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Let $X$ be a second countable metrizable space and $\mathcal{G}$ a subbase for the topology of $X$. Show that $\mathcal{G}$ generates $\mathcal{B}_X$, the Borel $\sigma$-algebra of $X$. Also show that this need not be true.

How to show that something generates something?

Idonknow
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    By definition, the $\sigma$-algebra generated by a set $\mathcal{C}$ is the smallest sigma-algebra $\sigma(\mathcal{C})$ containing $\mathcal{C}$. You know that $\mathcal{B}_X$ is, by definition, generated by the topology, let's call it $\tau$, i.e., $\mathcal{B}_X=\sigma(\tau)$. If you show that $\tau\subseteq\sigma(\mathcal{G})$, you conclude that $\mathcal{B}_X\subseteq\sigma(\mathcal{G})$, by the first phrase. The reverse inclusion should be easy. – Luiz Cordeiro Aug 06 '15 at 00:45
  • The inclusion $\mathcal{G} \subset \sigma(\tau)$ is due to the fact that $\mathcal{G}$ is a subbase for $\tau$, am I right? – Idonknow Aug 06 '15 at 03:05
  • Even more, $\mathcal{G}$ is a subset of $\tau$, and $\mathcal{B}_X=\sigma(\tau)$ is the smallest $\sigma$-algebra containing $\tau$, i.e., $\mathcal{G}\subseteq\tau\subseteq\sigma(\tau)=\mathcal{B}_X$, so $\mathcal{B}_X$ is a $\sigma$-algebra containing $\mathcal{G}$, from which follows that $\sigma(\mathcal{G})\subseteq\mathcal{B}_X$ (see first phrase of previous comment). This inclusion does not use the fact that $\mathcal{G}$ is a subbase. – Luiz Cordeiro Aug 06 '15 at 03:53

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