Show that $\bigcap_{n\in \mathbb{N^*}}\left(\bigcup_{k\geq n}A_k\right)=\bigcup_{n\in\mathbb{N^*}}A_n$ given that $A_k\subseteq A_{k+1}$

Question

Let $\mathbb{N^*}=\mathbb{N}\setminus \{0\}$

Given that $A_k\subseteq A_{k+1}$ $\forall k\in\mathbb{N^*}$

Show that

$$\bigcap_{n\in \mathbb{N^*}}\left(\bigcup_{k\geq n}A_k\right)=\bigcup_{n\in\mathbb{N^*}}A_n $$

How could I prove the equality?

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On part (a) of the same problem I've proven that $$\bigcup_{n\in \mathbb{N^*}}\left(\bigcap_{k\geq n}A_k\right)=\bigcup_{n\in\mathbb{N^*}}A_n $$ By proving that $\displaystyle A_n=\bigcap_{k\geq n}A_k$ (by induction on $k$ using $\displaystyle x\in A_k\implies x\in A_{k+1}$) so I was thinkin of doing something similar by letting $\displaystyle \bigcup_{k\geq n}A_k=B_n$ but I'm stuck.

Answer 1

Let $A$ the LHS and $B$ be the RHS. For each $n$, $$ \bigcup_{k\ge n} A_k \subset B $$ so that $$ A=\bigcap_n\bigcup_{k\ge n}A_k\subset B $$

Conversely, let $x\in B$. Then $x\in A_n$ for some $n$, so $x\in A_k$ for every $k\ge n$ by your assumption, so $x\in A$.