Krdytkyu

Let $X$ be a not empty set.

I must prove that the family $mathcal{A}$ defined as follows
$$mathcal{A}=bigg{Ainmathcal{P}(X);|;A;text{is finite or $A^c$ is finite}bigg}$$ 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 ${x_n}_{ninmathbb{N}}$ of elements of $X$ such that $x_ine x_j$ for $ine j$ and let $A_n$ be the set consisting of the single point $x_n$, in symbols $A_n={x_n}$ for all $ninmathbb{N}$. At this point I consider the sets $A_2,A_4dots ,A_{2k},dots$ where $kinmathbb{N}$; I observe that $A_{2k}inmathcal{A}$ for all $kinmathbb{N}$, but the union $$bigcup_{k=1}^{+infty}A_{2k}=bigg{x_2,x_4,dots,x_{2k},dotsbigg}$$
is infinite. On the other hand
begin{equation}
begin{split}
bigg[bigcup_{k=1}^{+infty}A_{2k}bigg]^c=Xsetminusbigcup_{k=1}^{+infty}A_{2k}=& Xcapbigg[bigcup_{k=1}^{+infty}A_{2k}bigg]^{c}=Xcapbigg[bigcup_{k=1}^{+infty}A_{2k+1}bigg]\
=&bigcup_{k=1}^{+infty}A_{2k+1}cuptilde{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 $$bigcup_{n=1}^{+infty} A_nnotinmathcal{A}.$$

it's correct? Thanks!

Your proof is correct.

A bit shorter:

Let $X$ be infinite.

Then we can write it as $X=Ycup Z$ where:

• 
  $Y$ and $Z$ are disjoint.


• 
  $Y$ is countably infinite.


• 
  $Z$ is infinite.

Then $Ynotinmathcal A$, but also $Y=bigcup_{yin Y}{y}$ 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.