# Examples of Lindelof Spaces that are not Hereditarily Lindelof

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.