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Can a similar result to this question be true for a topological space that has a pseudo basis "A collection of non-empty open sets (in a topological space X) such that every non-empty open set of X contains one of these is called a pseudo-basis"

Edit: Question: If $\mathcal B$ is a countable pseudo base for a topological space $(X,\tau)$, then $\sigma(\mathcal B)=\sigma(\tau)$, where $\sigma(*)$ is the sigma algebra generated by $(∗)$.

Badr Alharbi
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  • Can you explain what kind of "similar" result you are looking for? The statement in the question is false and replacing "base" with "pseudo-base" would make it even more false. – Nate Eldredge Nov 15 '17 at 21:05
  • Are you asking whether it is true for a topological space with a countable pseudo-base? – Nate Eldredge Nov 15 '17 at 21:06
  • @NateEldredge Thanks for your comment. This is the question: If $\mathcal{B}$ is a countable pseudo base for a topological space $(X, \tau)$, then $\sigma(\mathcal{B})=\sigma(\tau)$, where $\sigma()$ is the sigma algebra generated by $()$. – Badr Alharbi Nov 16 '17 at 20:00
  • Thanks for the update. Please use the edit button below your question to add this to the question itself. – Nate Eldredge Nov 16 '17 at 20:28

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Suppose $X$ is a second-countable Hausdorff space and that $x,y$ are two distinct limit points of $X$. Let $\{U_n\}$ be any countable basis for $X$ and let $V_n = U_n \setminus \{x,y\}$, so that $V_n$ is open, nonempty, and $V_n \subset U_n$. Then $\{V_n\}$ is a countable pseudo-basis. However, every set in the $\sigma$-algebra generated by $\{V_n\}$ either contains $\{x,y\}$ or is disjoint from it. In particular, this $\sigma$-algebra does not contain the Borel set $\{x\}$.

Nate Eldredge
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