Homework 5 [adapted from latex HW of Colin McKinney]

5.6A Show that if A ⊂ B, then \overline{A} ⊂ \overline{B}. Take x ∈ \overline{A}. By Theorem 17.5, every neighborhood U of x intersects A. Since A ⊂ B, U also intersects B, which by Theorem 17.5 implies that x ∈ \overline{B}.

5.6B Show \overline{A ∪ B} = \overline{A} ∪ \overline{B}.

A ⊂ \overline{A} and B ⊂ \overline{B}. Hence, \overline{A} ∪ \overline{B} is a closed set containing A ∪ B. Since \overline{A ∪ B} is the smallest closed set containing A ∪ B, \overline{A ∪ B} ⊂ \overline{A} ∪ \overline{B}. \overline{A} ∪ \overline{B} ⊂ \overline{A ∪ B} by 5.6C.

5.6C Show ∪\overline{A_{α}} ⊂ \overline{∪A_{α}}; give an example where equality fails.

Fix x ∈ ∪\overline{A_{α}}. This implies that x ∈ \overline{A_{α_{0}}} for some α_{0}. By Theorem 17.5, this implies that every neighborhood U of x intersects A_{α_{0}}. Since A_{α_{0}} ⊂ ∪A_{α}, we know that every neighborhood U of x intersects ∪A_{α}. By Theorem 17.5, then, x ∈ \overline{∪A_{α}}. Hence ∪\overline{A_{α}} ⊂ \overline{∪A_{α}}.

Specific counter-example where equality fails: ∪_{q∈Q}\overline{\{q\}} = ∪_{q∈Q}\{q\} = Q, but \overline{∪_{q∈Q}\{q\}} = \overline{Q} = R

8 A ∩ B ⊂ A. Hence \overline{A ∩ B} ⊂ \overline{A}. Similarly \overline{A ∩ B} ⊂ \overline{B}. Hence \overline{A ∩ B} ⊂ \overline{A} ∩ \overline{B}

Alternatively, \overline{A ∩ B} is the smallest closed set containing A ∩ B. A ⊂ \overline{A} and B ⊂ \overline{B}. Hence A ∩ B ⊂ \overline{A} ∩ \overline{B}. Thus, \overline{A} ∩ \overline{B} is a closed set containing A ∩ B. Hence \overline{A ∩ B} ⊂ \overline{A} ∩ \overline{B}.

Specific counter-example where equality fails: \overline{Q} ∩ \overline{Q^{c}} = R ∩ R = R, but \overline{Q ∩ Q^{c}} = \overline{∅} = ∅.

∩A_{α} ⊂ A_{α_{0}}. Hence \overline{∩A_{α}} ⊂ \overline{A_{α_{0}}} for all α_{0}. Hence \overline{∩A_{α}} ⊂ ∩\overline{A_{α}}

Take x ∈ \overline{A} − \overline{B}. Take U open such that x ∈ U. x ∉ \overline{B} implies there exists an open set V such the x ∈ V and V ∩ B = ∅. x ∈ U ∩ V, U ∩ V is open, x ∈ \overline{A} implies there exists a y ∈ (U ∩ V) ∩ A. Suppose y ∈ B. Then y ∈ B ∩ V, a contradiction. Hence y ∈ A − B. Thus y ∈ U ∩ (A − B). Hence x ∈ \overline{A − B} and \overline{A} − \overline{B} ⊂ \overline{A − B}.

Specific counter-example where equality fails: \overline{Q} − \overline{Q^{c}} = R − R = ∅, but \overline{Q − Q^{c}} = \overline{Q} = R or \overline{[0, 1]} − \overline{(0, 1)} = [0, 1] − [0, 1] = ∅, but \overline{[0, 1] − (0, 1)} = \overline{\{0, 1\}} = \{0, 1\}.

5.11 Show that the product of two Hausdorff spaces is Hausdorff.

Let A and B be two Hausdorff spaces. Take x_{1} × y_{1}, x_{2} × y_{2} in A × B such that x_{1} × y_{1} ≠ x_{2} × y_{2}. Then either x_{1} ≠ x_{2} or y_{1} ≠ y_{2}. Without loss of generality assume x_{1} ≠ x_{2}. Since A is Hausdorff, there exists U_{1}, U_{2} such that U_{1} is a neighborhood of x_{1} in A and U_{2} is a neighborhood of x_{2} in A, where U_{1} and U_{2} are disjoint. Then U_{1} × B will be a neighborhood of x_{1} × y_{1} in A × B, and U_{2} × B will be a neighborhood of x_{2} × y_{2} in A × B. Since U_{1} and U_{2} are disjoint, it follows that U_{1} × B and U_{2} × B are disjoint. Since we have found disjoint neighborhoods for two arbitrary points in A × B, by the definition of Hausdorff, A × B is Hausdorff.

5.12 Show that a subspace of a Hausdorff spaces is Hausdorff.

Let X be a Hausdorff space, and let Y be a subspace of X. Let x_{1} and x_{2} be elements of Y such that x_{1} ≠ x_{2}. Since X is Hausdorff, there exist disjoint neighborhoods U_{1} and U_{2} in X of x_{1} and x_{2}, respectively. Hence a set containing x_{1} in Y is V_{1} = U_{1} ∩ Y, which is open in Y by definition of the subspace topology on Y. Thus V_{1} is a neighborhood of x_{1} in Y. Similarly, a set containing x_{2} in Y is V_{2} = U_{2} ∩ Y, which is open in Y by the definition of the subspace topology on Y. Thus V_{2} is a neighborhood of x_{2} in Y. Now since V_{1} ⊂ U_{1} and V_{2} ⊂ U_{2}, and U_{1} and U_{2} are disjoint, it follows that V_{1} and V_{2} are disjoint. Thus, Y is Hausdorff.