Increasing Sequence of Sets induces Partition on Limit

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Theorem

Let $\sequence {S_n}_{n \mathop \in \N} \uparrow S$ 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 $\sequence {T_n}_{n \mathop \in \N}$ is a countable partition of $S$.


Proof

That $\sequence {T_n}_{n \mathop \in \N}$ partitions $S$, means precisely that:

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

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 is 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:

\(\ds T_m \cap T_l\) \(\subseteq\) \(\ds T_m \cap S_{m-1}\) Set Intersection Preserves Subsets
\(\ds \) \(\subseteq\) \(\ds \paren {S_m \setminus S_{m - 1} } \cap S_{m-1}\) Definition of $T_m$
\(\ds \) \(=\) \(\ds \O\) Set Difference Intersection with Second Set is Empty Set

Hence $T_m \cap T_l = \O$.

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

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

$\Box$


Proof of $(2)$

By Set Union Preserves Subsets and Set Difference is Subset, have that:

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

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


So let $s \in S$.

Then by definition of union, the set:

$N_s := \set {n \in \N: s \in S_n}$

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:

$s \in \displaystyle \bigcup_{n \mathop \in \N} T_n$

That is, by definition of subset:

$S \subseteq \displaystyle \bigcup_{n \mathop \in \N} T_n$

$\blacksquare$