M55: Exercise sheet 6

Alexandroff one-point compactification Here is a construction for making a topological space compact by adding a single new point. Let (X, 𝒯 ) be a topological space and p a point not in X. Write X^{∗} = X ∪ {p} and let 𝒯 ^{∗} consist of 𝒯 together with all subsets of X^{∗} of the form V ∪ {p} where X \ V is a closed, compact subset of X.

1. (a) Prove that 𝒯 ^{∗} is a topology on X^{∗}.

(b) Prove that 𝒯 is the topology on X induced by 𝒯 ^{∗}.

(c) Prove that (X^{∗}, 𝒯 ^{∗}) is compact.

2. Prove that if X is non-compact then X is dense in X^{∗}.

Deduce that if X is non-compact and connected then X^{∗} is connected.

3. Let (X, 𝒯 ) be the Sierpinski 2-point space. Write down the open sets in X^{∗}.

4. If X = (0, 1) with the metric topology, prove that X^{∗} is homeomorphic to the circle S^{1} = {(x, y) ∈ ℝ^{2} : x^{2} + y^{2} = 1}.

Definition. A topological space is locally compact if every point in X has a compact neighbourhood (thus, for each x ∈ X, there is an open set U and a compact set C with x ∈ U ⊂ C.)

5. Show that X^{∗} is Hausdorff if and only if X is Hausdorff and locally compact. On connectedness

6. Let A, B ⊂ X with A ∩ B ≠ ∅. Prove:

(a) If A, B are connected, so is A ∪ B.

(b) If A, B are path-connected, so is A ∪ B.

7. Let f : X → Y be continuous and X path-connected. Show that f(X) is pathconnected.

8. Show that X × Y is path-connected if and only if X and Y are path-connected.

April 17, 2003

Home page: http://www.maths.bath.ac.uk/~feb/math0055.html

M55: Exercise sheet 6—Solutions

1. (a) This is a matter of checking the axioms. First, note that ∅ ∈ 𝒯 ⊂ 𝒯 ^{∗} while X^{∗} = X ∪{p} and X \ X = ∅ which is trivially closed and compact. Thus X^{∗} ∈ 𝒯 ^{∗} also. Now let A, B ∈ 𝒯 ^{∗} and contemplate A ∩ B. There are two cases to consider: first suppose that p ∈ A ∩ B. Then p ∈ A and p ∈ B so that X^{∗} \ (A ∩ B) = (X^{∗} \ A) ∪ (X^{∗} \ B) is a union of closed, compact subsets of X and so is itself a closed compact subset of X. Thus, in this case, A ∩ B ∈ 𝒯 ^{∗}. Otherwise, p ∉ A ∩ B and so, without loss of generality, p ∉ A. Then A ∩ B = A ∩ (X ∩ B) which is a union of sets in 𝒯 since X ∩ B has closed (and compact) complement in X. Thus A ∩ B ∈ 𝒯 ⊂ 𝒯 ^{∗}. Finally, consider an arbitrary family {U_{α}}_{α∈I} of sets in 𝒯 ^{∗} and their union ⋃_{α∈I} U_{α}. Again there are two cases: if p ∉ ⋃_{α∈I} U_{α} then each U_{α} ∈ 𝒯 and so their union lies in 𝒯 ⊂ 𝒯 ^{∗}. If, on the other hand, p ∈ ⋃_{α∈I} U_{α}, then p ∈ U_{α_{0}}, say, and X \ ⋃_{α∈I} U_{α} = ⋂_{α∈I}(X \ U_{α}) is an intersection of closed subsets of X and so a closed subset of X. Moreover, since X \⋃_{α∈I} U_{α} ⊂ X \ U_{α_{0}} which last is a compact subset of X, we deduce that X \ ⋃_{α∈I} U_{α} is compact also (closed subsets of compact sets are compact) and so ⋃_{α∈I} U_{α} ∈ 𝒯 ^{∗} also. Thus 𝒯 ^{∗} is a topology on X^{∗}.

(b) X is open in X^{∗} so that U ⊂ X is open in the induced topology if and only if U is open in X^{∗}. However, the sets in 𝒯 ^{∗} that lie in X are precisely the elements of 𝒯 . Thus 𝒯 = 𝒯_{X}^{∗} as required.

(c) Let {U_{α}}_{α∈I} be an open cover for X^{∗}. Then p lies in some U_{α_{0}} so contemplate C = X^{∗}\U_{α_{0}} which is a closed compact subset of X. Now {X ∩ U_{α}}_{α≠α_{0}} is an open cover of C in (X, 𝒯 ) (each X ∩ U_{α} ∈ 𝒯 thanks to part (b)) and so there is a subcover X ∩ U_{α_{1}}, …, X ∩ U_{α_{k}}. Then U_{α_{0}}, …, U_{α_{k}} is a subcover of {U_{α}}_{α∈I} and X^{∗} is compact.

2. X is dense in X^{∗} if and only if \overline{X} = X^{∗} which holds if and only if X is not closed or equivalently {p} is not open. But this happens exactly when X \∅ = X is not compact (since X is certainly closed). If X is non-compact we now know that \overline{X} = X^{∗} so that, in this case, X^{∗} is connected when X is (the closure of a connected set is connected).

3. Let X = {a, b} with 𝒯 = {∅, {a}, {a, b}}. Then the closed sets of X are {∅, {b}, {a, b}} (and all these are compact since finite) so that 𝒯 ^{∗} = {∅, {a}, {a, b}, {p}, {a, p}, {a, b, p}}.

4. Define f : X^{∗} → S^{1} by

f(t) = \begin{cases}(cos 2πt, sin 2πt) & for t ∈ (0, 1); \\ (1, 0) & for t = p. \end{cases}

f is certainly a bijection, X^{∗} is compact and S^{1} is Hausdorff so that f is a homeomorphism as soon as it is continuous. Certainly f is continuous on (0, 1) so the issue is to see that f^{−1}(U) is open for U containing the point (1, 0). Moreover, we may assume that U = S^{1} ∩ (1 − ε, 1 + ε) × (−ε, ε) for some small ε (why?). In this case,

f^{−1}(U) =

\{p\} ∪ (((0, \frac{1}{2π} sin^{−1} ε) ∪ (1 − \frac{1}{2π} sin^{−1} ε)) ∩ ((0, \frac{1}{2π} cos^{−1}(1 − ε)) ∪ (1 − \frac{1}{2π} cos^{−1}(1 − ε), 1)))

so that (0, 1) \ f^{−1}(U) is an intersection

[\frac{1}{2π} sin^{−1} ε, 1 − \frac{1}{2π} sin^{−1} ε] ∩ [\frac{1}{2π} cos^{−1}(1 − ε), 1 − \frac{1}{2π} cos^{−1}(1 − ε)]

of compact intervals and so compact. Thus f^{−1}(U) ∈ 𝒯 ^{∗} and f is continuous.

5. First suppose that X^{∗} is Hausdorff. Then X is Hausdorff being a subspace of a Hausdorff space (see question 2, part (b)). Moreover, for x ∈ X, there are disjoint sets U, V ∈ 𝒯 ^{∗} with x ∈ U and p ∈ V. Now p ∉ U so that U ∈ 𝒯 and C = X^{∗} \ V is a closed, compact subset of X with x ∈ U ⊂ C. Thus X is locally compact. The converse is similar: let X is Hausdorff and locally compact. If x ≠ y ∈ X, there are open disjoint sets in 𝒯 (and so in 𝒯 ^{∗}) with x ∈ U and y ∈ V. Again, if x ∈ X, there is an open set U and a compact set C with x ∈ U ⊂ C. Then V = (X \ C) ∪ {p} ∈ 𝒯 ^{∗} contains p and is disjoint from U. Thus X^{∗} is Hausdorff.

6. (a) Let f : A ∪ B → {−1, 1} be continuous. We must show that f is constant. However, f|_{A} is continuous on A and so constant and similarly f|_{B} is constant. Since A ∩ B ≠ ∅, the values of f on A and B must coincide so that f is constant on A ∪ B.

(b) Fix x_{0} ∈ A ∩ B. Since both A and B are path-connected, there are paths in A or B from x_{0} to any point of A or B. Thus if x, y ∈ A ∪ B, there are paths γ_{0} from x to x_{0} and γ_{1} from x_{0} to y and so a path γ_{0} · γ_{1} in A ∪ B from x to y. Thus A ∪ B is path-connected.

7. Let y_{1}, y_{2} ∈ f(X). Then there are x_{1}, x_{2} ∈ X with y_{i} = f(x_{i}), i = 1, 2, and a path γ in X from x_{1} to x_{2}. Then f ∘ γ is a path in f(X) from y_{1} to y_{2}.

8. If X × Y is path-connected then so are X and Y by question 7 since they are the images of the continuous projections π_{1} and π_{2}. Conversely, suppose that X and Y are path-connected and let (x_{1}, y_{1}), (x_{2}, y_{2}) ∈ X × Y. There are paths γ_{1} in X from x_{1} to x_{2} and γ_{2} in Y from y_{1} to y_{2} and so we can define γ : [0, 1] → X × Y by γ(t) = (γ_{1}(t), γ_{2}(t)). It is clear that γ is continuous, since both γ_{i} are, so that γ is a path from (x_{1}, y_{1}) to (x_{2}, y_{2}).