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Let $X$ be a separable metric space and let $\mathfrak{M}$ be the $\sigma$-algebra generated by open balls in $X$. Show that $\mathfrak{M}$ contains all the open sets in $X$ and all the closed sets

My question is: In a metric space, isn't an open set same as an open ball. If it is then proof follows trivially for open balls. Taking the complement, all closed sets are also in $\mathfrak{M}$. Then what is the point of $X$ being separable? Am I missing something?

Please note I have already done the proof using the separability of $X$ but I want to know if the question is trivial enough

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Of course not. Here is an example: in $\mathbb R$, $V=(0,1)\cup(2,3)$ is open and not a ball.

If $X$ is not separable, there will be open sets which are not countable unions of balls, and so will not be in the $\sigma$-algebra generated by the open balls.

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In a metric space, every open ball is an open set, but certainly not the other way around. A trivial example is simply to take the union of two disjoint open balls (in, say $\mathbb R^2$ if you want to get a nice picture). Most certainly the union of two disjoint open balls is not an open ball, but it is an open set.

To solve your problem, think of closure properties of $\sigma$-algebras, of open sets and open balls (in separated metric spaces), and the relationship between open/close and set complementation.

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