# Prove if $A_1,A_2,...,A_N\in\mathcal{A}$ then $A_1\cap A_2\cap\ ...\ \cap A_N \in \mathcal{A}$

My definition of a $\sigma$-algebra i

A $\sigma$-algebra $\mathcal{A}$ on a set X is a familiy of subsets of X with the following properties

1. $X\in\mathcal{A}$

2. $A\in\mathcal{A}\Rightarrow A^c \in \mathcal{A}$

3. $(A_j)_{j\in \mathbb{N}}\subset\mathcal{A}\Rightarrow\bigcup_{j\in \mathbb{N}} A_j \in \mathcal{A}$

Let $\mathcal{A}$ be a $\sigma$-algebra. Prove if $A_1,A_2,...,A_N\in\mathcal{A}$ then $A_1\cap A_2\cap\ ...\ \cap A_N \in \mathcal{A}$

Now here is my approach

Since $A_1,A_2,...,A_N\in\mathcal{A}$ then $A_1^c,A_2^c,...,A_N^c\in\mathcal{A}$ by (1.), whereas $\bigcup_{j=1}^N A_j^c\in\mathcal{A}$ by (3.). By de Morgan we then get $(\bigcap_{j=1}^N A_j)^c\in\mathcal{A}$. Again by (2.) we finally have ${(\bigcap_{j=1}^N A_j)^c}^c = \bigcap_{j=1}^N A_j = A_1\cap A_2\cap\ ...\ \cap A_N \in \mathcal{A}$

Would this be valid? I’m uncertain about my second step, since I use (3.) as all the sets are subsets of $\mathcal{A}$, which my assertion doesn’t say anything about.

You are along the right kind of thinking, but you need to be little more careful. I think you have written correctly, but your logic is not visible. If it is the same as this answer then please ignore this answer.

$$$$\begin{split} \bigcap_{i=1}^N A_i & = \bigg(\bigcup_{i=1}^N A_i^c\bigg)^c\end{split}$$$$

So, we go along this logic:

1) Since $A_i$ are in the sigma algebra, so are their complements, namely $A_i^c$.

2) Since $A_i^c$ are in the sigma algebra, so is their union, namely $\bigg(\bigcup_{i=1}^N A_i^c\bigg)$.

3) Since $\bigg(\bigcup_{i=1}^N A_i^c\bigg)$ is in the sigma algebra, so is it's complement, namely $\bigcap_{i=1}^N A_i$.

Hence, we are done.