# Sigma-algebra without exhausting sequence?

Let me first state the definition of an Exhausting Sequence of Sets:

Let $X$ be a set, and let $\mathcal{S} \subset \mathcal{P}(X)$ be a collection of subsets of $X$. An exhausting sequence of sets (in $\mathcal{S}$) is a sequence $(S_n)_{n \in \mathbb{N}}$ in $\mathcal{S}$, subject to:

1. for all $n$, $S_n \subset S_{n+1}$
2. $\bigcup_n S_n = X$

Now I must be mistaken, because to me it seems that any $\sigma$-algebra on $X$ contains such a sequence. Take for example $S_n=X$ for all $n$. So here is my question: Is there a $\sigma$-algebra that does not have an exhausting sequence?

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Some definitions require a $\sigma$-algebra to have $X$ as an element. So no. You could also require that $X$ itself is not a member of the sequence, but even then I don't think there is a counterexample. – Asaf Karagila Oct 1 '12 at 8:29
Thank you sir, that clears things up. – Bas van Heiningen Oct 1 '12 at 8:40

Let $\langle A_k:k\in\Bbb N\rangle$ be any sequence of non-empty sets in the $\sigma$-algebra $\mathscr{S}$. For $n\in\Bbb N$ let $T_n=\bigcup_{k\ge n}A_k\in\mathscr{S}$, and let $C_n=X\setminus T_n\in\mathscr{S}$. Then $\langle C_n:n\in\Bbb N\rangle$ is an exhausting sequence in $\mathscr{S}$, even if you require that the sequence be non-trivial (i.e., not contain $X$ itself).