M55: Exercise sheet 4

1. Let A ⊂ X and B ⊂ Y be closed subsets of topological spaces X and Y. Show that A × B is closed (with respect to the product topology, of course).

2. Suppose X and Y are topological spaces and that y is a point of Y. Show that X is homeomorphic to the subspace X × {y} of X × Y.

3. A map f : X → Y of topological spaces is open if f(U) is open in Y for all U open in X. Show that the projection π_{1} : X × Y → X is open. Is the image of a closed set necessarily closed?

4. Show that the product topology is the unique topology on X × Y with the property that for any map h = (f, g) : Z → X × Y of a topological space Z, h is continuous if and only if f : Z → X and g : Z → Y are continuous.

5. Show that if X is Hausdorff then every singleton set {x}, x ∈ X, is closed.

6. Show that X is Hausdorff if and only if, whenever x ≠ y, there is an open set U with x ∈ U and y ∉ \overline{U}.

7. Is the Moore plane Hausdorff?

March 14, 2003

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

M55: Exercise sheet 4—Solutions

1. We have no choice but to show that (X × Y) \ (A × B) is open, that is, a union of basic open sets. Now (x, y) ∉ A × B if and only if either x ∉ A or y ∉ B. That is,

(X × Y) \ (A × B) = (X \ A) × Y ∪ X × (Y \ B).

Since X \A is open in X and Y \B is open in Y, this last is a union of basic (in fact, subbasic) open sets and so open.

2. Define f : X → X ×{y} by f(x) = (x, y). Now f is bijective with continuous inverse π_{1}|_{X×{y}}. For the continuity of f, note that π_{1} ∘ f = id_{X} and π_{2} ∘ f is constant so that both components of f are continuous whence f is continuous as a map into X × Y and thus as a map into X × {y} with the subspace topology.

3. Any open set U in X × Y is a union of basic open sets: U = ⋃_{α∈I} U_{α} × V_{α} and then

π_{1}(U) = π_{1}( ⋃_{α∈I} U_{α} × V_{α}) = ⋃_{α∈I} π_{1}(U_{α} × V_{α}) = ⋃_{α∈I} U_{α}

which is a union of open sets in X and so open. The converse is false: let C = {(x, y) ∈ ℝ^{2} : x > 0 and y = 1/x}. This is a closed subset of ℝ^{2} (use sequences!) but π_{1}(C) = (0, ∞) which is not open in ℝ.

4. We have already proved in lectures that the product topology has this property. For uniqueness, denote the product topology by 𝒯 and suppose that 𝒮 is another topology with the same property. First take Z = (X × Y, 𝒮) and h = id_{X×Y} = (π_{1}, π_{2}). By the property, since h = id_{X×Y} is continuous so are the π_{i} but this means, by the fact that 𝒯 also has the property that id_{X×Y} : (X × Y, 𝒮) → (X × Y, 𝒯 ) is continuous. Thus 𝒯 ⊂ 𝒮. Conversely, take Z = (X × Y, 𝒯 ) and once more h = id_{X×Y} = (π_{1}, π_{2}). We know that the π_{i} are continuous with respect to 𝒯 so the property gives that id_{X×Y} : (X ×Y, 𝒯 ) → (X ×Y, 𝒮) is continuous. Thus 𝒮 ⊂ 𝒯 so that 𝒮 = 𝒯 and we are done.

5. For y ≠ x, there are disjoint open sets U_{y}, V_{y} with x ∈ U_{y} and y ∈ V_{y}. In particular, x ∉ V_{y} or, otherwise said, V_{y} ⊂ X \ {x}. Thus

X \ {x} = ⋃_{y≠x} V_{y}

which is open so that {x} is closed.

6. Suppose X is Hausdorff and x ≠ y. Then there are disjoint open U, V with x ∈ U and y ∈ V. Since U ∩ V = ∅, U ⊂ X \ V which last is closed so that \overline{U} ⊂ X \ V. In particular, y ∉ \overline{U}. For the converse, if there is an open U with x ∈ U and y ∉ \overline{U} then put V = X \ \overline{U}. V is open (since \overline{U} is closed), disjoint from U (since U ⊂ \overline{U}) and certainly contains y. Thus X is Hausdorff.

7. Yes. Draw pictures to see that basic open sets build out of sufficiently small ε-balls will separate any two points.