Download Problem Sheet 1 Solutions

Problem Sheet 1 Solutions - Topics in Modern
Geometry
Adam Thomas
University of Bristol, 2016
1. Prove that arbitrary intersections and finite unions of closed sets of a topological
space (X, τ ) are are again closed.
Solution: Let (X, τ ) be a topological space and suppose {Vα | α ∈ A} ⊆ τ for any
set A. Then by one of De Morgan’s laws X \ ∩α∈A Vα = ∪α∈A (X \ Vα ). The sets
X \ Vα are open by the definition of a closed set. Hence the union of them is open
and so ∩α∈A Vα is closed. Similarly, X \ ∪ni=1 Vi = ∩ni=1 (X \ Vi ), which is again open
and hence ∪ni=1 Vi is closed.
2.* Show that a topological space is discrete if and only if every subset of X is closed.
Solution: Suppose X is a discrete topological space. Let U be a subset of X. Then
X \U is open since every subset of X is open. Conversely, suppose every subset of X
is closed. Let U be a subset of X. Then X \ U is closed and hence X \ (X \ U ) = U
is open.
3. Prove that the cofinite topology (a set U is defined to be open if X \ U is finite or
U is the empty set) on a set X defines a topology. Why did you not have to prove
anything if X is finite? (Hint: we’ve already seen the cofinite topology on finite
sets!)
Solution: Let X be a set and suppose τ = {U ⊆ X | X \ U is finite} ∪ {∅}.
Following on from Ex 1. (with a little more work), to show that τ is a topology it
is enough to show that 1. ∅ and X are closed, 2. arbitrary intersections of closed
sets are closed and 3. finite intersections of closed sets are closed. Note that by the
definition of τ , a subset V is closed if and only if X \ V is open if and only if X \ V
is finite or equal to X. Since X \ ∅ is X and X \ X = ∅ is finite we have Condition
1. The arbitrary intersection of finitely many sets is finite, and intersecting with X
does not change the set, so we have Condition 2. Similarly, the finite union of finite
sets is finite, and the union with X will always be X, so a finite union of closed sets
is again closed. Hence we have all three conditions.
4.* Let X = {0, 1, 2, 3}. Do the following collections of sets define a topology on X?
Prove your claims.
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(a) τ1 = {∅, {0}, X}
(b) τ2 = {∅, {0}, {1}, X}
(c) τ3 = {∅} ∪ {U ⊆ X : |U | ≥ 2}.
Solution: In all three cases we have ∅ and X in τi . Also note that since X is finite,
it suffices to show that U1 ∪ U2 ∈ τ and U1 ∩ U2 ∈ τ for all Ui ∈ τ . Moreover, the
intersection of any set U with ∅ or X is ∅ or U , respectively. Similarly, the union
of any set U with ∅ or X is U or X, respectively.
Hence τ1 is a topology immediately, and in fact {∅, A, X} is a topology for any
proper non-empty subset A.
The union {0} ∪ {1} = {0, 1} is not in τ2 and hence τ2 is not a topology on X.
The intersection {0, 1} ∩ {0, 2} = {0} is not in τ3 and hence τ3 is not a topology on
X.
5.* Give an example of two topologies, τ1 and τ2 , on a set X such that τ = {U | U ∈
τ1 or τ2 } is not a topology on X.
Solution: We can use Part (a) and (b) of Ex 4. as follows. Let X = {0, 1, 2, 3}
and let τ1 = {∅, {0}, X} and τ2 = {∅, {1}, X}. These are topologies by (a). Then
τ1 ∪ τ2 = {∅, {0}, {1}, X}. But this is not a topology by part (b).
6. Prove Lemma 2.4.
Solution: Let X is a topological space and Y ⊆ X. Firstly, suppose Y is endowed
with the subspace topology. We want to show that the inclusion map i : Y ,→ X
is continuous. Let U be an open set of X. Then the preimage of U is just U ∩ Y ,
which is an open set of Y by definition of the subspace topology. Secondly, suppose
that µ is a topology on Y strictly coarser than the subspace topology and that
i : Y ,→ X is continuous (with Y endowed with the topology µ). Then there exists
an open set in the subspace topology not contained in µ, let this be U . But U is
open in the subspace topology if and only if there exists an open set V of X such
that V ∩ Y = U . The preimage of the open set V under i is just V ∩ Y = U . Since
i is continuous U is contained in µ, a contradiction.
7.* Let X be a discrete topological space and let Y be any topological space. Prove
that every function f : X → Y is continuous.
Solution: Suppose U is an open set of Y . Then the preimage of Y is a subset of X.
Which is open by definition of the discrete topology. So f is continuous.
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