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Let us begin with a lemma ( see it in the Engelking's book):

If $X$ is a $T_1$ space and for every closed $F$ and every open $W$ that contains $F$ there exists a sequence $W_1$, $W_2$, ... of open subsets of $X$ such that $F\subset \cup_{i}W_i$ and $cl(W_i)\subset W$ for $i=$ 1, 2, ..., then the space $X$ is normal.

My question is this: Are there other certain classes of spaces which can be showed normality by this lemma?

I will give an example as following:

Every second-countable regular space is normal.

Proof: Every regular space with a countable base $B$ satisfies the condition in the lemma, because for any $x\in F$ there is a $U_x \in B$ such that $x \in U_x \subset cl(U_x) \subset W$, the family of the all $U_x$'s is countable and $F \subset \cup_{x\in F}U_x$.

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Your proof works for any Lindelof space. –  azarel Jul 11 '12 at 1:05
    
you can post your answer, so i can vote you:) –  Paul Jul 11 '12 at 1:09

2 Answers 2

up vote 2 down vote accepted

Every Lindelöf regular ($T_1$) space is normal: if $F$ is closed and $W$ is an open set containing it, for each $x \in F$ we can find open $U_x$, with $x \in U_x$ with closure inside $W$, by regularity applied to $x$ and $W$. Now as $F$ is Lindelöf, being a closed subset of a Lindelöf $X$, we can reduce the $U_x$ to a countable subcover of $F$ and the lemma applies directly.

This generalizes the second countable case (as second countable spaces are Lindelöf) and the previous answer by Loreaux (because we do not need local compactness, and locally compact plus Hausdorff implies regularity).

You can also use it to more easily show that an $F_{\sigma}$ subset of a normal space is normal:

Suppose $X$ is normal and $A = \cup_n F_n \subset X$ is an $F_\sigma$ in $X$, where all the $F_n$ are closed subsets of $X$. Then $A$ is normal (in the subspace topology). To apply the lemma, let $F$ be closed in $A$ and $W$ be an open superset of it (open in $A$). Let $O$ be open in $X$ such that $O \cap A = W$, and note that each $F \cap F_n$ is closed in $X$ and by normality of $X$ there are open subsets $O_n$ in $X$, for $n \in \mathbb{N}$ such that $$ F \cap F_n \subset O_n \subset \overline{O_n} \subset O $$ and define $W_n = O_n \cap A$, which are open in $A$ and satisfy that the $W_n$ cover $F$ (as each $W_n$ covers $F_n$ and $F = F \cap A = \cup_n (F \cap F_n)$) and the closure of $W_n$ in $A$ equals $$\overline{W_n} \cap A = \overline{O_n \cap A} \cap A \subset O \cap A = W$$ which is what is needed for the lemma.

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Any locally compact Hausdorff Lindelof space will do the trick. The proof is essentially the same as the one you gave for second countable regular spaces, just stripped to the bare bones.

Consider such a space $X$. Let $F$ be closed and $W\supset F$ be open. Then for each $x\in F$, there exists a compact set $K_x$ containing $x$ (local compactness). Let $F_x= F\cap K_x$, which is compact since $F$ is closed, $K_x$ is compact, and $X$ is Hausdorff. Then by the shrinking lemma (refer to Lee's Introduction to Topological Manifolds. Here the result is only stated for a single point, but it extends easily to compact sets), there exists an open set $U_x$ such that $F_x\subset U_x\subset \text{cl}(U_x)\subset W$. Since $X$ is Lindelof, the cover $\bigcup_{x\in F} U_x$ has a countable subcover $\bigcup_{n\in\mathbb{N}} U_{x_n}$ for some $x_n\in F$. Notice that $F$, $W$ and the $\{U_{x_n}\}_{n\in\mathbb{N}}$ satisfy the conditions of the lemma. Thus $X$ is normal.

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a Nice example. –  Paul Jul 11 '12 at 5:56

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