# When is a Lindelof Space Normal?

When is a Lindelof space normal? This question may seem strange since it is a well known basic result that Lindelof spaces are normal (see Theorem 3.8.2 in [1] and Theorem 16.8 in [6]). The proofs in both of these theorems require that the Lindelof space is also a regular space. We present an example showing that the requirement of regularity cannot be removed. The example defined below is a space that is Hausdorff, hereditarily Lindelof, not separable, not regular and not normal.

The example defined below makes it clear that we need to be careful in making the leap from the Lindelof property to normality of a space. We need to make sure that the space in question is also a regular space. The Lindelof property is a powerful property; it also implies paracompactness. But for this to happen, some separation axiom is required (it needs to be a regular space).

Another motivation for looking at examples such as the one defined here is that they are background information to the theory of S-spaces and L-spaces (see the remark given at the end of the post).

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The example is the Example 14.7 in [2]. The underlying set is the real number line $\mathbb{R}$. Let $\tau$ be the usual topology on $\mathbb{R}$. Let $X=\mathbb{R}$. We use the following collection of sets as a base for a new topology on $X$:

$\displaystyle \mathbb{B}=\left\{O-C: O \in \tau, C \subset \mathbb{R}, \lvert C \vert \le \omega \right\}$

The collection $\mathbb{B}$ needs to satisfy two conditions for being a base for a topology. One is that it covers the underlying set. The second is that whenever $B_1, B_2 \in \mathbb{B}$ with $x \in B_1 \cap B_2$, there is some $B_3 \in \mathbb{B}$ such that $x \in B_3 \subset B_1 \cap B_2$. Note that $\tau \subset \mathbb{B}$. This implies that $\mathbb{B}$ is a cover of $X$. It is also clear that the second condition is satisfied.

Let $\tau^{*}$ be the topology generated by the base $\mathbb{B}$. Another consequence of $\tau \subset \mathbb{B}$ is that the new topology on $X$ is finer than the usual topology, i.e., $\tau \subset \tau^{*}$. It is then clear that $(X, \tau^{*})$ is a Hausdorff space.

We denote the real line with the new topology $\tau^{*}$ by $X$ rather than $(X, \tau^{*})$ since the new topology $\tau^{*}$ is the primary focus.

In $X$, the complement of every countable set is open. So no countable set can be dense in $X$, making it not separable. Equivalently, every countable subset of $X$ is closed.

To see that $X$ is not regular, we first make the following observation.

Observation
Let $W \subset \mathbb{R}$ be a Euclidean open set. Let $A \subset \mathbb{R}$ be a countable set that is dense in $W$ (dense with respect to the Euclidean topology). Then for each $x \in A \cap W$, $x \in \overline{W-A}$ (with respect to the topology $\tau^*$).

To see that this observation is correct, let $x \in A \cap W$ and let $V-B$ be an open set containing $x$, where $V$ is a Euclidean open set and $B$ is a countable set. We can assume that $V \subset W$. Let $A_0=A-\left\{x \right\}$. Note that $x \in V-(A_0 \cup B) \subset V-B$ and that $V-(A_0 \cup B) \subset W-A_0$. Furthermore, for any $y \in V-(A_0 \cup B)$ with $y \ne x$, $y \in W-A$. Thus any open set containing $x$ contains many points of $W-A$. Hence, we have $x \in \overline{W-A}$.

To see that $X$ is not regular, we show that there is a closed set $C \subset X$ such that for any open set $O \subset X-C$, $\overline{O} \cap C \ne \varnothing$.

Let $C=\mathbb{Q}$ (the set of all rational numbers). Let $O$ be open in $X$ such that $O \subset X-C$. We assume that $O=W-A$ where $W$ is open in the Euclidean topology and $A$ is countable. Note that $O=W-A$ consisting of entirely of irrational numbers. Then $\mathbb{Q} \cap W \subset A$. Thus $A$ is dense in $W$ (with respect to the Euclidean topology). By the above observation, any point of $A$ is a member of $\overline{W-A}$ (with respect to the topology $\tau^*$). In particular, for each $x \in C=\mathbb{Q}$, $x \in \overline{W-A}$. Thus $X$ is not regular.

Any space that is not regular is also not normal. Thus $X$ is not normal. But we can also see this directly. Let $H=\mathbb{Q}$ and $K=\left\{x \right\}$ where $x$ is any irrational number. Then $H$ and $K$ are two disjoint closed sets that cannot be separated by disjoint open sets in $X$.

The space $X$ is Lindelof and also hereditarily Lindelof (both properties are piggy backed on the same properties of the Euclidean real line). To see that $X$ is Lindelof, suppose that a collection of open sets $O_a-A_a \in \mathbb{B}$ is a cover of $X$. The Euclidean open sets $O_a$ also form a cover of $X=\mathbb{R}$. Then pick countably many $O_a$ that also cover the real line (using the Lindelof property of the Euclidean real line). Then these countably many $O_a-A_a$ will cover $X$ except for countably many points. A countable subcover will result once we use open sets in the original cover to cover these countably many points. Using the same proof, it can be shown that every subspace of $X$ is also Lindelof.

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Remark

An S-space is any regular topological space that is hereditarily separable but not Lindelof. An L-space is any regular topological space that is hereditarily Lindelof but not separable. Elementary examples of hereditarily separable space that is not Lindelof and hereditarily Lindelof space that is not separable are quite easy to define (but they are not regular). The example we define above is an example of non-regular hereditarily Lindelof space that is not separable. Thus S-space and L-space are typically defined with the requirement of regularity.

It was shown in the early 1980’s that the existence of S-space is independent of the usual axioms of ZFC. This means that to prove the existence of an S-space or to prove the non-existence of S-space, one needs to assume axioms beyond those of ZFC. The L-space problem (whether an L-space can exist without assuming additional set-theoretic assumptions beyond those of ZFC) was not resolved until quite recently. For quite a long time, it was believed the L-space problem would have a similar solution (that its existence would be independent of ZFC). In 2005, Justin Tatch-Moore solved the L-space problem by constructing an L-space without assuming additional axioms.

For back ground information about S-space and L-space, see [3] and [4]. For Justin Moore’s proof, see [5].

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Reference

1. Engelking, R., General Topology, Revised and Completed edition, 1989, Heldermann Verlag, Berlin.
2. Hodel, R.,Cardinal Functions I, Handbook of Set-Theoretic Topology, (K. Kunen and J. E. Vaughan, eds), 1984, Elsevier Science Publishers B. V., Amsterdam, 1-61.
3. Mastros, S.,S and L Spaces, Master’s Thesis, 2009, University of Pittsburgh. Link
4. Roitman, J.,Basic S and L, Handbook of Set-Theoretic Topology, (K. Kunen and J. E. Vaughan, eds), 1984, Elsevier Science Publishers B. V., Amsterdam, 295-326.
5. Tatch-Moore, J.,A solution to the L space problem, Journal of the American Mathematical Society, 19 (2006), 3, 717—736. Link
6. Willard, S., General Topology, 1970, Addison-Wesley Publishing Company.

## 1 thought on “When is a Lindelof Space Normal?”

1. Very good example, Dan! 🙂

Two minor corrections…

1. When you say that $x \in \overbar{W – A}$ for each $x \in \mathbb{Q}$, I think you mean: for each $x \in \mathbb{Q} \cap W$.

2. When you say that any space that is not regular is also not normal, I think you mean: “Any HAUSDORFF space that is not regular is also not normal”. Right?

Sorry if I am missing something.

Cheers,
André Caldas.