Eksamen juni 2003

Solutions to the June 2003 exam

Problem 1

(1) Let 0 be the smallest element of S_{Ω} and n, the nth iterated immediate successor of 0. It is possible to construct such a sequence since all elements but the largest in a well-ordered set has an immediate successor [Ex 10.2.(a)]. (The uncountable ordered set S_{Ω} has no largest element since any section of it is countable while S_{ω} itself is uncountable.) Then 0 < 1 < 2 < ⋯ < n − 1 < n < ⋯ so the well-ordered subset {n | n ∈ Z_{+}} ⊂ S_{Ω} has the order type of Z_{+}. (This means that h: Z_{+} → S_{Ω} is recursively defined by h(n) = min(S_{Ω} − {h(1), …, h(n − 1)}) [Thm 8.4].)

(2) Put ω = sup Z_{+}, the least upper bound of Z_{+} in the well-ordered set S_{Ω} [Ex 10.1]. None of the elements of Z_{+} are immediate predecessors of ω, so an immediate predecessor of ω would be an upper bound for Z_{+}, smaller than the least upper bound; cf. [Ex 24.12.(c)]. (Also the smallest element 0 of S_{Ω} has no immediate predecessor for it has no predecessors at all!)

Problem 2

∂(A × B) =^{def} \overline{A × B} ∩ \overline{(X × Y) − (A × B)} =^{Hint} \overline{A × B} ∩ \overline{(X − A) × Y ∪ X × (Y − B)}

=^{[Ex 19.9]} (\overline{A}×\overline{B})∩(\overline{X − A}×Y ∪X×\overline{Y − B}) =^{[Ex 1.2]} ((\overline{A} ∩ \overline{X − A}) × (\overline{B} ∩ Y))∪((\overline{A} ∩ X) × (\overline{B} ∩ \overline{Y − B}))

=^{def} (∂A × \overline{B}) ∪ (\overline{A} × ∂B)

Problem 3

(1) [Ex 24.2] We must show that g(x) = 0 for some x. If g(x) = 0 for all x, then there is nothing to prove. If not, g assumes both positive and negative values because g(−x) = −g(x). Since S^{1} is connected [Thm 23.5], g also takes the value 0 at some point [Thm 24.3].

(2) It is not possible to imbed S^{1} in R for there do not exist injective continuous maps S^{1} ↪ R. Comment: The Borsuk–Ulam theorem says that for any continuous map f: S^{n} → R^{n}, n ≥ 1, there is a point x ∈ S^{n} such that f(x) = f(−x). Problem 4 [Ex 38.7] [2, 1]

(1) Let F: β(X) → {0, 1} be the extension [Thm 38.4] of the continuous function f: X → {0, 1} given by f(A) = 0 and f(X − A) = 1. Then \overline{A} ⊂ F^{−1}(0) and \overline{X − A} ⊂ F^{−1}(1) so these two subsets are disjoint; in other words \overline{X − A} ⊂ β(X) − \overline{A}.

(2) The inclusions

β(X) − \overline{A} =^{def} \overline{X} − \overline{A} ⊂^{[Ex17.8]} \overline{X − A} ⊂^{(1)} β(X) − \overline{A}

tell us that β(X) − \overline{A} = \overline{X − A}. In particular, \overline{A} is open (and closed).

(3) Since U ∩ X is a subset of U, it is clear that \overline{U ∩ X} ⊂ \overline{U} [Ex 17.6.(a)]. Conversely, let x be a point in \overline{U} and V any neighborhood of x. Then V ∩ U ≠ ∅ is nonempty for x lies in the closure of U, and hence (V ∩ U) ∩ X = V ∩ (U ∩ X) ≠ ∅ is also nonempty as X is dense. Thus every neighborhood V of x intersects U ∩ X nontrivially. This means that x ∈ \overline{U ∩ X}. We conclude that \overline{U ∩ X} = \overline{U}. From (2) (with A = U ∩ X) we see that \overline{U} is open (and closed).

(4) Let Y be any subset of β(X) contaning at least two distinct points, x and y. We shall show that Y is not connected. Let U ⊂ β(X) be an open set such that x ∈ U and y ∉ \overline{U}; such an open set U exists because β(X) is Hausdorff [Definition, p. 237]. Then Y = (Y ∩ \overline{U}) ∪ (Y − \overline{U}) is a separation of Y, so Y is not connected.

References

[1] Russell C. Walker, The Stone-Čech compactification, Springer-Verlag, New York, 1974, Ergebnisse der Mathematik und ihrer Grenzgebiete, Band 83. MR 52 #1595

[2] Nancy M. Warren, Properties of Stone-Čech compactifications of discrete spaces, Proc. Amer. Math. Soc. 33 (1972), 599–606. MR 45 #1123