Question

f: A->C and g: B->C. If A and B are distinct sets, prove g: A∪B->C

Answer 1

sorry prove h : A∪B->C

Answer 2

Ok...so basically you need to show
that if you have two functions f and g, their union is a function mapping to C, also.
Does the problem say anything else?
For example, how
h is formed from f and g?

Answer 3

it seems something is missing about h. It's asserting it's a function from A union B to C

Answer 4

but it doesn't describe how h relates to f and g.

Answer 5

actually h = g∪f but I don't get how h is related to f,g

Answer 6

I deliberately replaced f∪g: A∪B -> C, because I don't see how is it that the new function is defined as f∪g rather than just a function.

Answer 7

The way I understand see it is if f: A->C and g: B->C, then the new relation, say H, from A∪B to C is also a function. and define H = h.

Do you get what i'm trying to say?

Answer 8

*The way I understand it is if ...*

Answer 9

For example, f is the set of ordered pairs
(a, c)
where a is in A
and c is in C

It makes sense
if you view the functions as relations, in the ordered pair definition.
For example, f is the set of ordered pairs
(a, c) where a is in A and c is in C so that the a appears only once.
Oh, but if you think of them as relations/sets of ordered pairs, it does make sense to do a union.
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