# Increasing Sequence of Sets induces Partition on Limit

## 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:

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

$\blacksquare$