Here is my attempt at a solution for a proof about disjoint unions of sets $A$ and $B$. Can you please point out the mistakes? Thank you all.

Let $A$ and $B$ be any set. Prove:

  1. $A$ is the disjoint union of $A\setminus B$ and $A\cap B$
  2. $A\cup B$ is the disjoint union of $A\setminus B$, $A\cap B$, and $B\setminus A$.

Part 1

Suppose $A$ and $B$ are not disjoint. Then,

$$ \{x | x\in A\setminus B\text{ and }x\in A\cap B\} $$

Since $A\subseteq A\setminus B$ and $A\subseteq A\cap B$, $A\setminus B = A\cap B$. So, $A\setminus B\nsubseteq B$, or $A\setminus B\subseteq B^{C}$, and $A\cap B\subseteq B$.

Then, $B\cup B^{C} = \emptyset$, and $A$ and $B$ are disjoint.

Part 2

Suppose $A\cup B$ is not disjoint. Then,

$$ \{x | x\in A\setminus B\text{ and }x\in A\cap B\text{ and }x\in B\setminus A\} $$

$A\setminus B\subseteq B^{C}$, $A\cap B\subseteq A$, $A\cap B\subseteq B$, and $B\setminus A\subseteq A^{C}$. But since $B\cup B^{C} = \emptyset$ and $A\cap A^{C} = \emptyset$, $A\cup B$ is disjoint.

  • $\begingroup$ Each of these questions has two parts. You need to show that the sets on the right are disjoint (which you are attempting to do here) and show that their union is the set on the left. $\endgroup$ Sep 25 '11 at 18:16

In the first one, $A$ is not a subset of $A\setminus B$, but rather the other way around, that is $A\setminus B\subseteq A$ (Consider $A=\{0,1,2\}$ and $B=\{1\}$ as a counterexample to your statement).

Also $B\cap B^c=\varnothing$, and rather $B\cup B^c$ is everything.

You need to argue, however, $x\in A\setminus B$ then $x\notin B$, therefore $x\notin A\cap B$; and vice versa (that is $x\in A\cap B$ then $x\notin A\setminus B$). Then you need to show that $x\in A$ then either $x\in A\cap B$ or $x\in A\setminus B$ (which really boils down to the fact that either $x\in B$ or $x\notin B$).

In the second one the argument is completely unclear to me. Using the first part you can write $A=(A\setminus B)\cup (A\cap B)$ as a disjoint union, as well to apply the same argument on $B = (B\setminus A)\cup (B\cap A)$.

Now use the fact that $A\setminus B$ and $B\setminus A$ are disjoint to prove that the decomposition of $A\cup B=(A\setminus B)\cup(B\setminus A)\cup (A\cap B)$ is a disjoint union.

Lastly (after the $\LaTeX$ was fixed by cardinal) note that:

$$A\cup B=\{x\mid x\in A\ \mathbf{or}\ x\in B\}$$

While you wrote that this is "$x\in A\setminus B$ and $x\in A\cap B$ and $x\in B\setminus A$" which would be the intersection, which you can prove is empty.


You are using proof by contradiction unnecessarily. The disjointness arguments are much clearer when done directly. For example, $A \cap B \subseteq B$, and so $A \cap B$ must be disjoint from $A \setminus B$.

For the union, pick any element from $x \in A$. We want to show that it is in either $A \cap B$ or $A \setminus B$. This is a good place to use contradiction. Suppose $x$ is not in one of the sets (say $A \cap B$), since if it is, we are already happy. Since $x$ is an element of $A$ but not an element of $A \cap B$, it follows that $x$ is not an element of $B$. So, $x \in A$ and $x \notin B$, which means precisely that $x \in A \setminus B$.


Your statements 1. and 2. are so obvious that any "formal proof" actually weakens their credibility.

Ad 1. Any element $x\in A$ either does not belong to $B$ or is an element of $B$ as well.

Ad 2. For any element of $x\in A\cup B$ exactly one of the following is true: $x$ belongs solely either to $A$ or to $B$, or to both of them.

  • 7
    $\begingroup$ I completely disagree. Especially in introductory courses it is one of the most important things for a beginner: to actually write formal proofs for "obvious" theorems, and it is also one of the harder things a beginner is facing. $\endgroup$
    – Asaf Karagila
    Sep 25 '11 at 20:15
  • $\begingroup$ @Asaf Karagila: The beginner has to learn to write correct proofs for statements that are not obvious. These proofs should require playing around with the given and the definitions, but should not require ideas out of the blue. $\endgroup$ Sep 27 '11 at 14:16
  • 4
    $\begingroup$ No one said that the ideas should come out of the blue. I recall the hardest question in my freshman year was to prove from the axioms of the field that $x+0=x$. Not because it wasn't "obvious" but because it wasn't clear what was there the prove and how to approach it. The formal proof, however, made it clear that this is not obvious, and this requires proof. I think that this was one of the stepping stones in my studies when I realized that whatever is not given by definitions is not obvious. $\endgroup$
    – Asaf Karagila
    Sep 27 '11 at 14:24

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