Suppose $\mathscr{A}=\{A_i:i\in\mathbb{N}\}$ is a family of sets such that for all $i,j\in\mathbb{N}$, if $i\le j$ then $A_j\subseteq A_i$. (Such a family is called a nested family of sets).
a) Prove for every $k\in\mathbb{N}$, $\bigcap_{i=1}^{k}A_i=A_k$
b) Prove that $\bigcup_{i=1}^{\infty}A_i=A_1$
Proof of a):
Suppose $x\in\bigcap_{i=1}^{k}A_i$
iff $x\in A_i$ for all $i\in\mathbb{N}$.
iff $x\in A_j$ since $A_i\cap A_j= A_j$, $A_j \subseteq A_i$ and
$A_i=A_j$ since $j\in\mathbb{N}$iff $x\in A_k$ for all $k\in\mathbb{N}$, since $A_k=A_j\cap A_k$,
$A_k\subseteq A_j$ and $k\in\mathbb{N}$.Proof of b)
Suppose $x\in\bigcup_{i=1}^{\infty}A_i$. Then $x\in A_1$ since it is
contained in $\bigcup_{i=1}^{\infty}A_i$. Hence
$\bigcup_{i=1}^{\infty}A_i\subseteq A_1$.Now suppose $x\in A_1$ since $1\le i \le j$ for all
$i,j\in\mathbb{N}$, it follows $1\le i \le j$ for some
$i,j\in\mathbb{N}$. Since $A_j\subseteq A_i\subseteq A_1$, it follows
$x\in A_i$ for some $i\in\mathbb{N}$. Hence $x\in
> \bigcup\limits_{i\in\mathbb{N}}A_i$. Hence
$x\in\bigcup_{i=1}^{\infty}A_i$.Hence $\bigcup_{i=1}^{\infty}A_i=A_1$
Best Answer
I’m afraid that neither argument makes sense. In the first argument it is not true that $x\in\bigcap_{i=1}^kA_i$ iff $x\in A_i$ for all $i\in\Bbb N$. For example, let $A_1=\{0\}$, and let $A_i=\varnothing$ for all $i\ge 2$; then $\{A_i:i\in\Bbb N\}$ is a nest. Now take $k=1$: then
$$\bigcap_{i=1}^kA_i=\bigcap_{i=1}^1A_i=A_1=\{0\}\,,$$
so $0\in\bigcap_{i=1}^kA_i$, but clearly $0\notin A_2$ (and in fact $0\notin A_i$ whenever $i\ge 2$).
In the next line you talk about some $A_j$, but you never define it, so nothing that you say about it is meaningful. At the end you conclude that $x\in A_k$ for all $k\in\Bbb N$; this is just a repetition of what you said in the second line, but with a different name for the subscript, and it is every bit as false in general.
In the argument for (b) you start with an $x\in\bigcup_{i=1}^\infty A_i$ and conclude that it must be in $A_1$, since $A_1\subseteq\bigcup_{i=1}^\infty A_i$. Your reasoning here is exactly backwards: the fact that $A_1\subseteq\bigcup_{i=1}^\infty A_I$ tells you that if $x\in A_1$, then $x\in\bigcup_{i=1}^\infty A_i$, not the reverse.
Then when you try to prove the trivial implication that if $x\in A_i$, then $x\in\bigcup_{i=1}^\infty A_i$, you pick some unspecified positive integers $i$ and $j$ with $i\le j$ and claim that the fact that $A_j\subseteq A_i\subseteq A_1$ somehow proves that $x\in A_i$ for some $i\in\Bbb N$. This is a complete non sequitur, and in any case we already know that $x\in A_i$ for some $i\in\Bbb N$, since we assumed at the beginning that $x\in A_1$, and certainly $1\in\Bbb N$.
To prove (a), first observe that if $x\in\bigcap_{i=1}^kA_i$, then $x\in A_k$, so $\bigcap_{i=1}^kA_i\subseteq A_k$. Now suppose that $x\in A_k$, and let $i\in\{1,\ldots,k\}$; then $i\le k$, so $A_k\subseteq A_i$, and therefore $x\in A_i$. Thus, $x\in A_i$ for all $i\in\{1,\ldots,k\}$, and hence $x\in\bigcap_{i=1}^kA_i$. In other words, $A_k\subseteq\bigcap_{i=1}^kA_i$, and it follows that $\bigcap_{i=1}^kA_i=A_k$.
For (b) it’s immediate that if $x\in A_1$, then $x\in\bigcup_{i=1}^\infty A_i$, so $A_1\subseteq\bigcup_{i=1}^\infty A_i$. Now suppose that $x\in\bigcup_{i=1}^\infty A_i$; then there is some $i\in\Bbb N$ such that $x\in A_i$. And $i\ge 1$, so $A_i\subseteq A_1$, so $x\in A_1$. Thus, $\bigcup_{i=1}^\infty A_i\subseteq A_1$, and we conclude that $\bigcup_{i=1}^\infty A_i=A_1$.