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Consider an uncountable sequence of disjoint sets $S_a=\{a,a+1\}$, for $a\in [0,1)$. Each possesses the trivial topology and the corresponding Borel $\sigma$-algebra.

My questions is two fold:

I. (Half affirmation): $\cup_{a\in[0,1]} \mathcal{B}(S_a)\subset \mathcal{B}([0,2))$ but $\mathcal{B}(\cup_{a\in[0,1]} S_a)= \mathcal{B}([0,2))$ Is it true?

(the l.h.s. corresponds to a sigma algebra generated by a set of subsets which are sigma algebras for each $S_a$).

II. What should be spaces $S_a$ be for the latter equality to break down? In particular is it possible to have even $\mathcal{B}(\cup_aS_a)\subset \cup_{a\in[0,1]} \mathcal{B}(S_a)$?

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  • $\begingroup$ Do you mean strict inclusion by $⊂$? $\endgroup$ – user87690 Nov 7 '17 at 15:38
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The $σ$-algebra generated by $⋃_{a ∈ [0, 1]} \mathcal{B}(S_a)$ is the $σ$-algebra generated by all singletons, i.e. the countable-cocoutable algebra on $[0, 2)$.

$\mathcal{B}(⋃_{a ∈ [0, 1]} S_a) = \mathcal{B}([0, 2))$ holds trivially since $⋃_{a ∈ [0, 1]} S_a = [0, 2)$.

$\mathcal{B}(A) ⊆ \mathcal{B}(B)$ holds if and only if $A$ is Borel in $B$.

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