Increasing Sequence of Sets induces Partition on Limit

Theorem
Let $\left({S_n}\right)_{n \in \N} \uparrow S$ be an increasing sequence of sets with limit $S$.

Define $T_1 = S_1$, and, for $n \in \N$, $T_{n+1} = S_{n+1} \setminus S_n$, where $\setminus$ denotes set difference.

Then $\left({T_n}\right)_{n \in \N}$ is a countable partition of $S$.

Proof
That $\left({T_n}\right)_{n \in \N}$ partitions $S$, means precisely that:


 * $(1):\quad$ The $T_n$ are pairwise disjoint
 * $(2):\quad \displaystyle \bigcup_{n \in \N} T_n = S$

It is more convenient to prove $(1)$ and $(2)$ separately:

Proof of $(1)$
Let $l,m \in \N$ be such that $l < m$.

Then by Set Difference Subset, $T_l \subseteq S_l$.

As the $S_n$ form an increasing sequence of sets, it follows that also $T_l \subseteq S_{m-1}$ because $m-1 \ge l$.

Now compute as follows:

Hence $T_m \cap T_l = \varnothing$.

Reversing the roles of $m$ and $l$ leads to the same conclusion if $l > m$.

Hence, by definition, the $T_n$ are pairwise disjoint.

Proof of $(2)$
By Set Union Preserves Subsets and Set Difference Subset, have that:


 * $\displaystyle \bigcup_{n \in \N} T_n \subseteq \bigcup_{n \in \N} S_n = S$

To establish $(2)$, by Equality of Sets, it is now only required to show that $S \subseteq \displaystyle \bigcup_{n \in \N} T_n$.

So let $s \in S$.

Then by definition of union, the set:


 * $N_s := \left\{{n \in \N: s \in S_n}\right\}$

is nonempty.

By Well-Ordering Principle, $N_s$ contains a minimal element, $n$, say.

If $n = 1$, then $s \in S_1 = T_1$.

If $n > 1$, then, by minimality of $n$, $s \notin S_{n-1}$.

Hence, by definition of set difference, $s \in T_n = S_n \setminus S_{n-1}$.

By definition of set union, it follows that $s \in \displaystyle \bigcup_{n \in \N} T_n$.

That is, $S \subseteq \displaystyle \bigcup_{n \in \N} T_n$ by definition of subset.