Question

Prove the following DeMorgan's laws: if LaTeX: XX, LaTeX: AA and LaTeX: BB are sets and LaTeX: \{A_i: i\in I\} {Ai:i∈I} is a family of sets, then
LaTeX: X-(A\cup B)=(X-A)\cap (X-B)

LaTeX: X-(\cup_{i\in I}A_i)=\cap_{i\in I}(X-A_i)

Answer 1

• $X-(A\cup B)=(X-A)\cap(X-B)$

I'll assume the usual definition of set difference, $X-A=\{x\in X,x\not\in A\}$.

Let $x\in X-(A\cup B)$. Then $x\in X$ and $x\not\in(A\cup B)$. If $x\not\in(A\cup B)$, then $x\not\in A$ and $x\not\in B$. This means $x\in X,x\not\in A$ and $x\in X,x\not\in B$, so it follows that $x\in(X-A)\cap(X-B)$. Hence $X-(A\cup B)\subset(X-A)\cap(X-B)$.

Now let $x\in(X-A)\cap(X-B)$. Then $x\in X-A$ and $x\in X-B$. By definition of set difference, $x\in X,x\not\in A$ and $x\in X,x\not\in B$. Since $x\not A,x\not\in B$, we have $x\not\in(A\cup B)$, and so $x\in X-(A\cup B)$. Hence $(X-A)\cap(X-B)\subset X-(A\cup B)$.

The two sets are subsets of one another, so they must be equal.

• $X-\left(\bigcup\limits_{i\in I}A_i\right)=\bigcap\limits_{i\in I}(X-A_i)$

The proof of this is the same as above, you just have to indicate that membership, of lack thereof, holds for all indices $i\in I$.

Proof of one direction for example:

Let $x\in X-\left(\bigcup\limits_{i\in I}A_i\right)$. Then $x\in X$ and $x\not\in\bigcup\limits_{i\in I}A_i$, which in turn means $x\not\in A_i$ for all $i\in I$. This means $x\in X,x\not\in A_{i_1}$, and $x\in X,x\not\in A_{i_2}$, and so on, where $\{i_1,i_2,\ldots\}\subset I$, for all $i\in I$. This means $x\in X-A_{i_1}$, and $x\in X-A_{i_2}$, and so on, so $x\in\bigcap\limits_{i\in I}(X-A_i)$. Hence $X-\left(\bigcup\limits_{i\in I}A_i\right)\subset\bigcap\limits_{i\in I}(X-A_i)$.