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I know that supremum means least upper bound. If I have a sequence of events, $\{A_n\}_{n=1}^\infty$

then $$\limsup_{n\rightarrow \infty} A_n = \lim_{n\rightarrow \infty} \sup_{j\geq n} A_j$$

I'm having trouble understanding this statement:

"The supremum of a collection of elements in a partially ordered set is its least upper bound, so $\sup_{j\geq n} A_j$ should be a set and it should hold that $A_j \subset \sup_{j\geq n} A_j$ for all $j \geq n $. Because the supremum should also be the smallest upper bound it is not hard to see that $$\sup_{j\geq n} A_j= \bigcup_{j=n}^\infty A_j$$"

Why does $j \geq n $ matter? Also, I don't understand what set is partially ordered, and also why the supremum of $A_j$ is itself a set. Shouldn't it just be an element? And I also don't see how the supremum is the same thing as the union.

user130306
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2 Answers2

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I think you are conflating two similarly-notated notions. Given a set $A$ with an ordering, one might write $$\sup A$$ to denote the supremum of the set $A$.

Above, when one writes $$\sup_{j \ge n} A_j$$ it denotes something more like $$\sup \{A_j, A_{j+1}, A_{j+2}, \ldots, \}$$ which is the supremum of the collection of events $A_j, A_{j+1}, \ldots$, not something like "the supremum of the set $\bigcup_{j \ge n} A_j$." Of course, in order to talk about the supremum of something like $\{A_j, A_{j+1}, A_{j+2}, \ldots, \}$ (whose elements are themselves events/sets), you need an ordering of the elements. Here, we consider the partial order of set containment. So the least upper bound should be the smallest [in the partial order] set larger than any $A_j, A_{j+1}, \ldots$, which in terms of the set containment partial order, is $$\sup_{j \ge n} A_n \text{ is the smallest set such that } A_j \subset \sup_{j \ge n} A_j.$$

angryavian
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Probability events are (the measurable) subsets of the probability space $\Omega$, and the partial order among them is simply the relation 'being subset of'. So, for any countable set of events $B_i$, their supremum is just $\bigcup_i B_i$ (which is still measurable).

By the way, we also have $$\limsup_n A_n\ =\ \lim_n\sup_{j\ge n} A_j\ =\ \bigcap_n\bigcup_{j\ge n} A_j$$

Berci
  • 90,745