space is X homeomorphic to? (b) The range of F
is homeomorphic to one (or more) of the topological spaces in the previous problem.
Which one(s)?
Solution: This problem was badly designed, so I have dropped part (a).
For (b), the range is the closed square, [−1, 1] × [−1, 1], which is homeomorphic to the
disk, D (class D above).
5. Let A and B be two disjoint closed sets in a metric space (X, d). Define a continuous
function F : X → [−1, 1] such that F (x) = 1 for x ∈ A and F (x) = −1 for x ∈ B.
Using F , define disjoint open sets U ⊂ X and V ⊂ X such that A ⊆ U and B ⊆ V .
Solution: We know that if A ⊆ X, then d(x, A) is a continuous function from X to
the non-negative real numbers. Note that d(x, A) is defined by
d(x, A) = inf d(x, a).
a∈A
If A and B are disjoint closed sets, then d(x, A) > 0 if x ∈
/ A and d(x, B) > 0 if x ∈
/ B.
Thus, for any x ∈ X, d(x, A) + d(x, B) > 0 since x cannot be in both A and B. Define
F : X → [−1, 1] by
4 Introduction to Topology Professor M. Zuker
F (x) =
d(x, B) − d(x, A)
.
d(x, A) + d(x, B)
When x ∈ A, d(x, A) = 0 and d(x, B) > 0, so F (x) = 1. Similarly, F (x) = −1 if
x ∈ B. For x ∈
/ A ∪ B, −1 < F (x) < 1.
Let U = F −1 [(1/2, 1]] and V = F −1 [[0, 1/2)]. Since (1/2, 1] ∩ (1/2, 1] = ∅, U ∩ V = ∅.
U and V are both open because [0, 1/2) and (1/2, 1] are both open in [−1, 1]. Clearly
A ⊆ U and B ⊆ V .
Note: It is possible that A = U and/or B = V . This can happen if X is disconnected
and A and/or B are clopen sets.
6. Let A = {(x, y) ∈ R2 | x2 y 2 = 1}.
(a) Is A closed?
Solution: A is closed.
(b) How many connected components does A have in R2 ?
Solution: Four. They are Ai = A ∩ Qi for 1 ≤ i ≤ 4, where
•
•
•
•
Q1
Q2
Q3
Q4
= {(x, y) ∈ R2 | x > 0
= {(x, y) ∈ R2 | x < 0
= {(x, y) ∈ R2 | x < 0
= {(x, y) ∈ R2 | x > 0
and
and
and
and
y
y
y
y
> 0}
> 0}
< 0}
< 0}
(c) Show that the connected components are homeomorphic to one another by defining explicit homeomorphims. Hint: It suffices to show that one connected component is homeomorphic to all the others.
Solution: I define three homeomorphims: Fi : A1 → Ai , for 1 < i ≤ 4.
F1 (x, y) = (−x, y), F2 (x, y) = (−x, −y) and F3 (x, y) = (x, = yy). This demonstrates that F1 ' Fi for 1 < i ≤ 4.
(d) Compute the distance between every pair of connected components.
Solution: √
d(A1 , A2 ) = d(A1 , A4 ) = d(A2 , A3 ) = d(A3 , A4 ) = 0. d(A1 , A3 ) =
d(A2 , A4 ) = 2 2.
7. In R, define a countable number of disjoint closed sets, Cn , n ∈ Z+ such that d(Cm , Cn ) >
0 if m 6= n and such that for every > 0, there is a pair of closed sets, Cm and Cn such
that d(Cm , Cn ) < . Can this also be achieved if R is replaced by I?
Solution: There are many ways to define such sets. For example, let {an } be any
strictly increasing sequence of real numbers bounded above. Then an → a = supn {an }
as n → ∞. Let Cn = [a2n , a2n+1 ]. Then the distance between each pair, Cn and Cn+1
is a2n+2 − a2n+1 , which is strictly positive but converges to zero as n → ∞.PNote that
this is possible if R is replaced by I, since we can let a0 = 0 and an = ni=1 21i for
n > 0.
5 Introduction to Topology Professor M. Zuker
o
+
8. In R , let A =
cos(2πt),
sin(2πt) t ∈ R . Compute the interior of A and
the closure of A. Select some point (x, y) in the closure of A that is not in A and define
an explicit sequence of elements in A that converges to (x, y).
2
n
1+t
t
1+t
t
cos(2πt) and let g(t) = 1+t
sin(2πt).
Solution: int(A) = ∅. Why? Let f (t) = 1+t
t
t
Then A = {(f (t), g(t))} for t > 0. For any t > 0, consider the line that intersects
the curve (A) perpendicular to the tangent line at (f (t), g(t). Any neighborhood of
(f (t), g(t)) will contain points on this “normal line” that are not in A.
The closure of A is A ∪ S 1 . Note that no point of S 1 is in A. For any (x, y) =
(cos(2πt), sin(2πt)) ∈ S 1 , let (xn , yn ) = (f (t+n), g(t+n)) for n ∈ Z+ . Then (xn , yn ) →
(x, y) as n → ∞.
9. A complete metric space is a metric space in which every Cauchy sequence converges.
Let (X, d) be a complete metric space. Suppose that f : X → X is a function with the
property that d(f (x), f (y)) < rd(x, y) for all x, y ∈ X, where r is a fixed real number
< 1. Prove that f is a continuous function. Prove that f has at least one fixed point.
Hint: Start with any x1 ∈ X and define xn+1 = f (xn ).
Solution: To show that f is continuous, let x ∈ X and let > 0. Then if y ∈ B(x, ),
d(f (x), f (y)) ≤ rd(x, y) < r < . Thus f is continuous at x. Note that is used for δ
in this case.
Now let x1 ∈ X and define xn+1 = f (xn ). Observe that if N < m < n, then
d(xm , xn ) = d(f (xm−1 , xn−1 )
≤ rd(xm−1 , xn−1 )
= rd(f (xm−2 , f (xn−2 ))
≤ r2 d(xm−2 , xn−2 )
...
≤ rN d(xm−N , xn−N )
≤r
N
n−N
X−1
d(xi , xi+1 )
i=m−N
≤r
N
n−1
X
ri−1 d(x1 , x2 )
i=1
d(x1 , x2 )
1−r
N
= r c,
< rN
sum of geometric progression
0 ,x1 )
where c = d(x1−r
. Given any > 0, choose N ∈ Z+ such that crN < . Then
d(xm , xn ) < for m, n > N , so {xn } is a Cauchy sequence. It must converge to some
x ∈ X. d(f (x), xn ) ≤ rd(x, xn−1 ), which converges to zero as n → ∞. Thus xn also
converges fo f (x). Since the limit is unique, we deduce that f (x) = x, so f has at least
one fixed point.