If two sequences of sets have the same finite unions, then the infinite unions are also the same I am stuck in the proof of the following: 

If $\bigcup_{n=1}^k A_n = \bigcup_{n=1}^k B_n$ for every $k$, then $\bigcup_{n=1}^\infty A_n = \bigcup_{n=1}^\infty B_n$.

I can intuitively think this is true, because $k$ is arbitrary. But is there any formal way to prove this?  
 A: So if I read you correctly, you want to prove that $$\bigcup_{n=0}^\infty A_n = \bigcup_{n=0}^\infty B_n$$ Let's prove that $\bigcup_{n=0}^\infty A_n \subset \bigcup_{n=0}^\infty B_n$. The converse inclusion will hold by symmetry.
Take $x \in \bigcup_{n=0}^\infty A_n$ which mean that there exists $k$ integer such that $x \in A_k$. Hence $$x \in \bigcup_{n=0}^k A_n = \bigcup_{n=0}^k B_n$$ by hypothesis. So there is $n \le k$ such that $x \in B_n$ Finally $x \in \bigcup_{n=0}^\infty B_n$ which allows to conclude.
A: I would not say that it is trivial, and you definitely can't just say that because something's true for all $n$ that it must also be true in the limit. Instead, I would look at using the fact that when it comes to sets, $A = B$ is equivalent to $x \in A \iff x \in B$. So, looking at one side of things:
Consider an arbitrary $x \in \bigcup_{n=1}^\infty A_n$. Then $\exists k \in \mathbb{N}$ such that $x \in A_k$, and hence $x \in \bigcup_{n=1}^k A_n = \bigcup_{n=1}^k B_n$. And you can then use that to show that $x \in \bigcup_{n=1}^\infty B_n$. So then all you have to do is show that that works the other way, and hence the two sets are equal.
