Eksamen marts 2004

Solutions to the January 2004 exam

Problem 1 [Ex 30.16]

(1) The set D_{k} is countable for there is an injective map

D_{k} ↪ (\overbrace{Q × Q) × ⋯ × (Q × Q)}^{k} × \overbrace{Q × ⋯ × Q}^{k} = Q^{3k}

of D_{k} into a countable set [Cor 7.3]. As a countable union of countable sets, D = ⋃_{k∈Z_{+}} D_{k} is countable [Thm 7.5].

(2) The basis open sets in R^{I} are finite intersections ⋂_{j=1}^{k} π_{i_{j}}^{−1}(U_{i_{j}}) where i_{1}, …, i_{k} are k distinct points in I and U_{i_{1}}, …, U_{i_{k}} are k open subsets of R. Choose disjoint closed subintervals I_{j} such that i_{j} ∈ I_{j} and choose x_{j} ∈ U_{i_{j}} ∩ Q, j = 1, …, k. Then x(I_{1}, …, I_{k}, x_{1}, …, x_{k}) ∈ ⋂_{j=1}^{k} π_{i_{j}}^{−1}(U_{i_{j}}) for π_{i_{j}}x(I_{1}, …, I_{k}, x_{1}, …, x_{k}) = x_{j} ∈ U_{i_{j}} for all j = 1, …, k. This shows that any (basis) open set contains an element of x(D).

(3) Let D be a dense subset of R^{J} for some set J. Let f: J → P(D) be the map from the index set J to the power set of D given by f(j) = D ∩ π_{j}^{−1}(2003, 2004). For j and k two points of J we have

f(j) = f(k) ⟺ D ∩ π_{j}^{−1}(2003, 2004) = D ∩ π_{k}^{−1}(2003, 2004)

⟹ \overline{D ∩ π_{j}^{−1}(2003, 2004)} = \overline{D ∩ π_{k}^{−1}(2003, 2004)}

⟺ \overline{π_{j}^{−1}(2003, 2004)} = \overline{π_{k}^{−1}(2003, 2004)}

⟺^{[Thm 19.5]} π_{j}^{−1}[2003, 2004] = π_{k}^{−1}[2003, 2004] ⟺ j = k

and therefore f is injective. Thus cardJ ≤ cardP(D).

Here is a better proof (due to one of the students) that f is injective: Let j and k be two distinct points in J. Then f(j) ≠ f(k) for

f(j) − f(k) = (π_{j}^{−1}(2003, 2004) − π_{k}^{−1}(2003, 2004)) ∩ D

⊃ (π_{j}^{−1}(2003, 2004) ∩ π_{k}^{−1}(2002, 2003)) ∩ D ≠ ∅

since D is dense. Problem 2

(1) Any open subset of a locally compact Hausdorff space is a locally compact Hausdorff space [Thm 29.3].

(2) The open sets in ω(X − A) = (X − A) ∪ {ω} are of the form U, where U is open in X − A, or (X − A) − C ∪ {ω}, where C ⊂ X − A is compact. In the first case, f^{−1}(U) = U is open in X since U is open in X − A which is open in X. In the second case f^{−1}((X − A) − C ∪ {ω}) = (X − A) − C ∪ A = X − C is open since C is closed because it is a compact subset of a Hausdorff space.

(3) The universal property of quotients spaces implies that the map f factors through X/A since it sends A to one point [Thm 22.2]. The induced continuous map \overline{f}: X/A → ω(X − A) is clearly bijective.

(4) If X is compact, the quotient space X/A is also compact [Thm 26.5]. The map \overline{f} is now a continuous bijective map of a compact space onto a Hausdorff space, hence a homeomorphism [Thm 26.6]. In fact \overline{f} is a homeomorphism if and only if X/A is compact.

(5) Let p: X → X/A denote the quotient map. The quotient space X/A contains the infinite discrete closed subspace \{p(2n − \frac{1}{2}) | n ∈ Z\} so it is not compact [Thm 28.1]. [The notes [1] contain a similar argument showing that X/A = R/Z is not even locally compact.] On the other hand, The Alexandroff compactfication ω(X − A) is compact. These two spaces are therefore not homeomorphic.

(6) Let {U_{n}} be a countable collection of neighborhoods of A. Let U be the neighborhood of A that in the interval (2n − 1, 2n) equals U_{n} ∩ (2n − 1, 2n) with one point removed. Then U does not contain any of the U_{n}. Thus X/A is not first countable at the point p(A) (as shown in [1]).

(7) The Alexandroff compactification ω(X − A) is homeomorphic to the Hawaiian Earring [Example 1, p 436] ⋃_{n∈Z_{+}} C_{1/n} (as shown in [1]). Since the quotient space X/A is not first countable it can not embed into a first countable space [Thm 30.2] (as observed in [1]).

References

[1] Jesper M. Møller, General topology, http://www.math.ku.dk/ moller/e03/3gt/notes/gtnotes.dvi.