# Limit of increasing sequence of sets

## Definition

Given an increasing sequence of sets [ilmath](A_n)_{n=1}^\infty[/ilmath], we define the limit of the sequence as follows:

• $\lim_{n\rightarrow\infty}(A_n)=A$ where [ilmath]A[/ilmath] is the limit of [ilmath]A_n[/ilmath], and $A:=\bigcup_{n=1}^\infty A_n$

This may be written as:

• [ilmath]A_n\uparrow A[/ilmath]
• I do not like this notation, as [ilmath]\uparrow[/ilmath] only shows the notion of increasing, I prefer [ilmath]\nearrow[/ilmath] as this 'combines' (in a very vector-like sense) the [ilmath]\rightarrow[/ilmath] of limit and [ilmath]\uparrow[/ilmath] of increasing.
• [ilmath]A_n\nearrow A[/ilmath]
• I prefer this notation, however I always explicitly write $\lim_{n\rightarrow\infty}(A_n)=A$ myself, after letting [ilmath](A_n)_{n=1}^\infty[/ilmath] be an increasing sequence.

## Alec's definition

When I encountered this in a book() I didn't read on and formulated myself the definition of [ilmath]\lim_{n\rightarrow\infty}(A_n)=A[/ilmath] if:

• $\forall x\in A\exists N\in\mathbb{N}\forall n\in\mathbb{N}[n> N\implies x\in A_n]$ and
• $\forall n\in\mathbb{N}[A_n\subseteq A]$

Notice the first one alone is insufficient as any subset of some [ilmath]A_n[/ilmath] will satisfy it, so I formulated the second. The first also contains the increasing sequence idea as it requires after a certain index all sets contain [ilmath]x[/ilmath].

Proof that $\lim_{n\rightarrow\infty}(A_n)=A:=\bigcup_{i=1}^\infty A_n\iff$ those two conditions.

TODO: Be bothered, I've done this on some anonymous notepad, it isn't difficult