Show that $S_1 \cup S_2 = \overline{\overline{S_1}\cap \overline{S_2}}$.
So we want to show that the union of $S_1$ and $S_2$ is equal to the compliment of the intersection of the compliments of $S_1$ and $S_2$. I hope that makes sense!
I created some finite sets with some random values. Here's what I have so far but I'm stuck!
$$\begin{align*}
S_1 &= \{1, 2, 3, 4\}\\
S_2 &= \{2, 3, 4, 5, 6\}\\
S_1 \cup S_2 &= \{1, 2, 3, 4, 5, 6\}\\
\overline{S_1} &= \{5, 6\} &\text{(everything in $S_2$ but not in $S_1$)}\\
\overline{S_2} &= \{1\} &\text{(everything in $S_1$ but not in $S_2$)}
\end{align*}$$
I don't see a way to intersect $\overline{S_1}$ and $\overline{S_2}$, can someone point me in the right direction please? Am I on the right track, or way off?
Best Answer
Let $x\in S_1\cup S_2$ $\Rightarrow$ $x\in S_1$ or $x\in S_1$ $\Rightarrow$ $x\notin S_1^c$ or $x\notin S_1^c$ $\Rightarrow$ $x\notin S_1^c\cap S_2^c$$\Rightarrow$ $x\in (S_1^c\cap S_2^c)^c$.
Hence $S_1\cup S_2\subset (S_1^c\cap S_2^c)^c$.
Converse inclusion has the same proof
P.S. $A^c$ it's a complement of set $A$.