Monthly Archives: July 2014

A useful lemma for proving normality

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In this post we discuss a lemma (Lemma 1 below) that is useful for proving normality. In some cases, it is more natural using this lemma to prove that a space is normal than using the definition of normality. The proof of Lemma 1 is not difficult. Yet it simplifies some proofs of normality. One reason is that the derivation of two disjoint open sets that are to separate two disjoint closed sets is done in the lemma, thus simplifying the main proof at hand. The lemma is well known and is widely used in the literature. See Lemma 1.5.15 in [1]. Two advanced examples of applications are [2] and [3]. After proving the lemma, we give three elementary applications of the lemma. One of the applications is a characterization of perfectly normal spaces. This characterization is, in some cases, easier to use, e.g. making it easy to show that perfectly normal implies hereditarily normal.

In this post, we only consider spaces that are regular and T_1. A space X is regular if for each open set U \subset X and for each x \in U, there exists an open V \subset X with x \in V \subset \overline{V} \subset U. A space is T_1 if every set with only one point is a closed set.

Lemma 1
A space Y is a normal space if the following condition (Condition 1) is satisfied:

  1. For each closed subset L of Y, and for each open subset M of Y with L \subset M, there exists a sequence M_1,M_2,M_3,\cdots of open subsets of Y such that L \subset \bigcup_{i=1}^\infty M_i and \overline{M_i} \subset M for each i.

Proof of Lemma 1
Suppose the space Y satisfies condition 1. Let H and K be disjoint closed subsets of the space Y. Consider H \subset U=Y \backslash K. Using condition 1, there exists a sequence U_1,U_2,U_3,\cdots of open subsets of the space Y such that H \subset \bigcup_{i=1}^\infty U_i and \overline{U_i} \cap K=\varnothing for each i. Consider K \subset V=Y \backslash H. Similarly, there exists a sequence V_1,V_2,V_3,\cdots of open subsets of the space Y such that K \subset \bigcup_{i=1}^\infty V_i and \overline{V_i} \cap H=\varnothing for each i.

For each positive integer n, define the open sets U_n^* and V_n^* as follows:

    U_n^*=U_n \backslash \bigcup_{k=1}^n \overline{V_k}

    V_n^*=V_n \backslash \bigcup_{k=1}^n \overline{U_k}

Let P=\bigcup_{n=1}^\infty U_n^* and Q=\bigcup_{n=1}^\infty V_n^*. It is clear P and Q are open and that H \subset P and K \subset Q. We claim that P and Q are disjoint. Suppose y \in P \cap Q. Then y \in U_n^* for some n and y \in V_m^* for some m. Assume that n \le m. The fact that y \in U_n^* implies y \in U_n. The fact that y \in V_m^* implies that y \notin \overline{U_j} for all j \le m. In particular, y \notin U_n, a contradiction. Thus P \cap Q=\varnothing. This completes the proof that the space Y is normal. \blacksquare

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Spaces with nice bases

One application is that spaces with a certain type of bases satisfy condition 1 and thus are normal. For example, spaces with bases that are countable and spaces with bases that are \sigma-locally finite. Spaces with these bases are metrizable. The proof that these spaces are metrizable will be made easier if they can be shown to be normal first. The Urysohn functions (the functions described in Urysohn’s lemma) can then be used to embed the space in question into some universal space that is known to be metrizable. Using regularity and Lemma 1, it is straightforward to verify the following three propositions.

Proposition 2
Let X be a regular space with a countable base. Then X is normal.

Proposition 3
Let X be a regular space with a \sigma-locally finite base. Then X is normal.

Proposition 4
Let X be a regular space with a \sigma-discrete base. Then X is normal.

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A characterization of perfectly normal spaces

Another application of Lemma 1 is that it leads naturally to a characterization of perfect normality. Recall that a space X is perfectly normal if X is normal and perfect. A space X is perfect if every closed subset of X is a G_\delta set (i.e. the intersection of countably many open subsets of X). Equivalently a space X is perfect if and only if every open subset of X is an F_\sigma set, i.e., the union of countably many closed subsets of X. We have the following theorem.

Theorem 5
A space Y is perfectly normal if and only if the following condition holds.

  1. For each open subset M of Y, there exists a sequence M_1,M_2,M_3,\cdots of open subsets of Y such that M \subset \bigcup_{i=1}^\infty M_i and \overline{M_i} \subset M for each i.

Clearly, condition 2 is strongly than condition 1.

Proof of Theorem 5
\Longrightarrow
Suppose that the space Y is perfectly normal. Let M be a non-empty open subset of Y. Then M=\bigcup_{n=1}^\infty P_n where each P_n is a closed subset of Y. Using normality of Y, for each n, there exists open subset M_n of Y such that P_n \subset M_n \subset \overline{M_n} \subset M. Then consition 2 is satisfied.

\Longleftarrow
Suppose condition 2 holds, which implies condition 1 of Lemma 1. Then Y is normal. It is clear that condition 2 implies that every open subset of Y is an F_\sigma set. \blacksquare

The characterization of perfectly normal spaces in Theorem 5 is hereditary. This means that any subspace of a perfectly normal space is also perfectly normal. In particular, perfectly normal implies hereditarily normal. Thus we have the following theorem.

Theorem 6
Condition 2 in Theorem 5 is hereditary, i.e., if a space satisfies Condition 2, every subspace satisfies Condition 2. Therefore if the space Y is a perfectly normal space, then every subspace of Y is also perfectly normal. In particular, if Y is perfectly normal, then Y is hereditarily normal (i.e. every subspace of Y is normal).

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Normality is hereditary with respect to F_\sigma subsets

Normality is not a hereditary notion. Lemma 1 can used to show that normality is hereditary with respect to F_\sigma subspaces.

Theorem 7
Let Y be a normal space. Then every F_\sigma subspace of Y is normal.

Proof of Theorem 7
Let H be a subspace of Y such that H=\bigcup_{n=1}^\infty P_n where each P_n is closed subset of Y. Let L be a closed subset of H and let M be an open subset of H such that L \subset M. We need to find M_1,M_2,M_3,\cdots, open in H, such that L \subset \bigcup_{i=1}^\infty M_i and \overline{M_i} \subset M for all i (closure of M_i is within H).

Let U be an open subset of Y such that M=U \cap H. For each positive integer n, let H_n=P_n \cap L. Obviously H_n is closed in H. It is also the case that H_n is closed in Y. To see this, let p \in Y be a limit point of H_n. Then p is a limit point of P_n. Hence p \in P_n since P_n is closed in Y. We now have p \in H. The point p is also a limit point of L. Thus p \in L since L is closed in H. Now we have p \in H_n=P_n \cap L, proving that H_n is closed in Y.

Now we have H_n \subset U for all n. By Lemma 1, for each n, there exists a sequence U_{n,1},U_{n,2},U_{n,3},\cdots of open subsets of Y such that H_n \subset \bigcup_{j=1}^\infty U_{n,j} and \overline{U_{n,j}} \subset U for all j. Note that L=\bigcup_{n=1}^\infty H_n. Rename M_{n,j}=U_{n,j} \cap H over all n,j by the sequence M_1,M_2,M_3,\cdots. Then L \subset \bigcup_{i=1}^\infty M_i. It also follows that \overline{M_i} \subset M for all i (closure of M_i is within H). This completes the proof that the F_\sigma set H is normal. \blacksquare

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Reference

  1. Engelking, R., General Topology, Revised and Completed edition, Heldermann Verlag, Berlin, 1989.
  2. Gruenhage, G., Normality in X^2 for complete X, Trans. Amer. Math. Soc., 340 (2), 563-586, 1993.
  3. Nyikos, P., A compact nonmetrizable space P such that P^2 is completely normal, Topology Proc., 2, 359-363, 1977.

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\copyright \ 2014-2015 \text{ by Dan Ma} Revised April 14, 2015