Proving that ${\bigcup_{n=1}^{\infty} A_n}={\bigcup_{n=1}^{\infty} B_n}$

Question

I'm trying to give my interest more into set theory and I found one problem that seems to be interesting for me. Can anyone show me the solution to this problem?

$$\text{ {$A_i$}} - \text{infinitely many subsets. Let }B_n={\bigcup_{i=1}^{n} A_i}. \text{Prove that }{\bigcup_{n=1}^{\infty} A_n}={\bigcup_{n=1}^{\infty} B_n} $$

Answer 1

$$A_n\subset B_n=\bigcup_{i=1}^{n} A_i\implies \bigcup_{n=1}^{\infty}A_n\subset \bigcup_{n=1}^{\infty}B_n. $$.

Now conversely, if $x\in \bigcup_{n=1}^{\infty}B_n $ then there is $n_0$ such that, $$x\in B_{n_0} = \bigcup_{i=1}^{n_0}A_i$$

Hence there is $1\le i_0\le n_0\le n$ such that

$$x\in A_{i_0} \subset \bigcup_{n=1}^{\infty}A_n$$

Therefore, $$\bigcup_{n=1}^{\infty}A_n\supset \bigcup_{n=1}^{\infty}B_n. $$.

Answer 2

Suppose $x \in \bigcup_{i=1}^{\infty} A_i$. Then certainly $x \in A_n$ for some $n$, and it's immediate that $x \in \bigcup_{i=1}^n A_i = B_n$, so that $x \in \bigcup_{i=1}^{\infty} B_i$, giving $\bigcup_{i=1}^{\infty} A_i \subseteq \bigcup_{i=1}^{\infty} B_i$.

Conversely, suppose $x \in \bigcup_{i=1}^{\infty} B_i$. Then $x \in B_n$ for some $n$, or $x \in \bigcup_{i=1}^n A_i$. Thus, we have $x \in A_k$ for some $k$ between $1$ and $n$ inclusive, whence $x \in \bigcup_{i=1}^{\infty} A_i$. As a result, $\bigcup_{i=1}^{\infty} B_i \subseteq \bigcup_{i=1}^{\infty} A_i$, proving equality.