# Algebra generated by the set of singletons of an infinite countable set

Let $$X=\{x_i\}_{i=1}^{\infty}$$ be a coutable infinite set, $$C=\{\{x\}|x\in X\}$$, and $$\mathcal{A}(C)$$ the algebra of sets generated by $$C$$, prove $$\mathcal{A}(C)=\{A\subseteq X|\text{ A is finite } \lor \text{ A' is finite } \}$$.

I'm having trouble with the second half of the proof.

Let $$C^*=\{A\subseteq X|\text{ A is finite } \lor \text{ A' is finite } \}$$

• $$C^*\subseteq \mathcal{A}(C)$$:

Let $$A\in C^*$$, then either $$A=\{a_i\}_{i=1}^{m}=\bigcup_{i=1}^m\{a_i\}$$ such that $$a_i\in X$$ or $$A'=\{a'_i\}_{i=1}^{n}=\bigcup_{i=1}^n\{a'_i\}$$ such that $$a'_i\in X$$.

But $$\mathcal{A}(C)$$ is the smallest algebra in $$X$$ that contains C, this is that contains $$\{x_1\}$$, $$\{x_2\}$$, $$\{x_3\},...$$

so $$\{a_1\}$$, ..., $$\{a_m\}\in \mathcal{A}(C)$$ and $$\{a'_1\}$$, ..., $$\{a'_n\}\in \mathcal{A}(C)$$, but $$\mathcal{A}(C)$$ is an algebra, so it's closed under finite unions and $$A=\{a_i\}_{i=1}^{m}=\bigcup_{i=1}^m\{a_i\}\in\mathcal{A}(C)$$ or $$A'=\{a'_i\}_{i=1}^{n}=\bigcup_{i=1}^n\{a'_i\}\in\mathcal{A}(C)$$.

If $$A\in\mathcal{A}(C)$$ we're done; if $$A'\in\mathcal{A}(C)$$, then $$A\in\mathcal{A}(C)$$ because of Algebra axioms.

$$\blacksquare$$

Now, I'm missing $$\mathcal{A}(C)\subseteq C^*$$. I know that $$C\subseteq C^*$$ so i thought about proving that $$C^*$$ is an algebra of sets too, that way I'd have that both $$\mathcal{A}(C)$$ and $$C^*$$ are algebras and both contain $$C$$, thus $$\mathcal{A}(C)\subseteq C^*$$ given that $$\mathcal{A}(C)$$ is the smallest such algebra. However, I'm having trouble proving $$C^*$$ is closed under finite union, more particularly, in the case when I have $$A, B\in C^*$$ and $$A$$ is finite and $$B'$$ is finite.

Any ideas?

Note $$(A \cup B)'$$ is a subset of $$B'$$. Thus, if $$B'$$ is finite, then $$(A \cup B)'$$ is also finite.