# If we think of a U (set union) as a function what would be its domain and codomain?

We could think of a set union as a function that maps from several sets in A(the domain) to a set in B(the codomain). However, according to answers I have it an onto function. Could I get an explanation for the logic behind the reasoning in the answer?

1. We can think of ∪ (set union) and ∩ (set intersection) as functions. What are the domain and codomain of ∪? Is ∪ a 1-to-1 function? Is ∪ an onto function? Answer: Let U be the universal set, and P(U) be the powerset of U.

• The domain of ∪ is P(U) × P(U) and the codomain is P(U).

• Let A be any nonempty subset of U. Then, (A, ∅) and (∅, A) are distinct elements in the domain, yet ∪ maps them both to the same element A in the codomain. Hence, ∪ is not 1-to-1.

• For every set A in the codomain P(U), ∪ maps (A, ∅) to A. Hence, ∪ is onto.

• The logic behind the reasoning is in the answer you posted. What part(s) of it do you need clarification for? – Clive Newstead Aug 10 at 20:50
• I don't see the logic behind the definition of the domain and codomain. Why do we need to take the cartesian product of the power sets and map it to all the power sets in the codomain? Wouldn't the union only map to one power set that has the union of all the objects from the inputs. – BKS_headphonesman Aug 10 at 20:55

As you have said, we want $$\cup$$ to be a function that takes as input two sets and returns another set. First off though, arguing with the universal set is tricky. What I would rather do is restrict yourself to a smaller set of possible inputs. For example, consider only finite sets, or only countable sets, ... In general, choose a particular set $$X$$ and then one can talk about such function, which takes subsets $$A, B \subseteq X$$ and returns another subset $$C \subseteq X$$.

Now the domain of such a function should be the set containing all pairs of subsets of $$X$$ i.e. the cartesian product of $$P(X)$$ with itself. Since the function returns a subset of $$X$$ the co domain should be $$P(X)$$. Hence $$\cup: P(X) \times P(X) \rightarrow P(X)$$.

Since $$\emptyset \subseteq A$$ we have $$\emptyset \in P(X)$$. By the definition of the cartesian product of sets $$(\emptyset, A)$$ and $$(A, \emptyset)$$ are two distinct elements in $$P(X) \times P(X)$$. However, since $$A \cup \emptyset = A = \emptyset \cup A$$ for any $$A \in P(X)$$ the function is not injective i.e. not 1-1.

Finally, for any set $$A \in P(X)$$ we have $$A = \emptyset \cup A$$. Hence $$A$$ is the image of $$(\emptyset, A)$$ under the function. This makes the function by definition surjective i.e. onto.

• If the codomain is the powersets of U, how can that be a surjective function, when not all the powersets are guarantees to be unions of the set a and set b. The union will only map to one set. Do you mean that the domain is only one specific set? – BKS_headphonesman Aug 11 at 0:16
• Maybe you have a misconception regarding codomain and 8mage of a function. The com domain contains all the possible outputs, the image of a particular inout is a particular output. – G. Chiusole Aug 11 at 7:07

Apart from $$A\cup B=\{\,x\mid x\in A\lor x\in B\,\}$$ we can also consider the more general union $$\bigcup A=\{\,x\mid \exists y\colon x\in y\in A\,\}$$ (so if $$A$$ is a finite set of sets, $$A=\{A_1,A_2,\ldots, A_n\}$$, we have $$\bigcup A=A_1\cup A_2\cup \ldots \cup A_n$$).

Then $$\bigcup$$ can take as input any set ad produces an arbitrary set as output. Since the the class of all sets is not a set, speaking of domain set and range set is a bit problematic. The way in which $$\bigcup$$ is function-esque is that it is a class function. This just means that we have a predicate $$\Phi(x,y)$$ with the properties $$\forall x \exists y\colon \Phi(x,y)$$ and $$\forall x \forall y \forall z\colon \Phi(x,y)\land \Phi(x,z)\to y=z$$ (in other words, for every set $$x$$ there exists one and only one $$y$$ such that $$\Phi(x,y)$$; we interpret $$\Phi(x,y)$$ as "$$x$$ is mapped to $$y$$"). The $$\Phi$$ for $$\bigcup$$ is readily written down: $$\Phi(x,y)\equiv \forall z\colon ( z\in y\leftrightarrow \exists w\colon z\in w\land w\in x).$$