The mechanics of the union-intersection along a sequence of sets

elementary-set-theorylimsup-and-liminf

This question came about after I got stuck trying to understand this answer.

When $n\to \infty$ I don't understand why in a countable family of sets $(A_n)_{n\geq 0}$ the operation

$$\bigcap_{n\geq 0} \bigcup_{k \geq n} A_k $$

depends on whether the sequence is monotonous or not.

It is likely that I don't understand the symbols, or that I am just too dumb. Let's see… Should I read this as a composition of operations? If so, when I start from the inside, i.e.

$$\bigcup_{k \geq n} A_k$$

I necessarily end up with all the elements in all the sets combined into a single set. At that point this entire set containing all possible elements in any one of the sets $A_k$ is intersect-ioned with all of the sets $A_k,$ looking for elements that are in all sets. Is that it? If this is correct, then why does the order of the monotony of the sequence matter? Can I get an example?

Post-mortem:

'
For a finite collection of sets:

$$
\begin{align}
A_0&=\{a,b\}\\
A_1&=\{a,x\}\\
A_3&=\{a,x,y\}\\
A_4&=\{a,c,d,y\}\\[3ex]
\bigcap_{n\geq 0} \bigcup_{k \geq n} A_k &=\{a,b,x,y,c,d\}\cap\{a,x,y,c,d\}\cap\{a,x,y,c,d\}\\
&=\{a,x,y,c,d\}\\[3ex]
\bigcup_{n\geq 0} \bigcap_{k \geq n} A_k
&=\{a\}\cup\{a\}\cup\{a,y\}=\{a,y\}
\end{align}
$$

In pseudo-code $\bigcap_{n\geq 0} \bigcup_{k \geq n} A_k $:

for $n$ in $n=0$ to $\infty:$

for $k$ in $n$ to $\infty:$

$B_k \leftarrow A_k \cup A_{k+1} \cup A_{k+2} \cdots$

$B_0 \cap B_1 \cap B_2 \cap \cdots$

The pictorial intuition is that for an element $\color{blue}\spadesuit$ to be in $\bigcap_{n\geq 0} \bigcup_{k \geq n} A_k $ it would have to keep on appearing so that no matter how far we go into infinity it appears in some subset to as to be part of all $B_k$'s, and hence, in the intersection:

$$\{\cdots\},\cdots,\{\cdots\},\{\cdots\},\{\cdots,\color{blue}\spadesuit,\cdots\},\{\cdots\},\{\color{blue}\spadesuit,\cdots\},\cdots,\{\cdots\},\{\cdots,\color{blue}\spadesuit,\cdots\},\cdots$$

On the other hand, $\bigcup_{n\geq 0} \bigcap_{k \geq n} A_k $ is more demanding, requiring that $\color{blue}\spadesuit$ appears consistently after a certain point:

$$\{\cdots\},\cdots,\{\cdots\},\{\cdots\},\{\cdots,\color{blue}\spadesuit,\cdots\},\{\cdots,\color{blue}\spadesuit\},\{\color{blue}\spadesuit,\cdots\},\cdots,\{\color{blue}\spadesuit,\cdots\},\{\cdots,\color{blue}\spadesuit,\cdots\},\cdots$$

Best Answer

Let me suggest a way of thinking about

$$\bigcap_{n\ge 0}\bigcup_{k\ge n} A_k$$

and

$$\bigcup_{n\ge 0}\bigcap_{k\ge n} A_k$$

that may be helpful.

How does something get into either of these sets?

  • $x\in\bigcap\limits_{n\ge 0}\bigcup\limits_{k\ge n} A_k$ if and only if $x\in\bigcup\limits_{k\ge n}A_k$ for each $n\ge 0$, which is the case if and only if for each $n\ge 0$ there is a $k\ge n$ such that $x\in A_k$. And this is the same as saying that $x$ is in infinitely many of the sets $A_k$: no matter how far out in the sequence of sets you go, there is one further out that contains $x$.
  • $x\in\bigcup\limits_{n\ge 0}\bigcap\limits_{k\ge n}A_k$ if and only if there is an $n\ge 0$ such that $x\in\bigcap\limits_{k\ge n}A_k$, which is the case if and only if there is an $n\ge 0$ such that $x\in A_k$ for every $k\ge n$. This just says that $x$ is in every $A_k$ from some point on: it is in all but finitely many of the sets $A_k$.

To sum up:

  • $x\in\bigcap\limits_{n\ge 0}\bigcup\limits_{k\ge n} A_k$ if and only if $x$ belongs to infinitely many of the sets $A_k$, and
  • $x\in\bigcup\limits_{n\ge 0}\bigcap\limits_{k\ge n} A_k$ if and only if $x$ is in all but finitely many of the sets $A_k$ or, equivalently, $x$ is in every $A_k$ from some point on.
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