1st December 2004

Munkres §32

Ex. 32.1. Let Y be a closed subspace of the normal space X. Then Y is Hausdorff [Thm 17.11]. Let A and B be disjoint closed subspaces of Y. Since A and B are closed also in X, they can be separated in X by disjoint open sets U and V. Then Y ∩ U and V ∩ Y are open sets in Y separating A and B.

Ex. 32.3. Look at [Thm 29.2] and [Lemma 31.1]. By [Ex 33.7], locally compact Hausdorff spaces are even completely regular.

Ex. 32.4. Let A and B be disjoint closed subsets of a regular Lindelöf space. We proceed as in the proof of [Thm 32.1]. Each point a ∈ A has an open neighborhood U_{a} with closure \overline{U}_{a} disjoint from B. Applying the Lindelöf property to the open covering {U_{a}}_{a∈A} ∪ {X − A} we get a countable open covering {U_{i}}_{i∈Z_{+}} of A such that the closure of each U_{i} is disjoint from B. Similarly, there is a countable open covering {V_{i}}_{i∈Z_{+}} of B such that the closure of each V_{i} is disjoint from A. Now the open set ⋃ U_{i} contains A and ⋃ V_{i} contains B but these two sets are not necessarily disjoint. If we put U′_{1} = U_{1} − \overline{V_{1}}, U′_{2} = U_{2} − \overline{V_{1}} − \overline{V_{2}}, …, U′_{i} = U_{i} − \overline{V_{1}} − ⋯ − \overline{V_{i}}, … we subtract no points from A so that the open sets {U′_{i}} still form an open covering of A. Similarly, the open sets \{V′_{i}\}, where V′_{i} = V_{i} − \overline{U_{1}} − ⋯ − \overline{U_{i}}, cover B. Moreover, the open sets ⋃ U′_{i} and ⋃ V′_{i} are disjoint for U′_{i} is disjoint from V_{1} ∪ ⋯ ∪ V_{i} and V′_{i} is disjoint from U_{1} ∪ ⋯ ∪ U_{i}.

Ex. 32.5. R^{ω} (in product topology ) is metrizable [Thm 20.5], in particular normal [Thm 32.2]. R^{ω} in the uniform topology is, by its very definition [Definition p. 124], metrizable, hence normal.

Ex. 32.6. Let X be completely normal and let A and B be separated subspaces of X; this means that A ∩ \overline{B} = ∅ = \overline{A} ∩ B. Note that A and B are contained in the open subspace X − (\overline{A} ∩ \overline{B}) = (X − \overline{A}) ∪ (X − \overline{B}) where their closures are disjoint. (The closure of A in X − (\overline{A} ∩ \overline{B}) is \overline{A} − \overline{B} [Thm 17.4].) The subspace X − (\overline{A} ∩ \overline{B}) is normal so it contains disjoint open subsets U ⊃ A and V ⊃ B. Since U and V are open in an open subspace, they are open [Lemma 16.2].

Conversely, suppose that X satisfies the condition (and is a T_{1}-space). Let Y be any subspace of X and A and B two disjoint closed subspaces of Y. Since \overline{A} ∩ Y and \overline{B} ∩ Y are disjoint [Thm 17.4], \overline{A} ∩ B = \overline{A} ∩ (Y ∩ B) = (\overline{A} ∩ Y) ∩ (B ∩ Y) = ∅, and, similarly, A ∩ \overline{B} = ∅. By assumption, A and B can then be separated by disjoint open sets. If we also assume that X is T_{1} then it follows that Y is normal.

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