I received the following interesting point in ([1](https://mathoverflow.net/questions/286332/a-criterion-for-second-countability)). I could not find any reference or clear proof. Any suggestion? 


**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 $Y$.*



On the hereditary Lindelof topological spaces
If $X$ is hereditarily Lindelof then any open subset of $Y$ is Lindelof and therefore it is the union of countably many basic (for a given predetermined base for $Y$) open sets. Hence The $\sigma$-algebra generated by the base contains (so it is equal to) the Borel $\sigma$-algebra.


The other direction is not true. Let $X=\omega\_1$ with the topology of initial segments. A basis for this topology coincides with the topology itself (except perhaps for the empty and the whole sets). So trivially the Borel $\sigma$-algebra and the $\sigma$-algebra generated by any base coincide. The same is true for any $Y \subseteq X$. However $X$ is not Lindelof.


