Showing that if $A_{1},A_{2},...$ are all algebras then the union of all of them is an algebra

Question

I am not sure how to show this. It seems obvious but maybe its not. The help would be greatly appreciated!

Answer 1

We need to verify the axioms of an algebra. These are all of the form

$$\tag1 \forall x_1, \ldots ,x_n\in \mathcal A\colon \exists y_1,\ldots, y_m\in\mathcal A\colon \Phi(x_1,\ldots,x_n,y_1,\ldots,y_m)$$ (Verify that this is indeed the case for algebra-axioms; also investigate some other structures if they have thist property)

Actually, any axiom of this form holds for a nested union if it holds for all terms. In fact, let $\mathcal A_1\subseteq \mathcal A_2\subseteq\ldots $ be sets such that axiom $(1)$ holds for $\mathcal A:=\mathcal A_k$, $k\in\mathbb N$. Then $(1)$ also holds for $\mathcal A:=\bigcup_{k=1}^\infty \mathcal A_k$. Indeed, given $x_1,\ldots x_n\in\mathcal A$, we find $k_1,\ldots,k_n$ such that $x_i\in\mathcal A_{k_i}$. Then let $k=\max\{k_1,\ldots,k_n\}$ and notice that $x_1,\ldots ,x_n\in\mathcal A_k$, hence $\exists y_1,\ldots ,y_m\in\mathcal A_k\subseteq \mathcal A$ such that $\Phi(x_1,\ldots,x_n,y_1,\ldots,y_m)$.