Showing the union of minimal sigma algebras is not a sigma algebra This question is a problem from Allan Gut's 'Probability - A Graduate Course'. I'm new to the concept of sigma algebras and this question has stumped me.

Let $\Omega = \mathbb{R}$, set $I_k = [k-1,k)$ for $k \geq 1$. Let $\mathcal{A}_n = \sigma\{ I_k, k = 1, \dots , n \}$ (where $\sigma\{ A \}$ is the minimal sigma algebra on $A$). Show that $\bigcup_{n=1}^\infty \mathcal{A}_n$ is not a sigma algebra.

Thinking about the three main properties of sigma algebras: $\Omega$ is clearly an element of each $\mathcal{A}_n$, so it will be part of the union. For an element of $X \in \bigcup_{n=1}^\infty \mathcal{A}_n$, $X \in \mathcal{A_n}$ for some $n$, and thus its complement is part of the same algebra and is part of the union. So it must be something to do with a countable union of elements.
My idea was: $I_k \in \bigcup_{n=1}^\infty \mathcal{A}_n$, so construct $X = \bigcup_{k=1}^\infty I_{2k}$. If by way of a contradiction the union is a sigma algebra, then $X$ is an element of the union and thus $X \in \mathcal{A}_n$ for some $n$. Since a sigma algebra is a Dynkin system, we can construct $Y = \bigcup_{k > \lfloor n/2 \rfloor}I_{2k} \in \mathcal{A}_n$. It seems intuitively true that this set cannot be part of $\mathcal{A}_n$, since it is the minimal sigma algebra constructed from the first $n$ intervals, and there should be no way to construct this set from those intervals and their complements. But I don't know how to prove that.
If there is a way to generate a contraction with the set I have provided, or a more elegant argument I am missing, I would love to know.
 A: An auxiliary collection of $\sigma$-algebras might make things more clear.

Let $\mathcal B_n\subseteq\mathcal P(\mathbb R)$ such that $S\in\mathcal B_n$ iff $S\subseteq[0,n)$ or $S^c\subseteq[0,n)$.
Then evidently $\mathcal B_n$ is a $\sigma$-algebra and this with $I_1,\dots,I_n\in\mathcal B_n$.
From this we conclude that $\mathcal A_n\subseteq\mathcal B_n$ and consequently $\bigcup_{n=1}\mathcal A_n\subseteq\bigcup_{n=1}\mathcal B_n$.
Sets like $[0,\infty)=\bigcup_{k=1}^{\infty}I_{k}$ or  $X=\bigcup_{k=1}^{\infty}I_{2k}$ (if you like) are evidently not an element of one of the $\mathcal B_n$'s.
A: Given disjoint subsets $A_1,A_2,\cdots,A_n$ of  a set $\Omega$ you can explicitly write down all the sets in $\sigma (A_1,A_2,\cdots,A_n)$: Let $A_0=\Omega \setminus \bigcup_{i=1}^{n} A_i$. Then $\sigma (A_1,A_2,\cdots,A_n)$ consist of all possible unions of the stes $A_0,A_1,A_2,\cdots,A_n$ (including the empty union).
If $\bigcup_n \mathcal A_n$ is  a $\sigma-$ algebra then  $[0,\infty)$ must be in $\bigcup_n \mathcal A_n$ but it does not belong to any $\mathcal A_n$ from the explicit description of $\mathcal A_n$:  Note that any set in $\mathcal A_n$ contained in $[0,\infty)$ is necessarily bounded since it is  a finite union of the intervals $[k-1,k)$.
