# A useful lemma for proving normality

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