Download Question 1: a) Give the definition of a homeomorphism. [2] b) Find

Manifolds & Topology, MAT3009/3035
Test 1, Monday February 27 2012
Question 1:
a) Give the definition of a homeomorphism.
[2]
b) Find an explicit homeomorphism between the following spaces (all with standard
topology), or give an argument why there cannot be any:
(i) (−1, 1);
(ii) R;
(iii) ((−1, 1) × {0}) ∪ ({0} × (0, 1)).
[4]
c) Let S = {z ∈ C : |z| = 1} be the unit circle and let f : [0, 1) → S be defined
by f (x) = e2πix . Is f a homeomorphism? Justify your answer.
1
1
[4]
Question 2: Let (X, O) be a topological space and A ⊂ X.
a) Give the definition of a boundary point.
[2]
b) Assume now that X = (0, 1] with Euclidean topology. Take A = (0, 13 ) ∪ ( 23 , 1].
Identify ∂A.
[2]
c) Same question as b), but now when X has co-finite topology.
[3]
d) [for BSc] Is X with co-finite topology Hausdorff? Justify your answer.
[4]
e) [for MMath] Is X with co-finite topology separable? Justify your answer.
[4]
Question 3:
a) Give the definition of a (coverings-)compact set.
[2]
b) Show that in a metric space, an open ball is not compact.
[3]
c) [for BSc] Prove that in a Hausdorff space, every singleton (i.e., a set consisting
of one point) is closed.
[4]
d) [for MMath] Is the Sorgenfrey line R metrizable? Justify your answer.
[4]
Total
[30]
1
Solutions Test 1 (Manifolds & Topology)
Question 1: a) A homeomorphism is a continuous map with a continuous inverse. (In particular, a homeomorphism must be a bijection.
x
b) f : (−1, 1) → R, f (x) = 1−x
2 is a homeomorphism between the first two. [0, 1] is homeomorphic to neither : (−1, 1) nor R, because the first has a branch point which (when taken out)
divides the set in three components. No such point exists is in (−1, 1) or R.
c) f is a bijection, and continous but not a hoemomorphism, because the inverse is not continuous. Indeed, [0, 12 ) is open in [0, 1), but its image f ([0, 21 )), containing boundary point 1, is not
open in S1 .
Question 2: a) x is a boundary point of A if every neighbourhood of x intersects both A and
Ac .
b) ∂A = { 31 , 23 }.
c) ∂A = (0, 1].
d) X is not Hausdorff. For example, if x = 13 and y = 23 have neighbourhoods Ux and Uy respectively, then #(X \Ux ) < ∞ and #(X \Uy ) < ∞. Therefore Ux ∩Uy ⊃ X \((X \Ux )∪(X \Uy )) 6= ∅,
because X is an infinite set.
e) X is separable. Every infinite set is dense in X, in particular sets that are countably infinite.
Question 3:
a) A set A in a topological space is compact if every open open of A has a finite subcover. b)
n
) : n ∈ N} is an open cover without finite subcover,
Let B(x; r) be an open ball, then {B(x; r n+1
because if I only take the first N such balls, then I cover B(x; r NN+1 ) but not the entire ball
B(x; r).
Alternatively, in a Hausdorff space (so also in a metric space), every compact set is closed. If
r is small enough, B(x; r) is not closed as it won’t contain all its bounadary points. Therefore
B(x; r) is cannot be compact.
c) For every y 6= x, let Uy 3 x and Vy 3 y be disjoint open sets. Then ∪y6=x Vy is open and
{x} = (∪y6=x Vy )c is closed.
Alternatively, for every y 6= x, there is an open neighbourhood Vx disjoint from {x}, so y is
interior to {x}c . That means, {x}c contains only interior points, so it is open, and therefore {x}
is closed.
d) The Sorgenfrey line is separable (Q is a countable dense subset), but not second countable
(you need an uncountable basis of the topology to obtain all open sets of the form [x, ∞). Separable metric spaces are second countable, so the Sorgenfrey cannot have a metric.
General remarks:
• In definitions, get the quantifiers right and in the right order!
• In theorems (and hence proofs), get the direction of the implication right!
• Some linguistics: An open set U can have boundary points, only none of these boundary
points belongs to A. So A doesn’t contain boundary points.
A intersects B means A ∩ B 6= ∅. A is in B means A ⊂ B.
2
• To use a ball B(a; r) requires a metric, and hence doesn’t make sense in a topological space
without metric. In a topological space, an open set containing a is a neighbourhood of
a. If you have a metric, the ball B(a; r) is an example of a neighbourhood of a, but a has
many other neighbourhoods as well.
• Also the notion of bounded requires a metric.
• If A is not closed, that does not mean it is open. For example, in the Euclidean line, [0, 1),
Q and { n1 : n ∈ N} are examples of sets that are neither open nor closed.
3