Union Functions

Probability Level 3

Let $(f_i)_{i \in I}$ be a family of functions such that, $\forall i\in I, f_i:A_i \mapsto B_i$ . Is for every family, $\displaystyle \bigcup_{i\in I} f_i$ a function?

No Yes

1 solution

Paulo Guilherme Santos
Jun 14, 2015

No, it isn't. Here is an exemple of the falsity of these proposition: Consider $f_1=\{(1,1),(2,2)\}$ and $f_2=\{(1,2),(2,2)\}$ . It is clear that $\bigcup_{i\in\{1,2\}} f_i =\{(1,1),(2,2),(1,2),(2,2)\}$ . In this context we have that $(1,1)\in \bigcup_{i\in\{1,2\}} f_i$ and $(1,2)\in \bigcup_{i\in\{1,2\}} f_i$ , so $\bigcup_{i\in\{1,2\}} f_i$ is not a function.

As an addition to this solution, we can note that if all the $A_i$ 's are pairwise disjoint, then the union of the functions is indeed a function.

Prasun Biswas - 5 years ago

Union Functions

1 solution

0 pending reports