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Math 344-1,2: Kursobjekt

Aaron Greicius

Section 1.19 Regular and normal spaces

Definition 1.19.1. Separated by open sets.

Let \(X\) be a topological space. Subsets \(A,B\subseteq X\) are separated by open sets if there exist disjoint open sets \(U, V\) such that \(A\subseteq U\) and \(B\subseteq V\text{.}\)

Definition 1.19.2. Regular and normal spaces.

Let \(X\) be a topological space.
  • \(X\) is regular if \(X\) is \(T_1\text{,}\) and given any closed set \(C\) and any \(x\notin C\text{,}\) the sets \(\{x\}\) and \(C\) can be separated by open sets.
  • \(X\) is normal if \(X\) is \(T_1\text{,}\) and any two disjoint closed sets of \(X\) can be separated by open sets.

Remark 1.19.3. Separation axioms.

The properties of being \(T_1\text{,}\) Hausdorff, regular, or normal are called separation axioms as they articulate how various subsets can be separated by open sets. Below you find the descriptions of these properties using the Trennungsaxiom schema.
\begin{equation*} \begin{array}{r|l} \text{Name} \amp T_n \\ \hline T_1 \amp T_1 \\ \text{Hausdorff} \amp T_2 \\ \text{Regular} \amp T_3 \\ \text{Normal} \amp T_4 \end{array} \end{equation*}
A further remark about the use of “regular” and “normal” in the literature: some texts do not include the \(T_1\) condition in these notions, and describe the properties we define as “regular Hausdorff” and “normal regular”. There is considerably less ambiguity in the literature when using the \(T_n\) descriptions, but you should still be careful.

Remark 1.19.4. Trennungsaxiom implications.

It is easy to see that \(T_4\implies T_3\implies T_2\implies T_1\text{.}\) Furthermore, your argument for the first two implications will reveal why the \(T_1\) condition is needed.

Example 1.19.5. Elementary examples.

  1. Discrete spaces are regular and normal.
  2. An infinite space with the cofinite topology is neither regular nor normal.
Solution.
(1) is clear since all sets are both open and closed. (2) follows from the fact that in this space any two nonempty open sets intersect nontrivially.

Proof.

First observe that metric spaces are \(T_1\text{.}\)
Let \(A,B\) be disjoint closed subsets of \(X\text{.}\) For each \(a\in A\) we can find an \(\epsilon_a> 0 \) such that \(B_{\epsilon_a}(a)\cap B=\emptyset\text{.}\) Similarly for each \(b\in B\) we can find an \(\epsilon_b> 0\) such that \(B_{\epsilon_b}(b)\cap A=\emptyset\text{.}\) I claim the open sets
\begin{align*} U \amp =\bigcup_{a\in A}B_{\epsilon_a/2}(a)\\ V \amp =\bigcup_{b\in B}B_{\epsilon_b/2}(b) \end{align*}
are disjoint, proving \(A\) and \(B\) are separated by open sets. Indeed if \(x\in U\cap V\text{,}\) then \(x\in B_{\epsilon_a/2}(a)\cap B_{\epsilon_b/2}(b)\) for some \(a\in A\) and \(b\in B\text{.}\) Without loss of generality we may assume that \(\epsilon_a\geq \epsilon_b\text{.}\) But then we would have
\begin{align*} d(x,b) \amp \geq d(a,b)-d(a,x) \amp (\text{triangle ineq.}) \\ \amp \geq \epsilon_a-d(a,x) \amp (b\notin B_{\epsilon_a}(a))\\ \amp \geq \epsilon_a/2\\ \amp \geq \epsilon_b/2\text{.} \end{align*}
This is a contradiction since we assumed \(x\in B_{\epsilon_b/2}(b)\text{.}\)

Example 1.19.7. \(\R_K\) is not regular.

Prove: \(\R_K\) is Hausdorff, but not regular.
Solution.
Since the \(K\)-topology is finer than the standard topology on \(\R\text{,}\) \(\R_K\) is Hausdorff. To see that it is not regular, recall that \(K=\{1/n\colon n\in \Z_+\}\) is closed. I’ll show that we cannot separate \(0\) and \(K\) with open sets. Indeed, suppose we have open disjoint sets \(U, V\) containing \(0, K\) respectively. Since \(U\) is open, it must contain an open basis element \(B\) containing \(0\text{;}\) and since \(U\cap K=\emptyset\text{,}\) this basis element can be chosen of the form \((-\epsilon, \epsilon)-K\text{.}\) Now choose \(n\in \Z_+\) such that \(1/n\in (-\epsilon, \epsilon)\text{.}\) Since \(1/n\in K\) we can find a basis element of the form \((1/n-\delta, 1/n+\delta)\) that is contained in \(V\text{.}\) Furthermore, shrinking \(\delta\) if necessary, we can assume that \((1/n-\delta, 1/n+\delta)\subseteq (-\epsilon, \epsilon)\text{.}\) But then clearly \((1/n-\delta, 1/n+\delta)\cap \left((-\epsilon, \epsilon)-K\right)\ne\emptyset\text{,}\) and hence also \(U\cap V\ne \emptyset\text{.}\) Contradiction!

Proof.

  1. First, assume \(X\) is regular. Given \(x\) and \(U\) as described, we can separate \(\{x\}\) and the closed set \(C=X-U\) by open sets \(V\) and \(V'\text{.}\) It follows that \(\overline{V}\subseteq U\text{:}\) indeed, given any \(y\in X-U=C\text{,}\) the open set \(V'\) contains \(y\) and does not intersect with \(V\text{.}\)
    Conversely, assume for all \(x\in X\) and open sets \(U\ni x\) we can find an open set \(V\) satisfying \(x\in V\subseteq\overline{V}\subseteq U\text{.}\) Given any closed set \(C\) and element \(x\notin X\text{,}\) letting \(U=X-C\text{,}\) we see that there is an open set \(V\) satisfying \(x\in V\subseteq \overline{V}\subseteq U\text{.}\) The open sets \(V\) and \(V'=X-\overline{V}\) are then easily seen to separate \(x\) and \(C\text{.}\)
  2. This proof is very similar to the last. Essenially you need only replace \(x\) with a closed set \(A\) throughout.

Proof.

Remark 1.19.10. Locally compact and non-normal.

Yes, there are examples of locally compact Hausdorff spaces that are not normal. They are not particularly easy to build, however. Look it up!

Example 1.19.11. \(\R_\ell\) is regular.

Prove: \(\R_\ell\) is regular.
Solution.
Since \(\R_\ell\) has a finer topology than \(\R\text{,}\) it is \(T_1\text{.}\) We will use the equivalence (1) from Theorem 1.19.8. Given any \(a\in \R\) and open set \(U\ni a\text{,}\) we can find \(b\) such that \(V=[a,b)\subseteq U\text{.}\) But \(V=\overline{V}\) is closed! (Prove this.) Thus we have found an open \(V\) containing \(a\) such that \(\overline{V}\subseteq U\text{.}\)

Remark 1.19.12. Basis of clopen sets.

The argument from Example 1.19.11 is easily generalized to show that any \(T_1\)-space that has a basis consisting of sets that are both open and closed (i.e., “clopen” sets) is regular.

Proof.

  1. Assume \(Y\) is a subspace of the regular space \(X\text{.}\) We know already that subspaces of \(T_1\)-spaces are \(T_1\text{,}\) so \(Y\) is \(T_1\text{.}\) Next take any \(y\in Y\) and any set \(C\subseteq Y\) not containing \(y\) that is closed in \(Y\text{.}\) We have \(C=C'\cap Y\) for some closed set \(C'\subseteq X\text{.}\) Since \(y\notin C'\) and \(X\) is regular, we can find disjoint open sets \(U, V\) that separate \(y\) and \(C'\) in \(X\text{.}\) The disjoint open (in \(Y\)) sets \(U'=U\cap Y\) and \(V'=V\cap Y\) are then a separation of \(y\) and \(C\) in \(Y\text{.}\)
  2. We use (1) from Theorem 1.19.8. Given \(x=(x_i)_{i\in I}\) and open set \(U\) containing \(x\text{,}\) choose a basis element of the form \(\prod_{k=1}^nU_{i_k}\times \prod_{i\ne i_k}X_i\subseteq U\) containing \(x\text{.}\) Since each \(X_i\) is regular, we can find open sets \(V_{i_k}\subseteq X_{i_k}\) satisfying \(x\in \overline{V_{i_k}}\subseteq U_{i_k}\text{.}\) Letting \(V=\prod_{k=1}^nV_{i_k}\times\prod_{i\ne i_k}X_i\text{,}\) we have
    \begin{equation*} x\in \overline{V}=\prod_{k=1}^n\overline{V_{i_k}}\times\prod_{i\ne i_k}X_i\subseteq \prod_{k=1}^nU_{i_k}\times\prod_{i\ne i_k}X_i\subseteq U \text{.} \end{equation*}

Remark 1.19.14.

Interestingly, neither of the statements of Theorem 1.19.13 are true if “regular” is replaced with “normal”. It is worthwhile to examine what goes wrong in our arguments above if the element \(x\) is replaced with a closed set \(A\) throughout.

Example 1.19.15. \(\R_\ell\) is normal.

Prove: \(\R_\ell\) is normal.
Solution.
Let \(A, B\) be disjoint closed subsets of \(\R_\ell\text{.}\) We can find open coverings \(A\subseteq U=\bigcup_{a\in A}[a,x_a), B\subseteq V=\bigcup_{b\in B}[b, y_b)\) satisfying \([a,x_a)\cap B=[b,y_b)\cap A=\emptyset\) for all \(a\in A, b\in B\text{.}\) First observe that \([a,x_a)\cap [b,x_b)=\emptyset\) for all \(a\in A, b\in B\text{.}\) Indeed, otherwise we’d have \([a,x_a)\cap [b,x_b)=[\max(a,b),\min(x_a,y_b))\text{,}\) in which case either \(a\in [a,x_a)\cap [b,x_b) \) or \(b\in [a,x_a)\cap [b,x_b)\text{:}\) both are impossible since \(a\notin [b,x_b)\) and \(b\notin [a,x_a)\text{.}\) We conclude that \(U\cap V=\emptyset\text{,}\) proving that \(A\) and \(B\) can be separated by open sets.

Example 1.19.16. \(\R_\ell\times \R_\ell\) is not normal.

The space \(\R_\ell\times \R_\ell\) is not normal.
Solution.
See Munkres for the full proof. I’ll content myself with a discussion of some of the key elements. First, let \(D=\{(x,-x)\colon x\in\R\}\subseteq \R_\ell^2\text{.}\) This set is closed in \(\R^2\text{,}\) and hence closed in \(\R_\ell^2\text{,}\) since the left limit product topology is finer than the standard topology. Next, as we saw on a previous homework, \(D\) is discrete as a subspace of \(\R_\ell^2\text{:}\) if follows that every subset of \(D\text{,}\) being itself closed in \(D\text{,}\) is closed in \(\R_\ell^2\text{,}\) using transitivity. In particular the two sets
\begin{align*} A\amp=D\cap \Q^2=\{(x,-x)\colon x\in \Q\} \amp B\amp=D-A=\{(x,-x)\colon x\notin \Q\} \text{.} \end{align*}
By what we said above, the disjoint sets \(A\) and \(B\) are closed in \(\R_\ell^2\text{,}\) but as it turns out you cannot separate them by open sets. The proof is not terribly difficult, but will seem somewhat convoluted without knowing about the Baire category theorem. As such, I omit it. (Munkres doesn’t use this approach, choosing instead a proof by contradiction, but he outlines a proof in one of his exercises.)

Proof.

Since \(X\) is compact Hausdorff, it is locally compact Hausdorff, and hence regular by Theorem 1.19.9. We use the equivalent condition of normality from Theorem 1.19.8. To this end, let \(A\subseteq X\) be closed, and let \(U\) be any open set containing \(U\text{.}\) Since \(X\) is regular, for all \(a\in A\) we can find an open set \(V_a\) containing \(a\) such that \(\overline{V_a}\subseteq U\text{.}\) We have \(A\subseteq \bigcup_{a\in A}V_a\text{.}\) Since \(X\) is compact and \(A\) is closed, \(A\) is compact. Hence there is a finite subcovering \(A\subseteq V=\bigcup_{k=1}^nV_{a_k}\text{.}\) We have
\begin{equation*} \overline{V}=\bigcup_{k=1}^n\overline{V_{a_k}}\subseteq U\text{,} \end{equation*}
as desired.

Proof.

Assume \(X\) is regular and second countable. Let \(\mathcal{B}=\{B_n\colon n\in \Z_+\}\) be a countable basis for \(X\text{.}\) Let \(A\) and \(B\) be disjoint closed sets. By regularity, for each \(a\in A\) we can find disjoint open sets \(U, V\) containing \(a\) and \(B\) respectively. Since \(\mathcal{B}\) is a basis, we can find a basis element \(B_{n_a}\) such that \(a\in B_{n_a}\subseteq \overline{B_{n_a}}\subseteq X-B\text{.}\) The set of all such \(B_{n_a}\) is a countable subset of \(\mathcal{B}\text{,}\) giving rise to a countable cover \(A\subseteq \bigcup_{n=1}^\infty U_n\) satisfying \(\overline{U_n}\subseteq X-B\) for all \(n\text{.}\) Similarly, we can construct an open covering \(B\subseteq \bigcup_{n=1}^\infty V_n\) satisfying \(\overline{V_n}\subseteq X-A\) for all \(n\text{.}\) The corresponding open sets are not necessarily disjoint. However the sets \(U=\bigcup_{n=1}^\infty U_n'\) and \(V=\bigcup_{n=1}^\infty V_n'\text{,}\) where
\begin{align*} U_n'\amp =U_n-\bigcup_{k=1}^n\overline{V_k} \amp V_n'\amp =V_n-\bigcup_{k=1}^n\overline{U_k} \text{,} \end{align*}
still cover \(A\) and \(B\) respectively, are still open, and are disjoint.
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