Consider the the following set of numbers

A = { 0, 1, -1, 2, -2, 3, -3, 4, -4 }

If we want to partition this set, we need to form non-empty subsets that are all disjoint, and such that all elements in A are in one of the subsets.

Here’s one way to partition A:

A1	=	{ 0, 1 }
A2	=	{ -1, 2, -2 }
A3	=	{ 3, -3, 4, -4 }

Notice that we didn’t start with the index set

I = { 1, 2, 3 }

Instead, we formed the subsets first, and then indexed them. Of course, we should check to see if the conditions specified in the definition are met:

A₁ ∪ A₂ ∪ A₃ = A

A1 ∪ A2 ∪ A3	=	{ 0, 1 } ∪ { -1, 2, -2 } ∪ { 3, -3, 4, -4 }
	=	A

This means that every element from A was accounted for. Now we check to see if all the subsets are disjoint:

A1 ∩ A2	=	{ 0, 1 } ∩ { -1, 2, -2 }
	=	∅

A1 ∩ A3	=	{ 0, 1 } ∩ { 3, -3, 4, -4 }
	=	∅

A2 ∩ A3	=	{ -1, 2, -2 } ∩ { 3, -3, 4, -4 }
	=	∅

Both conditions have been met, so this is a legitimate partition of A.

There are other ways to partition the set

A = { 0, 1, -1, 2, -2, 3, -3, 4, -4 }

One such way to partition the set is with the following:

X1	=	{ 0, 4 }
X2	=	{ -1 }
X3	=	{ 1, 2 }
X4	=	{ -2, -3, -4 }
X5	=	{ 3 }

Notice that we didn’t use the letter A for the names of these subsets. The important thing is the content of the subsets themselves, not the name given to the subsets. As long as the two conditions are met, the collection of sets form a partition. We could also put every element in A into it’s own subset like so:

Y1	=	{ 0 }
Y2	=	{ 1 }
Y3	=	{ -1 }
Y4	=	{ 2 }
Y5	=	{ -2 }
Y6	=	{ 3 }
Y7	=	{ -3 }
Y8	=	{ 4 }
Y9	=	{ -4 }

Yet another partition we could form is just by taking the entire set itself:

Z = { 0, 1, -1, 2, -2, 3, -3, 4, -4 }

Since there are no other sets in this partition, it is trivially true that the union of all the sets form A. Furthermore, since there are no other sets in the partition, it is trivially true that all sets in the partition are disjoint.

As mentioned in the introduction, one thing we can do is split up the integers into two subsets: one containing the even integers, and a second subset containing the odd integers.

A	=	{ 0, 2, -2, 4, -4, 6, -6, … }
B	=	{ 1, -1, 3, -3, 5, -5, 7, -7, … }

Here, we didn’t use the same letter for the subset names, nor did we use indices. Again, the names of the subsets is not important. What’s important is knowing that all subsets are mutually disjoint, and that all elements from the original set are accounted for.

One way to partition the real numbers ℝ is into three subsets: the subset of all positive real numbers greater than 0, the subset of all negative real numbers less than 0, and finally the subset containing just the number 0.

X	=	{ x ∈ ℝ | x < 0 }
Y	=	{ 0 }
Z	=	{ x ∈ ℝ | x > 0 }

Some numbers that are in X include -π, -12.3372, and -√2.

Some numbers in Z include 1.2345, 12, and π * π.

As one more example, we could partition the positive rational numbers into two separate subsets.

A	=	{ a/b | a ∈ {0, 1, 2, 3, …} ∧ b ∈ {1, 2, 3, …} ∧ a/b < 1}
B	=	{ a/b | a ∈ {0, 1, 2, 3, …} ∧ b ∈ {1, 2, 3, …} ∧ a/b ≥ 1}