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.


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?

Thanks in advance.

  • 1
    $\begingroup$ Make subject lunes informative, not quotes ofvthe result you want until you run out of space. $\endgroup$ Sep 23 at 21:22
  • $\begingroup$ Done; thanks for the suggestion $\endgroup$ Sep 23 at 21:27

1 Answer 1


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


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .