# “Disjoint union” in set theory.

I just came across a term called "disjoint union". Define $\{ A_i : i \in I\}$ be a family of sets indexed by $I$. On wikipedia it defines disjoint union as as,

$$\bigcup_{i \in I} \{ (x,i): x \in A_i\}$$

and they say that

When one says that a set is the disjoint union of a family of subsets, this means that it is the union of the subsets and that the subsets are pairwise disjoint.

But this can't be right! If $A = \{1,2 \} = \{ A_1, A_2 \}$, then this means the disjoint union is

$$\{ (1,1) \} \cup \{ ( 2,1 ) \} \cup \{ (1, 2) \} \cup \{( 2,2) \}$$

However, $\{ (1,1) \} \cap \{ (1,2) \} = \{ \{1\}\}$, so they can't be pairwise disjoint!

Note that $(a,b) = \{ \{ a\}, \{a,b \} \}$

• @angryavian that's the definition of an ordered pair. – Kevin Carlson Sep 19 '14 at 3:52
• Doesn't $\{ (1,1) \} \cap \{ (1,2) \} = \{ \}$? – David McKnight Oct 31 '16 at 4:14

You haven't computed this correctly. If $A=\{A_1,A_2\}$ as you say then $\cup A=\{(x,1):x\in A_1, (y,2):x\in A_2\}$. So you'd get $\{(\{\},1),(\{\},2),(\{\{\}\},2)\}$, using the convention $1=\{\{\}\}$ and $2=\{\{\},\{\{\}\}\}$. So the images of $A_1$ and $A_2$ don't intersect after all. That said, this probably isn't what you meant: in the normal way of speaking, $A=\{1,2\}$ is not a family of sets at all-it's a set of numbers! If you had something like $A=\{\{1,2\},\{1\}\}$ a more reasonable family of sets indexed by $1,2$, the disjoint union would come out as $\{(1,1),(2,1),(1,2)\}$. Note that this is still disjoint. That said, it's sort of gross, so you'd often require that the set used to index $A$ not intersect any of the elements of $A$.
• How about switch my set $A$ to be $\{ a, b\}$? – Hawk Sep 19 '14 at 4:08
• OK, then how are you thinking of it as a family of sets, i.e. a set of sets with an indexing function? Or do you want the disjoint union of two copies of $A$? – Kevin Carlson Sep 19 '14 at 4:10
• Ah I see, so if $A = \{ a,b \}$, my disjoint union is just $(a,1) \cup (b,2).$ – Hawk Sep 19 '14 at 4:26
• Yes, or very nearly-it's $\{(a,1),(b,2)\}$. – Kevin Carlson Sep 19 '14 at 4:35