Subsets in Increasing Union

Theorem
Let $S_0, S_1, S_2, \ldots, S_i, \ldots$ be sets such that:
 * $S_0 \subseteq S_1 \subseteq S_2 \subseteq \ldots \subseteq S_i \subseteq \ldots$

that is, each set is contained in the next as a subset.

Let $S$ be the increasing union of $S_0, S_1, S_2, \ldots, S_i, \ldots$:
 * $\displaystyle S = \bigcup_{i \in \N} S_i$

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 Subsets 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 definition of subset.