Why does $\Sigma=\{A:|A|<\infty\;\text{or}\,|A^c|<\infty\}$ not a sigma-algebra. I don't see too much difference between this two sets
1. $$\Sigma=\{A:|A|<\infty\;\text{or}\,|A^c|<\infty\}$$
2 $$\Sigma=\{A:\text{either}\,A\,\text{or}\,A^c\,\text{is finite or countable}\}$$
Why does the first one is not a sigma-algebra, but the second one is?
My attempt: I can prove $\emptyset,\Omega\in\Sigma$ for both cases, and they are both closed under complement.  Whenever $A\in\Sigma,A=(A^c)^c\in\Sigma$. I think the only differences only occur for closed under countable union, but I cannot go further.
Thanks for any help.

If support set is finite, then it is a $\sigma$-algebra.
Otherwise, assume $x_i$, $i \in \mathbb N$ are different elements of support.
Take $A_i = \{x_{2i}\}$. Then each $A_i \in \Sigma$, but $\cup_{i=1}^\infty A_i \notin \Sigma$ because $A$ is infinite and also $A^c$ contains infinite set $\{x_{2i + 1} | i \in \mathbb N\}$, so it's infinite too.
