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In second countable topology(i.e. has countable bases), obviously all open sets are countable union of base elements. But the inverse seems not true.

The sorgenfrey line $R_l$ seems to be an example. The proof is analogous to sorgenfrey line is lindelof and sorgenfrey line is T6 perfectly normal. Given an open set $U = \bigcup B_\alpha$ for some base elements $B_\alpha \in \mathcal{B}$, partition it into its interior $I$ in the ordinary topology $R$ and $U-I$. Because $R_l$ is finer than the ordinary topology $R$ and $R$ is second countable, $I$ can be written as countable union of base elements both in $R$ and $R_l$ i.e. $I=\bigcup_i (c_i,d_i)$. For $U-I$, each $x \in U-I$ has to belong to a base element $[x, b_x) \subset U$ otherwise if $x \in (a,b) \subset U$, it will be in $I$. Replace $b_x$ with a rational number $b'_x$ by shrinking. For two distinct point $x, y \in U-I$, $[x,b'_x)$ and $[y,b'_y)$ have to be disjoint, otherwise either $x$ or $y$ will be in $I$. Therefore, the elements in $U-I$ is countable. Now, $U = I\bigcup (U-I) = \bigcup_i (c_i, d_i) \bigcup (\bigcup_{x\in U-I} [x, b'_x))$. Hence, all open sets in $R_l$ is countable union of base elements. However, $R_l$ is not second countable.

Is my proof correct roughly? And I would like more examples of topologies that are not second countable, but in which every open sets are countable union of base elements.

Edit: as the comments mentioned, there is a trivial example that if all open sets are considered base elements, every topology has this property. I think to eliminate this triviality, the original question should be reformulated as

For every possible collections of bases, all open sets can be written as countable union of bases in this topology, yet the topology is not second countable.

Clearly, $R_l$ is still valid as an example.

Alex Fan
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  • To add my initial thoughts, a topology in which every open set is compact might be a start point for the example and I do find this type of topology Noetherian topology. But I just started to learn general topology and can't find any way to modify it to non second-countable. – Alex Fan Nov 06 '20 at 02:00
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    Since a topology is a base for itself, every topology has a base with the property that every open set is the union not just of a countable family of base elements, but of a singleton set of base elements. – Brian M. Scott Nov 06 '20 at 03:57
  • @BrianM.Scott, True, I didn't think of this case. Is there any way to reformulate the question to eliminate this triviality? – Alex Fan Nov 06 '20 at 04:47
  • I thought about that a bit when I commented, and I couldn’t think of one. – Brian M. Scott Nov 06 '20 at 05:00
  • @BrianM.Scott how about for every fixed uncountable family of bases, open sets can be written as countable union of bases selected from the family. – Alex Fan Nov 06 '20 at 05:08
  • Your reformulation is just a complicated way of asking for a space which is hereditarily Lindelöf but not Lindelöf. – bof Nov 07 '20 at 01:35

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