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I have a question for an assignment that involves this term.

$$x\in \bigcup_{k=1}^{\infty}\bigcap_{n=k}^{\infty}A_{n}.$$

I have a faint idea of what it means, but perhaps someone can tell me if I am wrong.

first im considering the intersection sign by itself, assuming that n=k=1 at first, meaning a family of sets A1...A(inf) all joined by intersection such that its only true (ie. not the null set) if there is some common element in all the sets.

then i do the same thing but starting with n=k=2 so that you have an identical family of sets with the exception of it not including A1. again X must be in all the sets.

then repeat so you have an infinite family of families of sets, all one set smaller than the last. and take the union of all of them.

I'm having trouble though seeing what the point of the union is, It makes sense if the union and intersection signs were switched because then they both impose a condition, but having them this way, the union doesn't seem to do anything, or rule anything out?

if there is something obvious that I'm missing Id greatly appreciate your help.

Martin Sleziak
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Lincoln77
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4 Answers4

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This set is called $\lim\inf A_n$.

To understand what it means, you have to use the definitions of the union and intersection. Let $$x\in\bigcup_{k=1}^{\infty}\bigcap_{n=k}^{\infty}A_n$$ By definition of the union, there exists an integer $k$ such that $$x\in\bigcap_{n=k}^{\infty}A_n$$ Then by definition of the intersection: $$\forall n\geq k,\, x\in A_n$$

So $$x\in\bigcup_{k=1}^{\infty}\bigcap_{n=k}^{\infty}A_n\iff\exists k\in\mathbb{N},\, \forall n\geq k,\, x\in A_n$$ This means exactly that $x$ belongs to all the $A_n$ except finitely many.

If you permute the intersection and the union, you get $\lim\sup A_n$ which is the set of all $x$ that belong to an infinity of $A_n$.

You can check that $\lim\inf A_n\subset\lim\sup A_n$.

Augustin
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Traditionally, $\bigcup_{k=1}^{\infty}\bigcap_{n=k}^{\infty}A_{n}$ is called $\textit{the inf limit}$ of $(A_n)_n$ and it is sometimes noted $\liminf_n A_n.$ If you want any "interpretration" of this construction, I can give you this $$\bigcup_{k=1}^{\infty}\bigcap_{n=k}^{\infty}A_{n} = \{x : x \in A_n \,\,\forall n, \text{excepted for a finite number of}\,\, n \}.$$

C. Dubussy
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The union most certainly can do something. For example, if $A_i=[-i,\infty)$, then the union $$\bigcup_{k=1}^\infty\bigcap_{n=k}^\infty A_n$$ is equal to $\mathbb R$, even though none of the intersections is equal to $\mathbb R$.

This is because $\bigcap_{n=k}^\infty A_n$ is equal to $[-k, \infty)$, and even though each of these sets is bounded from below, their union is not.

mrp
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5xum
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$\bigcup_{k=1}^{\infty}\bigcap_{n=k}^{\infty}A_{n}$ is also known as $\liminf A_n$.

Imagine that you have a sequence $U$ of $0$ and $1$. Let $A_n=\{U_n=1\}$. $A_n$ is the event the $n$-th of $U$ element is 1.

$\bigcap_{n=k}^{\infty}A_{n} \neq \emptyset$ does mean that your sequence if full of $1$ from rank $k$. While $\bigcup_{k=1}^{\infty}\bigcap_{n=k}^{\infty}A_{n} \neq \emptyset$ means that from a certain rank, we dont know exactly which of them, the sequence is full of $1$.

Oppositely $\bigcap_{n=k}^{\infty}A_{n} = \emptyset$ means that there is at least one $0$ somewhere is the sequence after rank $k$. And $\bigcup_{k=1}^{\infty}\bigcap_{n=k}^{\infty}A_{n}=\emptyset$ that there will never be a rank such that, from it, there will be only $1$.

So the union is here to say "somewhere it exists a rank such that..." or "somewhere it does not exists ...".

nicomezi
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