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(1)I did not understand if this solution actually states that $\mathcal{A}$ is not a $\sigma$-algebra given the fact $X$ is finite. Does b) proves $\mathcal{A}$ is not $\sigma$-algebra ?

(2)Do $\sigma$-algebras need to have at least two sets that result from infinite unions to be considered $\sigma$-algebras?

Thanks in advance!

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(1) The solution shows two implications. First if $X$ is finite, then $\mathcal{A}$ is a $\sigma$-algebra. The reason is that every subset of a finite set is either finite or cofinite, and therefore $\mathcal{A}$ contains every subset of $X$, i.e. $\mathcal{A} = \mathcal{P}(X)$, the powerset of $X$. Powersets are always $\sigma$-algebras because they are closed under arbitrary set operations and therefore closed under complements and countable unions, the latter being the set operations that characterize $\sigma$-algebras.

Second, if $X$ is not finite, then $\mathcal{A}$ is not a $\sigma$-algebra. The reason is that in this case $\mathcal{A}$ is not closed under countable unions.

(2) It's not completely clear what you're asking here. Any set $A$ can be represented as an infinite union by $$A = \cup_{n=1}^{\infty}A_n, \ \ A_1 = A, \ \ A_n = \emptyset \ \ \text{if} \ \ n \geq 2.$$ What's distinctive of $\sigma$-algebras, again, is that they are closed under countable set operations. As shown in (1), this doesn't imply that every $\sigma$-algebra contains an infinite set.

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