April | 2012 | Dan Ma's Topology Blog

The Euclidean spaces $\mathbb{R}$ and $\mathbb{R}^n$ are both Lindelof and separable. In fact these two properties are equivalent in the class of metrizable spaces. A space is metrizable if its topology can be induced by a metric. In a metrizable space, having one of these properties implies the other one. Any students in beginning topology courses who study basic notions such as the Lindelof property and separability must venture outside the confine of Euclidean spaces or metric spaces. The goal of this post is to present some elementary examples showing that these two notions are not equivalent.

All topological spaces under consideration are Hausdorff. Let $X$ be a space. Let $D \subset X$ . The set $D$ is said to be dense in $X$ if every nonempty open subset of $X$ contains some point of $D$ . The space $X$ is said to be separable if there is countable subset of $X$ that is also dense in $X$ . All Euclidean spaces are separable. For example, in the real line $\mathbb{R}$ , every open interval contains a rational number. Thus the set of all rational numbers $\mathbb{Q}$ is dense in $\mathbb{R}$ .

Let $\mathcal{U}$ be a collection of subsets of the space $X$ . The collection $\mathcal{U}$ is said to be a cover of $X$ if every point of $X$ is contained in some element of $\mathcal{U}$ . The collection $\mathcal{U}$ is said to be an open cover of $X$ if, in addition it being a cover, $\mathcal{U}$ consists of open sets in $X$ .

Let $\mathcal{U}$ be a cover of the space $X$ . Let $\mathcal{V} \subset \mathcal{U}$ . If the collection $\mathcal{V}$ is also a cover of $X$ , we say that $\mathcal{V}$ is a subcover of $\mathcal{U}$ . The space $X$ is a Lindelof space (or has the Lindelof property) if every open cover of $X$ has a countable subcover.

The real $\mathbb{R}$ is Lindelof. Both the Lindelof property and the separability of $\mathbb{R}$ follows from the fact that the Euclidean topology on $\mathbb{R}$ can be generated by a countable base (e.g. one countable base consists of all open intervals with rational endpoints). Now some non-Euclidean (and non-metrizable) examples.

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Example 1 – A Lindelof space that is not separable
Let $X$ be any uncountable set. Let $p$ be a point that is not in $X$ , e.g., let $p=\left\{ X \right\}$ . Define the space $Y = \left\{p\right\} \cup X$ as follows. Let every point in $X$ be isolated, meaning any singleton set $\left\{ x \right\}$ is declared open for any $x \in X$ . An open neighborhood of the point $p$ is of the form $\left\{p\right\} \cup W$ where $X-W$ is a countable subset of $X$ .

It is clear that the resulting space $Y$ is Lindelof since every open set containing $p$ contains all but countably many points of $X$ . It is also clear that no countable set can be dense in $Y$ .

Even though this example $Y$ is Lindelof, it is not hereditarily Lindelof since the subspace $X$ is uncountable discrete space.

In a previous post, we showed that the space $Y$ defined in this example is a productively Lindelof space (meaning that its product with every Lindelof space is Lindelof).

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Remark

A space is said to have the countable chain condition (CCC) if there are no uncountable family of pairwise disjoint open subsets. It is clear that any separable space has the CCC. It follows that the space $Y$ in Example 1 does not have the CCC, since the singleton sets $\left\{ x \right\}$ (with $x \in X$ ) forms a pairwise disjoint collection of open sets, showing that the Lindelof property does not even imply the weaker property of having the CCC.

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Example 2 – A separable space that is not Lindelof
The example here is the Tangent Disc Space (Niemytzki’s Tangent Disc Topology in [2]). The underlying set is the upper half plane (the x-axis and the plane above the x-axis). In other words, consider the following set:

$\displaystyle . \ \ \ \ \ X=\left\{(x,y) \in \mathbb{R}^2: y \ge 0 \right\}$

Let $\displaystyle X_u=\left\{(x,y) \in \mathbb{R}^2: y>0 \right\}$ and $T=\left\{(x,0): x \in \mathbb{R} \right\}$ . The line $T$ is the x-axis and $X_u$ is the upper plane without the x-axis. We define a topology on $X$ such that $X_u$ as a subspace in this topology is Euclidean. The open neighborhoods of a point $p=(x,0) \in T$ are of the form $\left\{p \right\} \cup D$ where $D$ is an open disc tangent to the x-axis at the point $p$ . The figure below illustrates how open neighborhoods at the x-axis are defined.

It is clear that the points with rational coordinates in the upper half plane $X_u$ form a dense set in the tangent disc topology. Thus $X$ is separable. In any Lindelof space, there are no uncountable closed and discrete subsets. Note that the x-axis $T$ is a closed and discrete subspace in the tangent disc space. Thus $X$ is not Lindelof.

Though separable, the Tangent Disc Space is not hereditarily separable since the x-axis $T$ is uncountable and discrete.

The Tangent Disc Space is an interesting example. For example, it is a completely regular space that is an example of a Moore space that is not normal. For these and other interesting facts about the Tangent Disc Space, see [2].

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For the Lindelof property and the property of being separable, there are plenty of examples of spaces that possess only one of the properties. All three references indicated below are excellent places to look. The book by Steen and Seebach ([2]) is an excellent catalog of interesting spaces (many of them are elementary).
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Reference

Engelking, R., General Topology, Revised and Completed edition, 1989, Heldermann Verlag, Berlin.
Steen, L. A., Seebach, J. A.,Counterexamples in Topology, 1995, Dover Edition, Dover Publications, New York.
Willard, S., General Topology, 1970, Addison-Wesley Publishing Company.

We observe from the following statement two examples of Lindelof spaces that are not hereditarily Lindelof.

Any product space contains a discrete subspace having the same cardinality as the number of factor spaces.

Using the above observation, by choosing the factor spaces judiciously, the product of uncountably many spaces is a handy way of obtaining Lindelof spaces (in some cases $\sigma$ -compact spaces) that are not hereditarily Lindelof. For definition and basic information about product spaces, see this previous post.

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All spaces under consideration are at least Hausdorff. For each $\alpha \in A$ , let $X_\alpha$ be a space with at least two points. For each $\alpha \in A$ , fix two points $p_\alpha, q_\alpha \in X_\alpha$ . Then the product space $\displaystyle \prod \limits_{\alpha \in A} X_\alpha$ contains a discrete subspace $Y$ that has the same cardinality as the cardinality of the index set $A$ .

For each $\alpha \in A$ , define $y_\alpha \in \prod \limits_{\alpha \in A} X_\alpha$ by the following:

$\displaystyle (1) \ \ \ \ \ \ y_\alpha(\gamma)=\left\{\begin{matrix}p_\alpha&\ \gamma=\alpha\\{q_\alpha}&\ \gamma \ne \alpha \end{matrix}\right.$

Let $Y=\left\{ y_\alpha: \alpha \in A\right\}$ . It follows that $\lvert Y \lvert = \lvert A \lvert$ and that $Y$ is a discrete space.

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Whenever the index set $A$ is uncountable, the product space $\displaystyle \prod \limits_{\alpha \in A} X_\alpha$ contains an uncountable discrete subspace. Thus even if the product space $\displaystyle \prod \limits_{\alpha \in A} X_\alpha$ is Lindelof, one of its subspace $Y$ cannot be Lindelof. Taking the product of uncountably many factor spaces is a handy way to obtain Lindelof space that is not hereditarily Lindelof. Some examples are shown below.

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Examples

Let the index set $A$ be uncountable. To make the product space Lindelof, we can make every one of its factor $X_\alpha$ compact. Thus the product space $\displaystyle \prod \limits_{\alpha \in A} X_\alpha$ is compact and not hereditarily Lindelof.

Thus the product space $[0,1]^{\omega_1}$ , the product of $\omega_1$ many copies of the unit interval, is compact and not hereditarily Lindelof. Another example is $\left\{ 0,1 \right\}^{\omega_1}$ , the product of $\omega_1$ many copies of $\left\{ 0,1 \right\}$

Another way to make the product space $\displaystyle \prod \limits_{\alpha \in A} X_\alpha$ Lindelof is to make some of the factors compact such that the product of the remaining non-compact factors is Lindelof. Then the product space is essentially the product of a compact space and a Lindelof space, which is always Lindelof.

For example, let $X_0=\mathbb{R}$ and let $X_\alpha=[0,1]$ for all $\alpha$ with $0<\alpha<\omega_1$ . Then the product space $\displaystyle \prod \limits_{\alpha \in A} X_\alpha$ is Lindelof since it is essentially the product of a compact space and a Lindelof space. However, the product $\displaystyle \prod \limits_{\alpha \in A} X_\alpha$ is not hereditarily Lindelof.

In fact, the product space in the previous paragraph is $\sigma$ -compact (i.e. the union of countably many compact sets). To make the example not $\sigma$ -compact, simply make the first factor space a non-locally compact Lindelof space. For example, use the Sorgenfrey line or the space of the irrational numbers.

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Elementary Examples of Lindelof Spaces and Separable Spaces

Examples of Lindelof Spaces that are not Hereditarily Lindelof