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 $S_0, S_1, S_2, \ldots, S_i, \ldots$:

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

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:

$\exists x \in S: \map f x \ne \map g x$

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:

$\forall x \in S_k: \map f x = \map g x$

This contradicts the supposition that $\map f x \ne \map g x$.

Thus:

$\forall x \in S: \map f x = \map g x$

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

Hence the result.

$\blacksquare$