Is there any particular way to approach "finding an example" questions in elementary set theory?

Question

How should one approach answering questions like:

"Find an example of sets [a different number of sets] such that [some statement involving those sets]."

Guess and trial based on your intuition of the whole statement is the only way? If so, what about more complex statements?

So I'll give you an example: Find an example of sets $A$, $B$, and $C$ such that $(A ∪ B) △ C \neq (A △ C) ∪ (B △ C)$.

What's your way of solving this problem (except for guessing, if there's any that I'm missing)?

One short note: If you're considering that my question is too general then please only answer the example above and the way that you approached to solve it (if possible).

Thanks in advance.

Answer 1

Guess and trial based on your intuition of the whole statement is the only way?

It's not the only way, but it's the way you should approach such problems.

The point of posing such problems as exercises is not that students should train some kind of mechanical procedure for finding examples -- that would be a mostly useless skill.

On the other hand, developing enough intuition about how sets and logic work is a useful skill, and doing so is what such exercises is meant to give you an opportunity for.

Do not squander this opportunity by trying to find a mindless way of solving them instead of developing the intuition that is the point of the exercise.

Answer 2

Following the idea of John Griffin in comments, let us try to "prove" that $(A ∪ B) △ C = (A △ C) ∪ (B △ C)$. At a certain point we will get stuck and the part of our "proof" that breaks down will suggest us a counterexample. We should prove that $(A ∪ B) △ C \subseteq (A △ C) ∪ (B △ C)$ and $(A ∪ B) △ C \supseteq (A △ C) ∪ (B △ C)$.

• "Proof" of $(A ∪ B) △ C \supseteq (A △ C) ∪ (B △ C)$: Let $x \in (A △ C) ∪ (B △ C)$. Either $x \in A △ C$ or $x \in B △ C$, suppose $x \in A △ C$. There are two cases:

  1. if $x \in A \smallsetminus C$, then $x \in A \subseteq A \cup B$ and $x \notin C$, thus $x \in (A \cup B) △ C$;
  2. if $x \in C \smallsetminus A$, then $x \in C$ and $x \notin A$. Now, either $x \in B$ or $x \notin B$ (tertium non datur) and both are possible since we haven't supposed anything about the relationship between $x$ and $B$ (i.e. $B$ plays no role in this case). If $x \notin B$ then $x \notin A \cup B$ and hence $x \in (A \cup B) △ C$. But if $x \in B$ then $x \in A \cup B$ and so $x \notin (A \cup B) △ C$. We got stuck! We can't prove that $(A ∪ B) △ C \supseteq (A △ C) ∪ (B △ C)$.

Summing up, trying to "prove" that $(A ∪ B) △ C \supseteq (A △ C) ∪ (B △ C)$, we've found that the statement is false when there is an $x \in B \cup C$ such that $x \notin A$. This suggests an example of sets $A$, $B$, and $C$ such that $(A∪B)△C≠(A△C)∪(B△C)$: take $B = \{1\} = C$ and $A = \emptyset$, so that $1 \in B \cup C$ and $1 \notin A$; then, $A △ C = \{1\} = A \cup B$ while $B △ C = \emptyset $, therefore $(A \cup B) △ C = \emptyset \neq \{1\} = (A △ C) \cup (B △ C)$.