Mathematics Asked by hk1510 on November 1, 2021
$renewcommand{backslash}{setminus}$
The question is as is in the title. I am able to show that $A setminus (A setminus (A setminus B)) subseteq A setminus B$ but I am stuck on showing that $A setminus B subseteq A setminus (A setminus (A setminus B))$.
Am I doing something wrong? Any suggestions will help.
Here is what I have so far:
$textbf{Proof:}$
To show that $A backslash (A backslash (A backslash B)) = A backslash B$, we need to show that $A backslash (A backslash (A backslash B)) subseteq A backslash B$ and $A backslash B subseteq A backslash (A backslash (A backslash B))$.
We first show that $A backslash (A backslash (A backslash B)) subseteq A backslash B$.
Let $a in A backslash (A backslash (A backslash B))$.
$implies a in A cap (A backslash (A backslash B))^c$
$implies a in A cap (A cap (A backslash B)^c)^c$
$implies a in A cap (A cap (A cap B^c)^c)^c$
$implies a in A cap (A^c cup (A cap B^c))$
$implies a in A cap ((A^c cup A) cap (A^c cup B^c))$
$implies a in A cap (U cap (A^c cup B^c))$
$implies a in A cap (A^c cup B^c)$
$implies a in (A cap A^c) cup (A cap B^c)$
$implies a in emptyset cup (A cap B^c)$
$implies a in (A cap B^c)$
$implies a in (A backslash B)$
Next, we show that $A backslash B subseteq A backslash (A backslash (A backslash B))$.
Let $a in A backslash B$.
$implies a in A backslash (A backslash B)^c$
$implies a in A backslash (A backslash (A backslash B)^c)^c$
$implies a in A backslash (A cap (A backslash B))^c$
$implies a in A backslash (A^c cup (A backslash B)^c)$
$implies a in A backslash (A^c cup (A^c cup B))$
Note that you can reverse the whole thing, i.e. if in the first half you simply change every $Rightarrow$ into a $Leftrightarrow$, you're there!
In fact, with algebra it's pretty simple:
$$A setminus (A setminus (A setminus B)) = $$
$$A cap (A cap (A cap B^c)^c)^c =$$
$$ A cap (A^c cup (A cap B^c) =$$
$$ (A cap A^c) cup (A cap A cap B^c) =$$
$$ emptyset cup (A cap B^c) =$$
$$ Asetminus B$$
Answered by Bram28 on November 1, 2021
Too long for a comment, but you could simplify notation quite a bit by writing $C = A setminus B$ and showing that $A setminus (A setminus C) = C$ for any set $C subset A$.
Note that if $D subset A$, then $x in D$ if and only if $x notin A setminus D$, and $x notin D$ if and only if $x in A setminus D$
So, let $C subset A$. Then $$x in C iff x notin A setminus C iff x in A setminus (A setminus C)$$ by applying the preceding remark to $D = C$ and $D = A setminus C$, respectively.
Answered by Umberto P. on November 1, 2021
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