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Let $(X,\tau)$ be a topological space.

Assume for any arbitrary topological base $\mathcal{E}$ of $\tau$ we have that: the Borel sigma algebras coming form $\mathcal{E}$ and $\tau$ are the same. Can we conclude that $X$ is second countable ?!

This question is also asked when $X$ is a locally convex space. Please read the comments below.

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    $\begingroup$ What about non-countable discrete sets? $\endgroup$ – Wille Liou Nov 17 '17 at 19:51
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    $\begingroup$ @WilleLiou The Borel $\sigma$-algebra is the power-set, but the $\sigma$-algebra generated by the basis of singletons is the countable-co-countable $\sigma$-algebra. $\endgroup$ – Michael Greinecker Nov 17 '17 at 20:04
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    $\begingroup$ An example: Let $X$ be a non-countable set. We equip $X$ with the topology whose open sets consist of $\emptyset$ and those subsets $U\subseteq X$ with $|X\setminus U| < \infty$. Then any basis generates the same Borel $\sigma$-algebra but $X$ is not second countable. $\endgroup$ – Wille Liou Nov 17 '17 at 20:46
  • $\begingroup$ @WilleLiou It was really interesting. Thank you very much. $\endgroup$ – Ali Bagheri Nov 18 '17 at 5:16
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    $\begingroup$ I would suggest instead that the commenters post their answer as an answer, rather than as a comment. $\endgroup$ – Joel David Hamkins Nov 19 '17 at 14:58
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A counterexample to this question (and its locally convex version) is any non-metrizable (locally convex) space $X$, which is hereditarily Lindelof. The hereditary Lindelofness of $X$ implies that any open set is a countable union of basic open sets and this implies that the $\sigma$-algebra generated by any base of the topology coincides with the Borel $\sigma$-algebra.

As for an example of non-metrizable hereditarily Lindelof spaces, take any metrizable separable locally convex space with a weaker non-metrizable topology. Being a continuous image of a second-countable space, such space will have countable network and hence will be hereditarily Lindelof.

For example, you can take any infinite-dimensional separable Banach space endowed with the weak topology. It will be not metrizable but hereditarily Lindelof.

The function spaces $C_p(X)$ over cosmic spaces $X$ have countable network and hence are hereditarily Lindelof. The function spaces $C_k(X)$ with compact-open topology over $\aleph_0$-spaces $X$ have countable $k$-network and hence are hereditarily Lindelof.

The locally convex space $\mathbb R^\infty$, which is inductive limit of an increasing sequence of finite-dimensional spaces, has countable network and hence is hereditarily Lindelof and not metrizable.

So, there plenty of examples. But one can modify the question replacing the second countability by the hereditary Lindelofness:

Problem. Is a (locally convex) topological space hereditary Lindelof if the $\sigma$-algebra generated by any base of the topology coincides with the $\sigma$-algebra of Borel sets?

Remark. The example given by @Wille Liou in the comment to the original question is hereditarily Lindelof (even hereditarily compact), but not Hausdorff.

In fact, the hereditary Lindelofness admits the following characterization:

Theorem. A topological space $X$ is hereditary Lindelof if and only if for any subspace $Y\subset X$, the $\sigma$-algebra generated by any base of the topology of $Y$ coincides with the Borel $\sigma$-algebra of $Y$.

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  • $\begingroup$ Could you please give a bit more concerning the last Theorem. Indeed why the coincidences of $\sigma$-algebras implies the hereditary Lindelof property. $\endgroup$ – Ali Bagheri May 31 '18 at 20:10
  • $\begingroup$ @AliBagheri See Exercise 3.12.7(b) in "General Topology" of Engelking. $\endgroup$ – Taras Banakh Jun 1 '18 at 5:41
  • $\begingroup$ I could not get how this exercise helps us to find the result. Could you please write a bite more. $\endgroup$ – Ali Bagheri Jun 17 '18 at 16:20
  • $\begingroup$ @AliBagheri Have you read this Exercise 3.12.7(b)? Could you formulate what is proved in this exercise? $\endgroup$ – Taras Banakh Jun 17 '18 at 16:49

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