I can't seem to wrap my head around these two formulas. I don't know whether it's that I don't understand the union and intersection of elements of gamma being put into set A. Or either that I don't understand how the distributive property works in these kind of problems. If anyone could please help it would be greatly appreciated, thanks.
De Morgan’s formulas set theory
elementary-set-theory
Best Answer
Elements of $\Gamma$ aren’t being put into sets $A$: the sets $A$ in the expressions are elements of $\Gamma$: $\Gamma$ is simply a collection of sets.
It may help to work through the reasoning showing that one of these identities is true, illustrating it with a small example. Let’s take the first one,
$$X\setminus\bigcup_{A\in\Gamma}A=\bigcap_{A\in\Gamma}(X\setminus A)\;.\tag{1}$$
The lefthand side is the set of all members of $X$ that are not in any of the sets in the collection $\Gamma$.
Suppose that $x$ is such a member of $X$, i.e., that $x\in X\setminus\bigcup_{A\in\Gamma}A$; then $x$ is not in any of the members of $\Gamma$, so if $A\in\Gamma$, we know that $x\notin A$. But $x\in X$, so $x\in X\setminus A$. That’s true for every $A\in\Gamma$, so $x$ is in all of the sets $X\setminus A$ and is therefore in their intersection: $$x\in\bigcap_{A\in\Gamma}(X\setminus A)\;.$$
This shows that in general
$$X\setminus\bigcup_{A\in\Gamma}A\subseteq\bigcap_{A\in\Gamma}(X\setminus A)\;.$$
Now suppose that $x\in\bigcap_{A\in\Gamma}(X\setminus A)$. This just says that for every set $A$ in the collection $\Gamma$, $x\in X\setminus A$: that is, $x$ is in $X$ but not in $A$. This clearly implies that $x\in X$. It also implies that $x$ is not in the union of all of the sets $A$ in the collection $\Gamma$, since it’s not in any of those sets. That is, $x\in X$, but $x\notin\bigcup_{A\in\Gamma}A$, so $x\in X\setminus\bigcup_{A\in\Gamma}A$.
This shows that in general
$$\bigcap_{A\in\Gamma}(X\setminus A)\subseteq X\setminus\bigcup_{A\in\Gamma}A\;,$$
which completes the proof of $(1)$.
You might find it helpful to draw a Venn diagram for the small example that I used.