Every regular Lindelöf space is second countable.

Question

We know that any second countable space is a Lindelöf space but not conversely. Is the following can be taken as a modified converse?

Every regular Lindelöf space is second countable.

I have proved that every regular Lindelöf space is normal and also some other properties of Lindelöf spaces. Unable to prove the above.

Help Needed! Thanks in Advance.

Answer 1

$\Bbb R_\ell$ (real numbers with lower limit topology) is a regular Lindelof space which is not second countable. See Munkres - Topology Chapter 4.

Answer 2

Not only are regular Lindelöf not necessarily second-countable, we can show that even stronger assumptions have no bearing on the weight of the topological space. (The weight $w(X)$ of a topological space $X$ is the minimum cardinality of a base for $X$, so that "$X$ is second-countable" is equivalent to $w(X) \leq \aleph_0$.)

Note that even compact Hausdorff (and hence normal) spaces need not be second-countable. A prototypical example of this would be the closed ordinal space $\omega_1+1 = [ 0 , \omega_1]$, where $\omega_1$ denotes the least uncountable ordinal.

• As it is an order topology, it is Hausdorff (in fact, all order topologies are hereditarily normal).

• This space is compact basically because there cannot be infinite strictly decreasing sequences of ordinals.

  For an outline, suppose that $\mathcal U$ is an open cover of $[0,\omega_1]$. First pick an open set $U_1$ from the cover which contains $\omega_1$. By the definition of the topology there is an $\alpha_1 < \omega_1$ such that $( \alpha_0 , \omega_1 ] \subseteq U_1$. Now pick an open set $U_2$ from the cover which contains $\alpha_1$, so there is an $\alpha_2 < \alpha_1$ such that $( \alpha_2 , \alpha_1 ] \subseteq U_2$. We continue this in this fashion, noting that this process must end after finitely many steps because we are constructing a strictly decreasing sequence of ordinals ($\omega_1 > \alpha_1 > \alpha_2 > \cdots$).

• It is not second-countable because it has uncountably many isolated points (corresponding to the successor ordinals). So any base must include each of these uncountably many singletons.

The same argument can be made about any closed ordinal space corresponding to an uncountable ordinal (i.e., they are compact Hausdorff, but not second-countable). Moreover, an infinite ordinal $\beta$ has $|\beta|$-many successor ordinals below it, and each of these are isolated in the order topology. This means that the closed ordinal space $[0,\beta]$ cannot have a base of cardinality ${<}|\beta|$. From this it follows that there is no upper bound on the weight of compact Hausdorff spaces.