Is it the case that for all sets $A, B, C,$ and $D$, $(A \times B) \cup (C \times D) = (A \cup B) \times (C \cup D)$?

I think I have managed to work this out, however there is no solution provided with the question so I thought it best to check.

My working:

Let $(x,y) \in (A \times B) \cup (C \times D)$

$\Rightarrow x \in (A$ or $C)$ and $y \in (B$ or $D)$

$\Rightarrow (x, y) \in (A \cup C) \times (B \cup D)$

$\Rightarrow (A \times B) \cup (C \times D) \subseteq (A \cup C) \times (B \cup D)$

Now let $(x,y) \in (A \cup C) \times (B \cup D)$

$\Rightarrow x \in (A$ or $C)$ and $y \in (B$ or $D)$

$\Rightarrow (x, y ) \in (A \times B) \cup (A \times D) \cup (C \times B) \cup (C \times D)$

$\Rightarrow (A \cup C) \times (B \cup D) \nsubseteq (A \times B) \cup (C \times D)$

$\Rightarrow (A \times B) \cup (C \times D) \not= (A \cup C) \times (B \cup D)$

Is this correct?

Thank you.

• i suggest you to check what you have done for another time (even the title)
– user87543
Jul 28, 2014 at 12:42
• What you try to prove in the body is more sensible that what you wrote in the title (switched $B$ and $C$ on the right-hand side). But (as noted) even that is wrong in general. Jul 29, 2014 at 15:52

No.

Suppose $B = D = \emptyset$ and $A \neq \emptyset$ and $C \neq \emptyset$.

To see if something is true or not, it is always a good idea to check on trivial examples. For example:

• Is $[0,1]\times[0,1]\cup [1,2]\times[1,2]$ equal to $[0,2]\times[0,2]$?

Even better, just take a set with a small number of elements, for example:

• Is $\{x\}\times\{u\} \cup \{y\}\times\{v\}$ equal to $\{x,y\}\times \{u,v\}$?

The first part was good: you did indeed show that $$(x,y) \in (A \times B) \cup (C \times D) \implies (x,y) \in (A \cup C) \times (B \cup D)$$ which is the definition of $(A \times B) \cup (C \times D) \subseteq (A \cup C) \times (B \cup D)$.

The second part was not good. If you want to show that an inclusion does not hold, you need to produce a counterexample. All you have done is showing $(A \cup C) \times (B \cup D) \subseteq (A \times B) \cup (A \times D) \cup (C \times B) \cup (C \times D)$.

So if you want a counterexample, can you find an element of $(A \cup C) \times (B \cup D)$ which is neither in $A \times B$ nor in $C \times D$?

Usually if we want to show some set equality is incorrect, we can show this with a counterexample. You can think about what's going on here with the following example, and draw a picture: take $A=C=[0,1], B=D=[1,2].$ $A\times B=[0,1]\times [1,2],$ $C\times D=[0,1]\times [1,2].$ If you draw their union, it's a square with unit side length. But $(A\cup B)\times (C\cup D)$ is a square of side length two, namely $[0,2]^2$.

You're right that $(A\times B)\cup(C\times D)\subseteq(A\cup C)\times(B\cup D)$, but your proof is slightly lacking.

Instead of "$x\in(A\text{ or }C)$" you should write "$x\in A$ or $x\in C$" -- natural language allows more freedom in where you can put "or" and expect to be understood, but once you start using symbols you're expected to be more rigorous.

Then it looks like everything of interest actually happens in the step from $(x,y)\in(A\times B)\cup(C\times D)$ to ($x\in A$ or $x\in C$) and ($y\in B$ or $y\in D$). Where you should go, doing things step by step is to ($x\in A$ and $y\in B$) or ($x\in C$ and $y\in D$).

You're also right that the opposite inclusion doesn't always hold, but instead of a series of $\Rightarrow$-connected formulas you should just show a counterexample.

The way you're writing it there looks like you're claiming that $(A\cup C)\times(B\cup D)\not\subseteq(A\times B)\cup(C\times D)$ necessarily holds, which isn't the case. For example, with $A=B=C=D=\{1\}$ equality does hold (both sides are $\{\langle 1,1\rangle\}$).