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I am learning this wiki page, which uses sequence of sets in the definition

Suppose that ${\displaystyle \{A_{n}\}_{n=1}^{\infty }}$ is a sequence of sets. The two equivalent definitions are as follows.

Using union and intersection, define

$\liminf_{n \rightarrow \infty} A_n = \bigcup_{n \ge 1} \bigcap_{j \geq n} A_j$

...

The sequence ${A_n}$ is said to be nondecreasing if each $A_n ⊂ A_{n+1}$

the simplest example of a (monotonic increasing) sequence I can imagine is the Natural number $\{0, 1, 2, …\}$

I assume this $\{\{0\}, \{0, 1\}, ..., \{0, 1, 2, …\}\}$ is an nondecreasing sequence of sets.

limit infimum is defined as

$\liminf_{n \rightarrow \infty} A_n = \bigcup_{n \ge 1} \bigcap_{j \geq n} A_j$

to understand this easily, I would like to consider $\bigcap_{j \geq 1} A_j$ first

so, is $\bigcap_{j \geq 1} A_j$ equal to

$\{0\}$

or

$\{\{0\}\}$

I think it is the last one, and I need a double-check

JJJohn
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    You have a sequence ${ A_n }_n$ of sets, meaning that the $A_i$ are the elements of the sequence, that is itself a set. The intersection of all the sets of the sequence - if it is different from the empty set - will be the "smallest" set of the sequence, and thus will be an element of the sequence. – Mauro ALLEGRANZA Aug 01 '19 at 13:40
  • $\bigcap_{j\geq 1} A_j$ in this context is referring to ${0}\cap {0,1}\cap {0,1,2}\cap \cdots$ which will be ${0}$ – JMoravitz Aug 01 '19 at 13:42
  • Consider for simplicity the first two elements of the sequence : ${ 0 }$ and ${ 0,1 }$. The intersection of the two is the "usual" intersection, i.e. the set of the elements belonging to both : ${ 0 }$. – Mauro ALLEGRANZA Aug 01 '19 at 13:44
  • If $A_n={0,1,\dots,n}$ then $$\bigcap_{i\geq n} A_i = A_n.$$ That's true for any monotonic increasing sequence of sets. – Thomas Andrews Aug 01 '19 at 14:37

1 Answers1

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The $$\bigcap _{j\ge 1} A_j =\{0\}$$ because the only member which is common to all $A_j$ is $0$

Note that $\{0\}$ is not a member of $A_j$ so it is not a member of the intersection.

Thomas Andrews
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Mohammad Riazi-Kermani
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