Let $(X_i\mid i\in I)$ be a family of subsets of $X$. $$\bigcup_{i\in I}(X\setminus X_i)=X\setminus\bigcap_{i\in I} X_i$$ $$\bigcap_{i\in I}(X\setminus X_i)=X\setminus\bigcup_{i\in I} X_i$$
My attempt:
To prove these formulas, I show that for any $x\in X$, $x$ either belongs to sets in both sides, or belongs to neither of them.
- $\bigcup_{i\in I}(X\setminus X_i)=X\setminus\bigcap_{i\in I} X_i$
For any $x\in X$, there are only two cases.
a. $\forall i\in I$, $x\in X_i$
Then $x\notin X\setminus X_i\space\forall i\in I$ and $x\in \bigcap_{i\in I} X_i$. It follows that $x\notin \bigcup_{i\in I}(X\setminus X_i)$ and $x\notin X\setminus\bigcap_{i\in I} X_i$.
b. $\exists i'\in I$, $x\notin X_{i'}$
Then $x\in X\setminus X_{i}'$ and $x\notin\bigcap_{i\in I} X_i$. It follows that $x\in \bigcup_{i\in I}(X\setminus X_i)$ and $x\in X\setminus\bigcap_{i\in I} X_i$. $$\tag*{$\blacksquare$}$$
- $\bigcap_{i\in I}(X\setminus X_i)=X\setminus\bigcup_{i\in I} X_i$
Approach 1:
For any $x\in X$, there are only two cases.
a. $\forall i\in I$, $x\notin X_i$
Then $x\in X\setminus X_i\space\forall i\in I$ and $x\notin \bigcup_{i\in I} X_i$. It follows that $x\in \bigcap_{i\in I}(X\setminus X_i)$ and $x\in X\setminus\bigcup_{i\in I} X_i$.
b. $\exists i'\in I$, $x\in X_{i'}$
Then $x\notin X\setminus X_{i}'$ and $x\in\bigcup_{i\in I} X_i$. It follows that $x\notin \bigcap_{i\in I}(X\setminus X_i)$ and $x\notin X\setminus\bigcup_{i\in I} X_i$. $$\tag*{$\blacksquare$}$$
Approach 2:
Let $Y:=X$ and $Y_i:=X\setminus X_i$ for all $i\in I$. Then $(Y_i\mid i\in I)$ be a family of subsets of $Y$.
We have $\bigcup_{i\in I}(Y\setminus Y_i)=Y\setminus\bigcap_{i\in I} Y_i$ from 1.
\begin{align}\bigcup_{i\in I}(Y\setminus Y_i)&=Y\setminus\bigcap_{i\in I} Y_i\\\iff Y\setminus \bigcup_{i\in I}(Y\setminus Y_i)
&= Y\setminus (Y\setminus\bigcap_{i\in I} Y_i)\\\iff Y\setminus \bigcup_{i\in I}(Y\setminus Y_i)
&=\bigcap_{i\in I} Y_i \\\iff X\setminus \bigcup_{i\in I}(X\setminus (X\setminus X_i))
&=\bigcap_{i\in I} (X\setminus X_i)\\\iff X\setminus \bigcup_{i\in I} X_i
&=\bigcap_{i\in I} (X\setminus X_i)\end{align} $$\tag*{$\blacksquare$}$$
Does this proof look fine or contain gaps? Do you have suggestions? Many thanks for your dedicated help!
Best Answer
Direct proof.
x in X - $\cap_i$X$_i$ iff x not in $\cap_i$X$_i$ iff
some i with x not in X$_i$ iff some i with x in X - X$_i$
iff x in $\cup_i$(X - X$_i$).
That is the set version of the logical theorem:
not (for all x, P(x)) iff exists x with not P(x).
The second equation is a direct result of the first.
Simply exchange X$_i$ and X - X$_i$ and
take the complement of both sides.