Exercise 2.3 (De Morgan's Laws)

Question

Let $(A_n)_{n\geq 1}$ be a sequence of sets. Show that
\begin{enumerate}
\item $\left(\bigcap_{n=1}^\infty A_n\right)^c = \bigcup_{n=1}^\infty A_n^c$
\item $\left(\bigcup_{n=1}^\infty A_n\right)^c = \bigcap_{n=1}^\infty A_n^c$.
\end{enumerate}

Toğrul Topal · Accepted Answer

extbf{First law.} \begin{itemize}
\item[$\supseteq$] If $a \in \bigcup_{n=1}^\infty A_n^c$, then $a\in A_n^c$ for some $n\in\mathbb{N}$. In particular, $a
ot \in \bigcap_{n=1}^\infty A_n$, so we must have $a\in (\bigcap_{n=1}^\infty A_n )^c$. This shows that $\bigcup_{n=1}^\infty A_n^c\subseteq (\cap_{n=1}^\infty A_n )^c$. 
\item[$\subseteq$] Pick an arbitrary $a\in(\bigcap_{n\in\mathbb{N}} A_n)^c$. By the laws of negation in logic, we see that there is no $n\in\mathbb{N}$ such that $a \in A_n$. In other words, for all $n\in\mathbb{N}$ we have $a 
ot\in A_n$. Therefore, $a\in\bigcup_{n\in\mathbb{N}} A_n^c$.
\end{itemize}

ewline

extbf{Second law.} Follows similarly.

Exercise 2.3 (De Morgan's Laws)

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Exercise 2.3 (De Morgan's Laws)

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