Union of Mappings which Agree is Mapping

From ProofWiki
Jump to navigation Jump to search

Theorem

Let $A, B, Y$ be sets.

Let $f: A \to Y$ and $g: B \to Y$ be mappings.

Let $X = A \cup B$.

Let $f$ and $g$ agree on $A \cap B$.


Then $f \cup g: X \to Y$ is a mapping.

Family of Sets

Let $Y$ be a set.

Let $\family {A_i}_{i \mathop \in I}$ be a family of sets indexed by $I$.

Let $\family {f_i: A_i \to Y}$ be a family of mappings indexed by $I$.


Let $X = \ds \bigcup_{i \mathop \in I} A_i$.

Let $f = \ds \bigcup_{i \mathop \in I} f_i : X \to Y$ where $\ds \bigcup_{i \mathop \in I} f_i$ is the union of relations.


Let for all $i, j \in I$, $f_i$ and $f_j$ agree on $A_i \cap A_j$.


Then:

$f : X \to Y$ is a mapping


Proof

By definition, $f \cup g$ is a relation whose domain is $X = A \cup B$.

Let $\tuple {x, y_1} \in f \cup g$ and $\tuple {x, y_2} \in f \cup g$.

At least one of the following must be true:

$(1): \quad \tuple {x, y_1} \in f, \tuple {x, y_2} \in f$
$(2): \quad \tuple {x, y_1} \in g, \tuple {x, y_2} \in g$
$(3): \quad \tuple {x, y_1} \in f, \tuple {x, y_2} \in g$
$(4): \quad \tuple {x, y_1} \in g, \tuple {x, y_2} \in f$

Because $f$ and $g$ are mappings, $(1)$ and $(2)$ imply that $y_1 = y_2$.

If $(3)$ holds, then $y_1 = \map f x$ and $y_2 = \map g x$.

But then $x \in A \cap B$.

So by hypothesis:

$y_1 = \map f x = \map g x = y_2$

Similarly if $(4)$ holds.

Thus in all cases:

$\tuple {x, y_1}, \tuple {x, y_2} \in f \cup g \implies y_1 = y_2$

and so by definition $f \cup g: X \to Y$ is a mapping.

$\blacksquare$


Sources

Retrieved from "https://proofwiki.org/w/index.php?title=Union_of_Mappings_which_Agree_is_Mapping&oldid=578406"