Download Problem Sheet 2 Solutions

Problem Sheet 2 Solutions - Topics in Modern
Geometry
Adam Thomas
University of Bristol, 2016
1. Let X be an infinite set and endow it with the cofinite topology τ . Is (X, τ )
Hausdorff? Is it compact?
Solution: We will prove that X is not Hausdorff but it is compact.
Suppose X is Hausdorff and let x, y ∈ X with x 6= y. Then there exist open sets U ,
V with x ∈ U , y ∈ V and U ∩ V = ∅. Since U and V are open and non-empty, we
see that X \ U and X \ V are both finite and hence the union (X \ U ) ∪ (X \ V ) is
finite. But (X \ U ) ∪ (X \ V ) = X \ (U ∩ V ) = X \ ∅ = X, which is infinite. This
is a contradiction and so we conclude that X is not a Hausdorff space.
Now we will show that X is compact. Let C = {Uα }α∈A be an open cover of X.
Let Uβ1 for β1 ∈ A, be any non-empty subset. Non-empty open subsets have finite
complements. So let X \ Uβ1 = x2 , . . . , xn . Then for each xi , we can pick βi ∈ A
such that xi ∈ Uβi (we can do this since {Uα }α∈A is a cover of X). Therefore,
{Uβi }ni=1 is a finite subcover of C.
2.* Prove that [0, 1] and (0, 1) are not homeomorphic.
Solution: By the Heine-Borel Theorem, a subset of R is compact if and only if it is
closed and bounded. Therefore [0, 1] is compact but (0, 1) is not. Since compactness
is a topological property we see that [0, 1] and (0, 1) are not homeomorphic.
3.* Does there exist an uncountable dense subset U of R such that R\U is uncountable?
Solution: Yes. One such example is U = Q ∪ [0, 1]. Then since Q is dense we
certainly have that U is dense. Moreover, [0, 1] is uncountable so U is uncountable.
Finally, R \ U is the set of irrational numbers less than 0 or greater than 1, which
is uncountable.
4.* Show that a subgroup of a topological group is again a topological group when
endowed with the subspace topology.
Solution: Suppose (G, τ, ·) is a topological group and let H be a subgroup of G.
Then m : G × G → G with m(g1 , g2 ) = g1 · g2 , and inversion, ι : G → G are
continuous. First, we claim that an open set of H × H is the intersection of an open
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set of G × G with H × H. To prove the claim take an open set of H × H, which
by definition of the product topology, is the union of sets of the form U1 × U2 for
U1 , U2 ∈ µ. By definition of µ, there exists subsets V1 , V2 ∈ τ with Ui = Vi ∩ H.
Thus U1 × U2 = (V1 × V2 ) ∩ (H × H) and the claim follows.
We need to show that (H, µ, ·) is a topological group, where µ = {U ∩ H | U ∈ τ }.
The things we need to check are that the restriction of multiplication, mH : H×H →
H, and the restriction of inversion, ιH : H → H are continuous. (These maps are
well-defined since H is a subgroup) Take an open set of H, call it U ∈ µ. Then
by definition of the subspace topology U = V ∩ H for some open set V ∈ τ . Thus
−1
m−1 (V ) is open in G×G. Now, m−1
H (U ) = m (V )∩(H×H) and hence by the claim,
−1
−1
−1
mH (U ) is open in H × H. Similarly, ι (V ) is open and hence ι−1
H (U ) = ι (V ) ∩ H
is open in H. Therefore, both mH and ιH are continuous.
5.* Prove Tychonoff’s theorem for a finite number of spaces i.e. prove the following.
Suppose that X1 , . . . , Xn are compact topological spaces. Then the product
X=
n
Y
Xi
i=1
is a compact topological space when endowed with the product topology.
Solution: We prove that X × Y is compact and the result follows by induction. Let
C = {Cα }α∈A is an open cover of X × Y . For each point (x, y) ∈ X × Y there
exists an open set C(x,y) with (x, y) ∈ C(x,y) ∈ C. A basis of the product topology
is given by products of an open set in X with an open set in Y . Hence we can find
x ∈ Ax,y ⊆ X with A(x,y) open in X and y ∈ Bx,y ⊆ Y with B(x,y) open in Y such
that A(x,y) × B(x,y) ⊆ C(x,y) .
Now let y be fixed. Then {A(x,y) | x ∈ X} is an open cover of X. So by compactness
of X there exists a finite subcover {A(x1 (y),y) , . . . , A(xny (y),y) } of X. Now,
By :=
ny
\
B(xi (y),y)
i=1
is an open set since it is a finite intersection of open sets. And moreover, {By |
y ∈ Y } is an open cover of Y . By compactness of Y , we have a finite subcover
m
S
{Vy1 , . . . , Vym } of Y . Therefore, by construction, the set {A(xk (yj ),yj ) ×B(xk (yj ),yj ) |
j=1
k = 1, . . . , ny } is a cover of X × Y . Hence
m
S
{C(xk (yj ),yj ) | k = 1, . . . , ny } is a finite
j=1
subcover of C.
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