# Mapping on 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 $S_0, S_1, S_2, \ldots, S_i, \ldots$:

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

For each $i \in \N$, let $f_i: S_i \to T$ be a mapping such that:

- $\forall j < i: f_i \restriction_{S_j} = f_j$

where $f_i \restriction_{S_j}$ denotes the restriction of $f_i$ to $S_j$.

Then there is a **unique** mapping:

- $f: S \to T$

which extends each $f_i$ to $S$.

## Proof

Suppose $f: S \to T$ and $g: S \to T$ are both extensions of $f_i$ to $S$ for all $i \in \N$ such that $f \ne g$.

We have been given the domain and codomain of $f$.

Suppose there exists any $g: S \to T$ which is different from $f$.

This must be because $f$ and $g$ do not agree throughout their entire domain.

Thus:

- $\exists x \in S: f \left({x}\right) \ne g \left({x}\right)$

Then by definition of set union:

- $\exists S_k \subseteq S: x \in S_k$.

From the definition, both $f$ and $g$ are extensions of $S_k$.

But this means they must agree on $S_k$.

That is:

- $\forall x \in S_k: f \left({x}\right) = g \left({x}\right)$

This contradicts the supposition that $f \left({x}\right) \ne g \left({x}\right)$.

Thus:

- $\forall x \in S: f \left({x}\right) = g \left({x}\right)$

and $f$ has been shown to equal $g$.

Hence the result.

$\blacksquare$

## Sources

- 1996: H. Jerome Keisler and Joel Robbin:
*Mathematical Logic and Computability*... (previous) ... (next): Appendix $\text{A}.4$: Composition and Restriction: Problem $\text{A}.4.1$