Is $(A × B) ∪ (C × D) = (A ∪ C) × (B ∪ D)$ true for all sets $A, B, C$ and $D$?

Is $(A \times B) \cup (C \times D) = (A \cup C) \times (B \cup D)$ true for all sets $A, B, C$ and $D?$

I tried to wrap my head around this, but I have absolutely no idea what is going on here. How could I possibly go about trying to prove this? Or disprove it?

Any help or guidance would be appreciated.

• What examples did you try? How about $A=\{a\},B=\{b\},C=\{c\},D=\{d\}$? – bof Apr 15 '15 at 8:41
• dave, please do not vandalize your post. – André 3000 Apr 15 '15 at 11:59

No, take $A=D=\{1\}$ and $B=C=\{2\}$.

$$\begin{array}{c} \begin{array}{c|cc} 2&\bullet&\bullet\\ 1&\bullet&\bullet\\ \hline &1&2\\ \end{array}\\\\ (A\cup C)\times(B\cup D) \end{array} \qquad\qquad \begin{array}{c} \begin{array}{c|cc} 2&\bullet&\\ 1&&\bullet\\ \hline &1&2 \end{array}\\\\ (A\times B)\cup(C\times D) \end{array}$$

• I took the liberty of adding a diagram; feel free to delete it if you prefer. – Brian M. Scott Apr 15 '15 at 14:32
• @BrianM.Scott no it's great (and very esthetic), thank you very much :). – Surb Apr 15 '15 at 14:33

Let $A=[0,1]=B$ and $C=[1,2]=D$

Here the first figure represents $(A\times B) \cup (C\times D)$ while the second one represents $(A\cup C) \times (B\cup D).$

Comment: I thought this problem looked familiar--I encountered it some time ago at the beginning of a real analysis class (surveying proof techniques, basic set theory, etc.). Although you may have encountered this problem elsewhere, it is problem 1.9.25 in Witold Kosmala's book A Friendly Introduction to Analysis. There, it occurs as a review problem where you are supposed to label the statement as true or false. "If a statement is true, prove it. If not, (i) given an example of why it is false, and (ii) if possible, correct it to make it true, and then prove it." I thought I would provide you with what I vaguely recall having answered with (both the counterexample and corrected statement with a proof).

Original claim: If $A,B,C,D$ are any sets, then $(A\times B)\cup(C\times D)=(A\cup C)\times(B\cup D)$.

[False] Counterexample. Let $A = \{ 1 \}, B = \{ 2 \}, C = \{ 3 \}, D = \{ 4 \}$. Then

• $A \cup C = \{1,3\}$;
• $B \cup D = \{2,4\}$;
• $(A\cup C) \times (B \cup D) = \{(1,2),(1,4),(3,2),(3,4)\}$;
• $A \times B = \{(1,2)\}$;
• $C\times D = \{(3,4)\}$;
• $(A \times B) \cup (C \times D) = \{(1,2),(3,4)\}$.

From this example, it is clear that $(A \times B) \cup (C \times D) \subseteq (A\cup C) \times (B \cup D)$, but there is not mutual subset inclusion, for $(A\cup C) \times (B \cup D) \nsubseteq (A \times B) \cup (C \times D)$.

Corrected claim: $(A \times B) \cup (C \times D) \subseteq (A\cup C) \times (B \cup D)$.

[True] Proof. Denote $\phi \colon (A \times B) \cup (C \times D)$ and $\sigma \colon (A\cup C) \times (B \cup D)$. Suppose $(p,q) \in \phi$. Then $(p,q)\in A\times B$ or $(p,q) \in C \times D$. Thus, $p \in A$ or $p \in C$; i.e., $p \in A \cup C$. Also, $q \in B$ or $q \in D$; i.e., $q \in B \cup D$. Thus, $(p,q) \in (A \cup C) \times (B \cup D)$, and, consequently, $(A \times B) \cup (C \times D) \subseteq (A\cup C) \times (B \cup D)$. $\blacksquare$

Actually, the equality is very rarely true. It's simple to see what is happening if you imagine $A,B,C,D$ as intervals in $\mathbb R^3$. For example, take $A=B=[0,1]$ and $C=D=[1,2]$. Then, $(A\cup C)\times (B\cup D) = [0,2]\times [0,2]^2$, i.e. the whole square.

On the other hand, $(A\times B)\cup (C\times D) = [0,1]^2\cup [1,2]^2$.

It is not always true. The actual truth is (A × B) ∪ (C × D) ⊆ (A ∪ C) × (B ∪ D) To proof that (A × B) ∪ (C × D) ≠ (A ∪ C) × (B ∪ D), you can take A,B as null sets, then it would be easy.