Prove that in every algebra the union of any chain of subalgebras is a subalgebra.
Let we have an $\left \langle A, \Omega \right \rangle$ and a chain $A_1 \subset A_2 \subset ... \subset A_n$ of subalgebras. This means that $\exists x \in A_2: x \notin A_1$ and both $A_1$ and $A_2$ is closed under operation in $\Omega$.
For example: $4\mathbb{Z} \subset 2\mathbb{Z}$ and from that we easy see $4\mathbb{Z} \cup 2\mathbb{Z} = 2\mathbb{Z}$. And If we have $3\mathbb{Z} \nsubseteq 2\mathbb{Z}$ so $3*2 = 6 \notin 2\mathbb{Z} \cup 3\mathbb{Z}$
But I have no idea how to show it for any subalgebras with chain. Thanks for any help.
Probably you want to take an upwards-infinite chain, as $\bigcup\limits_{i=1}^n A_i = A_n$ is not an interesting case.
To show that the union $U = \bigcup\limits_{i=1}^\infty A_i$ is a subalgebra, you need to check that is is closed under operations of $\Omega$. Indeed, each element $u \in U$ lies in some $A_{k(u)}$, and the $n$-element set $\{u_1, \ldots, u_n\}$ lies in the subalgebra $A_{K}$ for $K = \max\limits_{i=1}^n k(u_i)$, and $A_K$ is invariant under all operations of $\Omega.$ See also direct limit.