[Math] Proof on disjoint union of sets $A$ and $B$

elementary-set-theory

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.

Best Answer

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.

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