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I received the following interesting point in (1). 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 $Y$.

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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.

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  • $\begingroup$ Thanks a lot. It will be great if you hint me concerning a weaker topological condition that would equivalent to coincidness the $\sigma$-algebras $\endgroup$
    – ABB
    Commented Mar 10, 2020 at 9:42

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