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This might be an easy question, but I don't really know what to use.

Let $(S,\mathcal{S},d)$ be a seperable metric space, and let furthermore $(S,\mathcal{S})$ be standard borel

That is there exist a injective map so $\varphi:$ $(S,\mathcal{S} )\mapsto(\mathbb{R} ,\mathcal{B} (\mathbb{R} ))$ where $\varphi(S)\in \mathcal{B} (\mathbb{R} )$, $\varphi\in\mathcal{M}(\mathcal{S})$ and $\varphi^{-1}$ is $\mathcal{B} (\mathbb{R})-\mathcal{S}$-measurable.

Can I without further assumptions know my metric space contains all open (or closed) sets?

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I don't understand what standard Borel means nor what $\mathcal B(\mathbb R)$ and $\mathcal M ( \mathcal S)$ stand for but in general no: the trivial sigma algebra does not contain all Borel sets. If you want the sigma algebra to contain all Borel sets it will have to contain the Borel sigma algebra. –  Matt N. Dec 1 '12 at 20:04
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Standard Borel, according to some papers I read, means S is a complete separable metric space. Hope that's what he also means. –  Gautam Shenoy Dec 1 '12 at 20:20
    
@MattN.: But the trivial $\sigma$-algebra is not standard Borel: no injective map $\varphi$ satisfying the above conditions exists. $\mathcal{B}(\mathbb{R})$ is the Borel $\sigma$-algebra on $\mathbb{R}$, and I suppose $\mathcal{M}(\mathcal{S})$ means that $\varphi$ is measurable, i.e. $\varphi^{-1}(A) \in \mathcal{S}$ for each $A \in \mathcal{B}(\mathbb{R})$. –  Nate Eldredge Dec 1 '12 at 20:51

1 Answer 1

up vote 4 down vote accepted

You cannot.

Take $S = \mathbb{R}$, $d$ the Euclidean metric, and $\mathcal{S}$ defined as follows. Let $E \subset \mathbb{R}$ be any non-Borel set such that $E$ and $\mathbb{R} \setminus E$ both have cardinality continuum. Then there is a bijection $\varphi : \mathbb{R} \to \mathbb{R}$ which maps $(0,1)$ onto $E$ and $\mathbb{R} \setminus (0,1)$ onto $\mathbb{R} \setminus E$. Take $\mathcal{S} = \{ \varphi^{-1}(B) : B \in \mathcal{B}(\mathbb{R})\}$. This is clearly a $\sigma$-algebra and $(S, \mathcal{S})$ is standard Borel, since by construction $\varphi$ is a measurable bijection from $(S, \mathcal{S})$ to $(\mathbb{R}, \mathcal{B}(\mathbb{R}))$. But since $E$ is not Borel, we have $(0,1) = \varphi^{-1}(E) \notin \mathcal{S}$, even though $(0,1)$ is open in $(S,d)$.

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Thanks! I guess I just haven't thought enough about what implications being standard Borel actually has. –  Henrik Dec 1 '12 at 22:13

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