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Suppose I have a sequence of sets $\{A_n\}$.

Now, we know that an algebra is close under finite union intersection while a sigma-algebra is close under infinite countable intersection.

Finite Intersection means that for Every $N$ $$ \bigcup_{k=1}^N A_k $$ Is well defined and it's a set in the algebra.

My question is: why this does no imply the Infinite union intersection ?

Intuitively, for every $N$ this set is in the algebra $$ \bigcup_{k=1}^N A_k \cup A_{N+1} $$ And I can proceed in such a way for Every $N$. I can do partial union as in the series. Why this does not imply that the infinite union is in the sigma algebra?

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Take for example the algebra $\mathcal A$ generated by the closed set. It's an algebra, but a countable union of closed can be not closed. Indeed, $$\bigcup_{n\in\mathbb N}[0,\frac{1}{n}]=[0,1[\notin \mathcal A$$

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Example: let $\mathcal{A}$ be the subset of the power set of $\mathbb N$ be defined by:

$A \in \mathcal{A}$ iff $A$ is finite or $ \mathbb N \setminus A$ is finite. $\mathcal{A}$ is an algebra.

Put $A_n=\{2n\}$ for $n \in \mathbb N$.

Then $\bigcup_{k=1}^N A_k \in \mathcal{A}$ for each $N$, but $\bigcup_{k=1}^{\infty} A_k \notin \mathcal{A}$.

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