Reference- "Knowing the Odds: An introduction to probability" by John B. Walsh 2011 edition. Ex-1.1
Show that the following class is a field, but not $\sigma$-field-
All finite subsets of $\mathbb{R}$ together with their complements
My arguments- Let class be $\mathcal{C}$. Since the there are finite subsets in $\mathcal{C}$, let $(-\infty, 1] \in \mathcal{C}$. Therefore, $(1, -\infty) \in \mathcal{C}$ because complements also exist.
Since it's not given as a condition, I think I have the right not to include $\Phi$ and $(-\infty, \infty)$. So, I can't even see $ \mathcal{C}$ as a field too, as union will produce $(-\infty, \infty) \notin \mathcal{C}$. [I must have been wrong here with something very elementary]
Even if I consider $\Phi, (-\infty, \infty) \in \mathcal{C}$, Taking the countable unions of all elements in $\mathcal{C}$ (union of all 4 elements), I will again get $\mathbb{R} \in \mathcal{C}$ which shows that countable union is in $\mathcal{C}$.
More generally, for any interval $I \in \mathcal{C}, I^C \in \mathcal{C}$. Considering $\mathbb{R}, \Phi \in \mathcal{C}$, $I\cup I^C = \mathbb{R} \in \mathcal{C}$. Now no matter hom many finite intervals are there in $\mathcal{C}$, their union will be $\mathbb{R} \in \mathcal{C}$ and hence contians all countable unions. Hence, it should be a $\sigma$-field.
I might have some impreciseness about the concepts and arguments. Kindly help me in correcting those.
A $\sigma$-algebra should be closed under countable unions. Does $$ \bigcup_{n=0}^{\infty}\{n\} $$ belong to the class?