# Union of Functions Theorem

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## Theorem

Let $X$ be a set.

Let $\set {X_i: i \in \N}$ be an exhausting sequence of sets on $X$.

For each $i \in \N$, let $g_i: X_i \to Y$ be a mapping such that:

$g_{i + 1} \restriction X_i = g_i$

where $g_{i + 1} \restriction X_i$ denotes the restriction of $g_{i + 1}$ to $g_i$.

Then:

$\displaystyle \bigcup \set {g_i: i \in \N}$

is a mapping from $X$ to $Y$.

### Corollary

For each $i \in \N$, let $g_i : X_i \to Y$ be invertible.

Then $\displaystyle \bigcup \left\{{g_i: i \in \N}\right\}$ is invertible and:

$\displaystyle \left({\bigcup \left\{{g_i: i \in \N}\right\} }\right)^{-1} = \bigcup \left\{{g_i^{-1}: i \in \N}\right\}$

## Proof

By definition, $\displaystyle g = \bigcup \set {g_i: i \in \N}$ is a relation whose domain is $X$.

Aiming for a contradiction, suppose $g$ is not a mapping.

Then for some $x \in X$ and $i, h \in \N$:

$(1): \quad x \in X_i, \map {g_i} x \ne \map {g_{i + h} } x$

Let $k \in \N$ be the smallest such that:

$\map {g_i} x \ne g_{i + k}$

where $x$ and $i$ are the same as in $(1)$.

Then:

$\map {g_{i + k - 1} } x = g_i$

But then:

$g_{i + k} \restriction X_{i + k - 1} = g_{i + k - 1}$

From this contradiction it follows that our supposition that $g$ is not a mapping must be false.

That is:

$\displaystyle \bigcup \set {g_i: i \in \N}$

is a mapping from $X$ to $Y$.

$\blacksquare$