Download and x ∈ U y ∈ V ˆ N = N∪{∞} (d) Let a, b:ˆN

INTRODUCTION TO TOPOLOGY (MA30055)
SEMESTER 2 MATHEMATICS: PROBLEM SHEET 5: SOLUTIONS
Let X be a topological space. Then X is said to be:
• a T1 -space if, for any x 6= y ∈ X , there is an open set U with y ∈ U and x 6∈ U ;
• a T2 - or Hausdorff space, if for any x 6= y ∈ X , ∃ disjoint open U , V with x ∈ U and
y ∈V.
Recall also the topology on N̂
= N∪{∞} whose closed subsets are either finite or contain ∞.
1. True or false? (Give reasons.)
(a) X is T1 if and only if ∀ x
∈ X , the point {x} is closed.
(b) Any T1 -space is Hausdorff.
(c) Any countable Hausdorff space is discrete
(d) Let a,
b: N̂ → X be continuous with a(n) = b(n) for all n ∈ N. Then a(∞) = b(∞).
(e) As in (d) but assuming also X is Hausdorff.
Solution:
(a) True. If {x} is closed, then U
= X \ {x} is open and contains every y 6= x, fulfilling the
requirement for T1 . Conversely, if the topology is T1 , then for every y
with y
∈ Uy and x 6∈ Uy . Then X \ {x} =
S
y6=x
6= x there is an open Uy
Uy is a union of open sets, hence open.
Thus {x} is closed.
(b) False. The cofinite topology on an infinite set is
T1 by part (a) above, but is not Hausdorff
since two cofinite sets, i.e., two non-empty open sets, are never disjoint.
(c) False. The topology on Q induced from the Euclidean topology on R is Hausdorff, because
it is induced from a metric
d (for x, y ∈ Q, let δ = d(x, y)/2 so that U = Bδ (x) and
V = Bδ (y) are disjoint open sets containing x and y rspectively). However, the topology on
Q is not discrete since singletons are not open.
(d) False. Take
X to have more than one point and the trivial topology. Then any functions
a, b: N̂ → X are continuous, and they can be chosen to agree for n ∈ N and disagree at ∞.
(e) True. This is a reformulation of uniqueness of limits in Hausdorff spaces. Here is a proof of
the contrapositive. Suppose
a, b: N̂ → X are continuous. If a(∞) 6= b(∞) then they can
be separated by disjoint open sets U , V . Now a∗ (U ) and b∗ (V ) are open and contain ∞, so
they must be cofinite; in particular a∗ (U ) ∩ N and b∗ (V ) ∩ N have nonempty intersection (the
union of their complements is finite), i.e., ∃n
a(n) 6= b(n).
∈ N with a(n) ∈ U and b(n) ∈ V ; in particular
2. Prove that X is Hausdorff if and only if the diagonal ∆X
:= {(x1 , x2 ) ∈ X × X : x1 =
x2 } is a closed subspace of the product space X × X . [Hint: it is possible to write down the
definition of what it means for ∆X to be closed using a basic open set in X
Solution: The diagonal ∆ is closed if and only if, for all (x1 , x2 )
basic open set W
× X .]
∈ (X × X) \ ∆, there is a
= U × V with (x1 , x2 ) ∈ W and W ∩ ∆ = ∅, where U, V are open in
X . In other words, if x1 6= x2 in X , then there are open sets U, V with x1 ∈ U and x2 ∈ V
such that
(U × V ) ∩ ∆ = {(x, x) | x ∈ U, x ∈ V } = ∅.
But this set is empty precisely when
U ∩ V is empty, i.e. U and V are disjoint. Thus ∆ is
closed if and only if X is Hausdorff.
3. Prove that if X1 and X2 are Hausdorff, so is the product space X1
× X2 .
Solution:
If X and X2 are Hausdorff spaces, consider distinct points (x1 , x2 ) and (y1 , y2 ) in X1 × X2 .
They are distinct, so either
x1 6= y1 or x2 6= y2 . First suppose that x1 6= y1 . Then
there are disjoint open sets
U1 3 x1 and U2 3 y1 , because X1 is Hausdorff. Hence
U1 × X2 3 (x1 , x2 ) and U2 × X2 3 (y1 , y2 ) are disjoint open sets in X × X2 . If instead
x2 6= y2 , then we use a similar argument, relying on the fact that X2 is Hausdorff. Thus
X1 × X2 is Hausdorff. This argument can easily be adapted to arbitrary products: two points
differ if they differ at one of their coordinates, and finding disjoint open sets for this coordinate
gives disjoint open sets in the product.
→ X1 and f2 : Y → X2 be continuous maps. A space Z with continuous maps
4. Let f1 : Y
p1 : X1 → Z and p2 : X2 → Z is called a pushout of f1 , f2 iff p1 f1 = p2 f2 and for any
other space
with qi
W with such maps qi : Xi → W there is a unique continuous map h: Z → W
= hpi .
(a) Show that the pushout is a quotient of the disjoint union X1
(b) Suppose
`
X2 . [Hint: it is a coequalizer.]
X1 , X2 are open subspaces of a space V and Y = X1 ∩ X2 with fi the
inclusions of Y into Xi . Show that the union Z
= X1 ∪ X2 (with inclusion maps pi : Xi → Z
the inclusion maps) is a pushout. What if X1 and X2 are not open?
Solution:
`
X1
X2 be the disjoint union with inclusions ι1 : X1 → X1
`
X2 and ι2 : X2 →
`
`
X1 X2 . The maps fi : Y → Xi determine continuous maps gi = ιi fi : Y → X1 X2 .
`
Let p: X1
X2 → Z be the coequalizer of g1 and g2 , and pi = pιi : Xi → Z .
(a) Let
We claim that Z with these two maps pi is a push-out. Indeed,
p1 f1 = pι1 f1 = pg1 = pg2 = pι2 f2 = p2 f2 ,
so the first property of pushout is satisfied precisely because the first property of a coequalizer
is satisfies. Draw a diagram!
For the second (universal) property, suppose qi : Xi
→ W satisfy q1 f1 = q2 f2 . First of all,
by the universal property of the disjoint union, we may patch q1 and q2 to give a continuous map
q: X1
`
X2 → W (with qιi = qi ). Add this to your diagram!
qg1 = qg2 , there is a unique continuous map h: Z → W with q = hp (by the
`
universal property of the coequalizer p: X1
X2 → Z ). It follows that qi = hpi as required.
Now since
Since coequalizers are quotient maps, we are done.
(b) Since all maps are inclusions of subspaces, it is clear that p1 f1
= p2 f2 (they both include
the intersection into the union). It remains to check that the union has the universal property of
a pushout.
So suppose qi : Xi
that q1 |X1 ∩X2
→ W are continuous maps with q1 f1 = q2 f2 . The latter condition means
= q2 |X1 ∩X2 . However, X1 and X2 form an open cover of X1 ∪ X2 and
this is the patching condition for the existence of a continuous map
h: X1 ∪ X2 → W with
qi = h|Xi = hpi .
If X1 and X2 are not open, patching may not apply (it does if X1 and X2 are both closed, but
not in general). By the construction in (a), the pushout is the quotient X1
`
X2 /X1 ∩ X2 (i.e.
X1 ∩ X2 are identified). The underlying point set
`
of this space is |X2 | ∪ |X2 |, but equipped with the quotient topology (from X1
X2 ) rather
by the equivalence relation where points in
than the subspace topology (from V ).
5. Let X be the set {0, 1} with the discrete topology and let Z
in X , with the (infinite) product topology.
= X N be the set of sequences
(a) Define
f : Z → R: (xn )n∈N
X xn
7
→
2n
n∈N
Show that f is continuous. Is f injective? What is its image?
(b) Define
g: Z → R: (xn )n∈N
X 2xn
7
→
3n
n∈N
Show that g is continuous. Is g injective? The image of g is called the Cantor set. Show that it
is a closed subset of [0, 1].
Solution:
(a) The map
f takes a sequence x = (x1 , x2 , . . .) ∈ Z to the real number with binary
expansion 0.x1 x2 x3 . . .. Hence the image of f is [0, 1] (noting that 1
not injective, because, e.g., 1/2
To show that
of
= 0.111 . . .), but f is
= 0.100 . . . = 0.0111 . . ..
f is continuous at x ∈ Z we must show, for any basic open neighbourhood V
f (x), that f ∗ (V ) contains a basic open neighbourhood of x. That is, for any open ball
Bε (f (x)) around f (x) = 0.x1 x2 . . . ∈ R, there is a basic open neighbourhood
B(x1 , . . . , xk ) = {y = (yn ) ∈ Z | yi = xi ∀i ≤ k}
of x such that f
(B(x1 , . . . , xk )) ⊂ Bε (f (x)).
More simply still, for any ε
> 0, there is a k such that, for any y ∈ Z with yi = xi for i ≤ k ,
we have |f (y) − f (x)|
< ε. But with this assumption on y we have
∞
∞
X
X
|xn − yn |
1
1
|f (x) − f (y)| ≤
≤
= k
n
n
2
2
2
n=1
n=k+1
and so we succeed by choosing k large enough that 1/2k
< ε.
(b) The continuity of g follows by a similar argument to above, with the last estimate being
∞
∞
X
1
2|xn − yn | X 2 =
.
|g(x) − g(y)| ≤
n
n
k
3
3
3
n=1
n=k+1
The map g takes a sequence (x1 , x2 , . . .) to the real number with ternary expansion 0, y1 y2 . . .,
where yn
= 2xn . As for binary, ternary expansions are not always unique. However, coin-
cidences only occur when one expansion contains the digit
or
1, e.g. 0.022 . . . = 0.100 . . .
0.122 . . . = 0.200 . . . and hence the two expressions cannot come from applying g to a
sequence (xn ). Thus g is injective.
The Cantor set
C (the image of g ) consists of all real numbers in [0, 1] which have a ternary
expansion in which all the digits are either 0 or 2. In other words.
C = [0, 1] \
[
Tn ,
n∈N
where
Tn is the set of real numbers whose nth ternary digit must be 1. Note, e.g. 1/3 =
0.100 . . . = 0.022 . . ., so 1/3 6∈ T1 . Thus T1 = (1/3, 2/3). Similarly, T2 = (1/9, 2/9)∪
(4/9, 5/9) ∪ (7/9, 8/9). In general, Tn is a union of open intervals of length 1/3n , and
hence open. Hence C is the complement of a union of open sets, hence closed.
GKS, 17/3/17