Would someone be able to help me understand the (concept of) "union of intersections" and the "intersection of unions". More specifically, how to approach (prove) the following equality of two sets $X$ and $Y$;
$$ X=\bigcup_{n=1}^{\infty}\bigcap^\infty_{j=n} A_{j} \text{ and } Y=\bigcap^\infty_{n=1}\bigcup_{j=n}^{\infty} A_{j} $$
Your feedback is much appreciated as always.
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$\begingroup$As pointed out in the comments, the sets are not necessarily equal. Consider the following counterexample :
For even $n$, define $A_n = \{0\}$ and for odd $n$, define $A_n = \{1\}$.
That is, $A_1 = A_3 = A_5 = \cdots = \{1\}$ and $A_2 = A_4 = A_6 = \cdots = \{0\}$.
Let us look at the set $X$ first.
Note that for any given $n \in \mathbb{N}$, the set $\displaystyle\bigcap_{j=n}^\infty A_j$ is empty. This can be easily observed if you notice that $A_n \cap A_{n+1} = \varnothing$. (Why?)
Thus, $X = \displaystyle\bigcup_{n=1}^\infty\varnothing = \varnothing$.
Now, let us look at $Y$.
Note that for any given $n \in \mathbb{N}$, the set $\displaystyle\bigcup_{j=n}^\infty A_j$ equals $\{0, 1\}$.
This is also easy to prove. Note that $A_j \subset \{0, 1\}$ for each $j \ge n$ and hence, $\displaystyle\bigcup_{j=n}^\infty A_j \subset \{0, 1\}$.
To see that the equality follows, note that $A_n \cup A_{n+1} = \{0, 1\}$. (Why?)
Thus, we get that $X \neq Y$.
The next natural question to ask would be whether $X \subset Y$ is true in general?
To see this, one could first try to observe what sort of elements are there in $X$.
Observe that there is a big union outside. Thus, if $x \in X$, then $x \in \displaystyle\bigcap_{j=n_0}^\infty A_j$ for some $n_0 \in \mathbb{N}$.
This then gives us that $x \in A_j$ for every $j \ge n_0$.
Thus, in a more intuitive phrasing: $x$ must belong to every set after a point.
Let us now see if this is enough for $x$ to belong to $Y$ as well.
Note that there is a big intersection on the outside in the definition of $Y$. Thus, for $x$ to belong to $Y$, $x$ must belong to $\displaystyle\bigcup_{j=n}^\infty A_j$ for every $n \in \mathbb{N}$. Does our given $x$ have this property?
Well, yes, it does! Given any $n \in \mathbb{N}$, know that $x$ belongs to $A_{\max\{n, n_0\}}$. (How?)
Thus, $x$ will belong to $\displaystyle\bigcup_{j=n}^\infty A_j$, no matter the $n$. (How?)
In turn, this gives us that $x \in \displaystyle\bigcap_{n=1}^\infty\bigcup_{j=n}^\infty A_j = Y.$
So, what we have shown is that if $x \in X$, then $x \in Y$. This is precisely what it means for $X \subset Y$ and thus, we are done.
Some more intuition: If you try to characterise the elements that belong to $Y$, you'll observe that $y \in Y$ iff $y$ belongs to infinitely many $A_i$s.
However, as we saw, the condition that an element belongs to $X$ was stricter, it required that $x$ must belong to all $A_i$s after a point. That is, $x$ must belong to all but finitely many $A_i$s.
This goes well with the sort of counterexample that we saw, to begin with.
Suppose $x \in X$. Then $x \in \displaystyle\bigcap_{j=n}^\infty A_j $ for at least one $n = 1, 2, 3, \ldots$. Thus there exists a natural number $n = N$ such that $x \in A_j$ for every natural number $j \geq N$, which implies that $x \in \displaystyle\bigcup_{j=n}^\infty A_j$ for every $n = 1, 2, 3, \ldots$, and so $x \in Y$. Thus $X \subset Y$.
Conversely, suppose that $y \in Y$. Then $y \in \displaystyle\bigcup_{j = n}^\infty A_j$ for every $n = 1, 2, 3, \ldots$. Let $n \in \mathbb{N}$ be arbitrary. Then $y \in \displaystyle\bigcup_{j = n}^\infty A_j$, and so $y \in A_j$ for at least one natural number $j \geq n$. Thus for every natural number $n$, there exists a natural number $m \geq n$ such that $y \in A_m$.
As has been shown in the other answer. We cannot prove the inclusion $Y \subset X$ in general.
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