Definition 9 (Union). X ∪ Y = {x : x ∈ X or x ∈ Y}, where the “or” is inclusive.
13Unions
Given two sets X and Y, their union is the set containing those objects which belong to at least one of them. We write it X ∪ Y. So the inclusion criterion for X ∪ Y is x ∈ X or x ∈ Y. Where intersection asks for both, union asks for at least one. (Note that we use “or” in the inclusive sense: if x belongs to both X and Y, then we say it is true that “x ∈ X or x ∈ Y”.)
For example, {7, 8, 9} ∪ {8, 9, 10} = {7, 8, 9, 10}. In Figure 7, X ∪ Y is the shaded region: everything inside either circle, the overlap included.
Where intersection only ever made a set smaller, union only ever makes it bigger.
Proposition 15 (each set is a subset of the union). For any sets X and Y, X ⊂ X ∪ Y and Y ⊂ X ∪ Y.
Suppose x ∈ X. Then we immediately have x ∈ X or x ∈ Y; hence x satisfies the inclusion criterion for X ∪ Y; hence x ∈ X ∪ Y. Since x was chosen arbitrarily, it follows that X ⊂ X ∪ Y.
The proof that Y ⊂ X ∪ Y is exactly analogous.
The remaining facts about union are the mirror images of the ones we proved about intersection in Section 12, with “or” in place of “and”. Rather than prove them for you, I will let you do them yourself. Each one can be done by the method of demonstrating equivalent inclusion criteria, and in each case the fact about logic you need is the one about “or” corresponding to the fact about “and” we used before.
Proposition 16 (union commutes). For any sets X and Y, X ∪ Y = Y ∪ X.
Exercise 3. Prove Proposition 16 (union commutes).
Proposition 17 (union is associative). For any sets X, Y and Z, (X ∪ Y) ∪ Z = X ∪ (Y ∪ Z).
Exercise 4. Prove Proposition 17 (union is associative).
From now on, we can write X ∪ Y ∪ Z with no brackets, and we can ignore the difference between X ∪ Y and Y ∪ X, exactly as we did for intersection. The next two propositions are the analogues of Propositions 12 (intersecting a set with itself) and 13 (intersecting with the empty set).
Proposition 18 (uniting a set with itself). For any set X, X ∪ X = X.
Exercise 5. Prove Proposition 18 (uniting a set with itself).
Proposition 19 (uniting with the empty set). For any set X, X ∪ ∅ = X.
Exercise 6. Prove Proposition 19 (uniting with the empty set).
Proposition 20 (union respects inclusion). For any sets X, Y and Z, if X ⊂ Y, then X ∪ Z ⊂ Y ∪ Z.
Exercise 7. Prove Proposition 20 (union respects inclusion).
Now for something which has no analogue among the propositions about intersection. Proposition 15 (each set is a subset of the union) tells us that X ∪ Y is big enough to contain both X and Y. The next proposition tells us that it is no bigger than it has to be: any set which contains both X and Y already contains their union.
Proposition 21 (the union is the smallest set containing both). For any sets X, Y and Z, if X ⊂ Z and Y ⊂ Z, then X ∪ Y ⊂ Z.
Suppose that X ⊂ Z and Y ⊂ Z. Let x ∈ X ∪ Y. It follows that either (i) x ∈ X, or (ii) x ∈ Y.
Case (i): since x ∈ X and X ⊂ Z, we have x ∈ Z.
Case (ii): since x ∈ Y and Y ⊂ Z, we have x ∈ Z.
Hence x ∈ Z, from which we conclude that X ∪ Y ⊂ Z.
That proof needed a move we had not used before. To show that an arbitrary element of X ∪ Y lies in Z, we only know that it lies in X or in Y, and not which. So we take the two cases in turn and show that the conclusion follows in each of them. Since one case or the other must hold, the conclusion follows either way. This is called proof by cases, and it is what disjunctions generally demand of us. (If you have taken logic, you may recognise this as an informal version of the disjunction elimination inference rule.)
Finally, union gives us another way of saying that one set is a subset of another, just as intersection did in Proposition 14 (inclusion via intersection).
Proposition 22 (inclusion via union). For any sets X and Y, X ⊂ Y if and only if X ∪ Y = Y.
We again prove both sides of the biconditional.
(i) Left to right: suppose that X ⊂ Y. We want to show that X ∪ Y = Y. We have Y ⊂ X ∪ Y immediately from Proposition 15 (each set is a subset of the union). Further, if x ∈ X ∪ Y then by the definition of union we have either (a) x ∈ X or (b) x ∈ Y. In case (a), since X ⊂ Y we have x ∈ Y. In case (b), of course we also have x ∈ Y. Hence x ∈ Y either way, and so we conclude that X ∪ Y ⊂ Y. By Proposition 1 (double inclusion) it follows that X ∪ Y = Y.
(ii) Right to left: suppose that X ∪ Y = Y. We want to show that X ⊂ Y. Proposition 15 (each set is a subset of the union) implies that X ⊂ X ∪ Y, hence by substitution of identicals (that is, X ∪ Y = Y) we have X ⊂ Y.