# Subsets in Increasing Union

## Theorem

Let $S_0, S_1, S_2, \ldots, S_i, \ldots$ be a nested sequence of sets, that is:

$S_0 \subseteq S_1 \subseteq S_2 \subseteq \ldots \subseteq S_i \subseteq \ldots$

Let $S$ be the increasing union of $S_0, S_1, S_2, \ldots, S_i, \ldots$:

$\displaystyle S = \bigcup_{i \mathop \in \N} S_i$

Then:

$\forall s \in S: \exists k \in \N: \forall j \ge k: s \in S_j$

## Proof

Let $k \in \N$.

Let $j \ge k$.

Then by as many applications as necessary of Subset Relation is Transitive, we have:

$S_k \subseteq S_j$

Now $s \in S$ means, by definition of set union, that:

$\exists S_k \subseteq S: s \in S_k$

Then from above:

$j \ge k \implies S_k \subseteq S_j$

it follows directly that:

$\forall s \in S: \exists k \in \N: \forall j \ge k: s \in S_j$

from the definition of subset.

$\blacksquare$