A family of sets that is not a $\sigma$-algebra

Question

Let $X$ be a not empty set.

I must prove that the family $\mathcal{A}$ defined as follows $$\mathcal{A}=\bigg\{A\in\mathcal{P}(X)\;|\;A\;\text{is finite or $A^c$ is finite}\bigg\}$$ it is not a $\sigma$-algebra if $X$ is infinite.

Obviously $\mathcal{A}$ is an algebra. To show that it is not a $\sigma$-algebra I reasoned as follows: fixed a sequence $\{x_n\}_{n\in\mathbb{N}}$ of elements of $X$ such that $x_i\ne x_j$ for $i\ne j$ and let $A_n$ be the set consisting of the single point $x_n$, in symbols $A_n=\{x_n\}$ for all $n\in\mathbb{N}$. At this point I consider the sets $A_2,A_4\dots ,A_{2k},\dots$ where $k\in\mathbb{N}$; I observe that $A_{2k}\in\mathcal{A}$ for all $k\in\mathbb{N}$, but the union $$\bigcup_{k=1}^{+\infty}A_{2k}=\bigg\{x_2,x_4,\dots,x_{2k},\dots\bigg\}$$ is infinite. On the other hand \begin{equation} \begin{split} \bigg[\bigcup_{k=1}^{+\infty}A_{2k}\bigg]^c=X\setminus\bigcup_{k=1}^{+\infty}A_{2k}=& X\cap\bigg[\bigcup_{k=1}^{+\infty}A_{2k}\bigg]^{c}=X\cap\bigg[\bigcup_{k=1}^{+\infty}A_{2k+1}\bigg]\\ =&\bigcup_{k=1}^{+\infty}A_{2k+1}\cup\tilde{A}, \end{split} \end{equation} where $\tilde{A}$ it is what remains if $X$ is not countable, if $X$ is countable $\tilde{A}=\emptyset$. Therefore $\cup_k A_{2k}$ and $[\cup_k A_{2k}]^c$ are both infinite, then $$\bigcup_{n=1}^{+\infty} A_n\notin\mathcal{A}.$$

it's correct? Thanks!

Answer 1

Your proof is correct.

A bit shorter:

Let $X$ be infinite.

Then we can write it as $X=Y\cup Z$ where:

• $Y$ and $Z$ are disjoint.
• $Y$ is countably infinite.
• $Z$ is infinite.

Then $Y\notin\mathcal A$, but also $Y=\bigcup_{y\in Y}\{y\}$ where the RHS is a countable union of sets that are elements of $\mathcal A$.

So $\mathcal A$ is not closed under countable unions, hence is not a $\sigma$-algebra.