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 $$S_0, S_1, S_2, \ldots, S_i, \ldots$$:
 * $$S = \bigcup_{i \in \N} S_i$$

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 $$f$$, and so if there exists any $$g: S \to T$$ which is different from $$f$$, this must be because they 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.