Show that $\left(\bigcup_{i \in I}A_{i}\right )^{c}=\bigcap_{i \in I}A_{i}^{c}$ I'm learning for a test and I think there will be some task similar to this one. I'd like to know if I did it correct and if not can you please say how to do it correctly?

Let $\Omega$ be a set, let $I$ be an index-set and let $A \subseteq
\Omega$ and $A_{i} \subseteq \Omega$ for all $i \in I$.
Show that $$\left(\bigcup_{i \in I}A_{i}\right )^{c}=\bigcap_{i \in
I}{A_{i}}^{c}$$

We defined this in our readings:
Complement: $A^{c} \equiv \bar{A}= \left\{x \in \Omega; (x \notin A)\right\}$
and $$\overline{\bigcup_{i \in I}A_{i}}=\bigcap_{i \in
I}\overline{A_{i}}$$
I did it like this:
$$\left(\bigcup_{i \in I}A_{i}\right)^{c}=\overline{\bigcup_{i \in I}A_{i}}=\bigcap_{i \in
I}\overline{A_{i}}=\bigcup_{i \in I}{A_{i}}^{c}$$
 A: Here is a basic beginner's guide on how to solve this question.
So you want to show $(\cup_{i\in I} A_{i})^{c} = \cap_{i \in I} A_{i}^{c}$.
Notice that the thing on either side of the $=$ is a set.  So you are trying to show two sets are equal.  If you want to show $B = C$ for sets $B$ and $C$, the standard way to approach this is to show $B \subseteq C$ and $C \subseteq B$.  If you show they are subsets of each other, then that's good enough to show they are equal.
Okay, so we need to show then two statements:


*

*$(\cup_{i\in I} A_{i})^{c} \subseteq \cap_{i \in I} A_{i}^{c}$

*$\cap_{i \in I} A_{i}^{c} \subseteq (\cup_{i\in I} A_{i})^{c}$.
I'll show you how to do 1, and you will have to do 2 yourself, but the way to do 2 is very similar to the way to do 1 so if you understand what I'm about to say, then you'll be able to do 2.  Feel free to ask me questions in the comments about your attempt.
Okay, so I want to prove $(\cup_{i\in I} A_{i})^{c} \subseteq \cap_{i \in I} A_{i}^{c}$.  To show this, I will need to let $x$ be any element of the left hand side $(\cup_{i\in I} A_{i})^{c}$, and I will need to show it is in the right hand side $\cap_{i \in I} A_{i}^{c}$.
Alright, then let $x \in (\cup_{i\in I} A_{i})^{c}$ be any arbitrary element.  To show it is in $\cap_{i \in I} A_{i}^{c}$, let's start by inspecting what it means for $x$ to be in $(\cup_{i\in I} A_{i})^{c}$.
Since $x \in (\cup_{i\in I} A_{i})^{c}$, that means $x$ is not in $\cup_{i\in I} A_{i}$.  Okay, but if $x$ is not in that union, then that means it is not in any $A_{i}$ for any $i$ (since if it was on one of them, it would be in the union).  But if $x$ is not in $A_{i}$ for any $i$, then that means $x$ is in $A_{i}^{c}$ for every $i$.  But if $x \in A_{i}^{c}$ for every $i$, then that means $x \in \cap_{i \in I} A_{i}^{c}$.  But this is what we wanted to show!  So we are done with this part.
Now you have to prove statement 2.  If you have any questions, feel free to ask me in the comments.
A: Ok, let's make things a little more clear: you have a set $\Omega$ and a family of subsets of $\Omega$ indexed by $I$: $\{A_i\}_{i\in I}$.
You want to prove that $\left(\bigcup_{i\in I}A_i\right)^c=\bigcap_{i\in I}A_i^c$.
You don't actually need any other $A\subseteq \Omega$, I think...
Let's introduce some notation: $U:=\bigcup_{i\in I}A_i$, $V:=\bigcap_{i\in I}A_i^c$. Thus, the claim is that $U^c=W$.
We say that $x\in U^c$ iff $x\notin U$, i.e., if $x\notin A_i$, for every $i\in I$ (from the fact that $y\in U$ if there is some $i_0\in I$ - not nec. unique - such that $y\in A_{i_0}$; recall then how quantifiers behave under negation). This means exactly that $x\in A_i^c$ for every $i\in I$, thus, by definition of the intersection, you get $x\in V$. Hence: $U\subseteq V$. Every implication is actually an equivalence (an iff), thus $V\subseteq U$, too. 
A: By definition we have $$\bigcup_{i \in I}A_{i}=\{x\in\Omega:\exists i\in I(x\in A_i)\}$$
and $$\bigcap_{i \in I}{A_{i}^c}=\{x\in\Omega:\forall i\in I(x\notin A_i)\}$$
By the definition of equality of sets we have $$\left(\bigcup_{i \in I}A_{i}\right )^{c}=\bigcap_{i \in I}{A_{i}}^{c}$$ if and only if $$\left(\bigcup_{i \in I}A_{i}\right )^{c}\subseteq\bigcap_{i \in
I}{A_{i}}^{c} \text{ and }\bigcap_{i \in
I}{A_{i}}^{c}\subseteq\left(\bigcup_{i \in I}A_{i}\right )^{c}$$


*

*Let $x\in\left(\bigcup_{i \in I}A_{i}\right )^{c}$. By the defintion of compliment of a set we have $x\notin \bigcup_{i \in I}A_{i}$. Since $x\notin \bigcup_{i \in I}A_{i}$, by the definition of $\bigcup_{i \in I}A_{i}$ and the axiom of specification we have $x\in\Omega$ and $x\notin A_i$ for all $i\in I$. Since $x\in\Omega$ and $x\notin A_i$ for any $i\in I$, we have $x\in\Omega$ and $x\in A_i^c$ for all $i\in I$. Therefore by the definition of $\bigcap_{i \in
I}{A_{i}}^{c}$ we have $x\in\bigcap_{i \in
I}{A_{i}}^{c}$. We thus have $x\in\bigcap_{i \in
I}{A_{i}}^{c}$ whenever $x\in\left(\bigcup_{i \in I}A_{i}\right )^{c}$. Hence $\left(\bigcup_{i \in I}A_{i}\right )^{c}\subseteq\bigcap_{i \in
I}{A_{i}}^{c}$.

*Now suppose conversely that we have $x\in\bigcap_{i \in
I}{A_{i}}^{c}$. By the definition of $\bigcap_{i \in
I}{A_{i}}^{c}$ we have $x\in\Omega$ and $x\in A_i^c$ for all $i\in I$. In other words, we have $x\in \Omega$ and $x\notin A_i$ for any $i\in I$. Again using the axiom of specification and the definition of $\bigcup_{i \in I}A_{i}$ we have $x\notin\bigcup_{i \in I}A_{i}$. Since $x\notin\bigcup_{i \in I}A_{i}$, we have $x\in\left(\bigcup_{i \in I}A_{i}\right )^{c}$. Since $x\in\left(\bigcup_{i \in I}A_{i}\right )^{c}$ whenever $x\in\bigcap_{i \in I}{A_{i}}^{c}$, we have $\bigcap_{i \in
I}{A_{i}}^{c}\subseteq\left(\bigcup_{i \in I}A_{i}\right )^{c}$.


Since $\left(\displaystyle\bigcup_{i \in I}A_{i}\right )^{c}\subseteq\displaystyle\bigcap_{i \in
I}{A_{i}}^{c}$ and $\displaystyle\bigcap_{i \in
I}{A_{i}}^{c}\subseteq\left(\displaystyle\bigcup_{i \in I}A_{i}\right )^{c}$, we have $\left(\displaystyle\bigcup_{i \in I}A_{i}\right )^{c}=\displaystyle\bigcap_{i \in I}{A_{i}}^{c}$.
A: What you did is correct. But since you asked in a comment how to prove the result without using the proposition provided by your professor (which does all the work for you!), you can do it as follows. You should prove that the elements of $\left(\bigcup_{i \in I}A_{i}\right )^{c}$ are precisely the elements of $\bigcap_{i \in I}{A_{i}}^{c}$, because in that case the sets are equal. So you can start by saying suppose that $x\in \left(\bigcup_{i \in I}A_{i}\right )^{c}$, and then you will have to show that $x$ is also an element of $\bigcap_{i \in
I}{A_{i}}^{c}$. And after that you have the do the same in the opposite direction. I'll show how to do one direction. Then you can try the other one yourself.
So we start by saying suppose that $x\in \left(\bigcup_{i \in I}A_{i}\right )^{c}$. This means by definition that $x\neq \bigcup_{i \in I}A_{i}$. We have, also by definition, that
$$\bigcup_{i \in I}A_{i} = \{x\in \Omega:x\in A_i\text{ for some }i\in I\}$$ and since $x$ is not an element of this set, there can be no $A_i$ that contains $x$ (for otherwise $x$ would be an element of the set). Therefore, for all $i$ we have that $x\notin A_i$, which, by definition, means that $x\in A_i{}^c$ for all $i$. Now, since, again by definition,
$$\bigcap_{i \in I}{A_{i}}^{c} = \{x\in\Omega : x\notin A_i \text{ for all } i\in I\}$$
it follows that $x\in \bigcap_{i \in I}{A_{i}}^{c}$.
So we have now proved that all elements of $\left(\bigcup_{i \in I}A_{i}\right )^{c}$ are also elements of $\bigcap_{i \in I}{A_{i}}^{c}$. If you can now show that the opposite statement also holds, then you have proven that the sets are equal. This second part of the proof is very similar to the part that I have shown you, so it would be a good exercise to try it yourself.
