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Homelimit of sequence of sets

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# limit of sequence of sets

Recall that $\limsup$ and $\liminf$ of a sequence of sets $\{A_{i}\}$ denote the limit superior and the limit inferior of $\{A_{i}\}$, respectively. Please click here to see the definitions and here to see the specialized definitions when they are applied to the real numbers.

Theorem. Let $\{A_{i}\}$ be a sequence of sets with $i\in\mathbb{Z}^{{+}}=\{1,2,\ldots\}$. Then

1. for $I$ ranging over all infinite subsets of $\mathbb{Z}^{{+}}$,

$\limsup A_{i}=\bigcup_{{I}}\bigcap_{{i\in I}}A_{i},$ 2. 3. $\liminf A_{i}\subseteq\limsup A_{i}$.

Proof.

1. We need to show, for $I$ ranging over all infinite subsets of $\mathbb{Z}^{{+}}$,

$\displaystyle\bigcup_{{I}}\bigcap_{{i\in I}}A_{i}=\bigcap_{{n=1}}^{\infty}% \bigcup_{{i=n}}^{\infty}A_{k}.$ (1) Let $x$ be an element of the LHS, the left hand side of Equation (1). Then $x\in\bigcap_{{i\in I}}A_{i}$ for some infinite subset $I\subseteq\mathbb{Z}^{{+}}$. Certainly, $x\in\bigcup_{{i=1}}^{{\infty}}A_{i}$. Now, suppose $x\in\bigcup_{{i=k}}^{{\infty}}A_{i}$. Since $I$ is infinite, we can find an $l\in I$ such that $l>k$. Being a member of $I$, we have that $x\in A_{l}\subseteq\bigcup_{{i=k+1}}^{{\infty}}A_{i}$. By induction, we have $x\in\bigcup_{{i=n}}^{{\infty}}A_{i}$ for all $n\in\mathbb{Z}^{{+}}$. Thus $x$ is an element of the RHS. This proves one side of the inclusion ($\subseteq$) in (1).

To show the other inclusion, let $x$ be an element of the RHS. So $x\in\bigcup_{{i=n}}^{{\infty}}A_{i}$ for all $n\in\mathbb{Z}^{{+}}$ In $\bigcup_{{i=1}}^{{\infty}}A_{i}$, pick the least element $n_{0}$ such that $x\in A_{{n_{0}}}$. Next, in $\bigcup_{{i=n_{0}+1}}^{{\infty}}A_{i}$, pick the least $n_{1}$ such that $x\in A_{{n_{1}}}$. Then the set $I=\{n_{0},n_{1},\ldots\}$ fulfills the requirement $x\in\bigcap_{{i\in I}}A_{i}$, showing the other inclusion ($\supseteq$).

2. Here we have to show, for $I$ ranging over all subsets of $\mathbb{Z}^{{+}}$ with $\mathbb{Z}^{{+}}-I$ finite,

$\displaystyle\bigcup_{{I}}\bigcap_{{i\in I}}A_{i}=\bigcup_{{n=1}}^{\infty}% \bigcap_{{i=n}}^{\infty}A_{k}.$ (2) Suppose first that $x$ is an element of the LHS so that $x\in\bigcap_{{i\in I}}A_{i}$ for some $I$ with $\mathbb{Z}^{{+}}-I$ finite. Let $n_{0}$ be a upper bound of the finite set $\mathbb{Z}^{{+}}-I$ such that for any $n\in\mathbb{Z}^{{+}}-I$, $n<n_{0}$. This means that any $m\geq n_{0}$, we have $m\in I$. Therefore, $x\in\bigcap_{{i=n_{0}}}^{{\infty}}A_{i}$ and $x$ is an element of the RHS.

Next, suppose $x$ is an element of the RHS so that $x\in\bigcap_{{k=n}}^{\infty}A_{k}$ for some $n$. Then the set $I=\{n_{0},n_{0}+1,\ldots\}$ is a subset of $\mathbb{Z}^{{+}}$ with finite complement that does the job for the LHS.

3. The set of all subsets (of $\mathbb{Z}^{{+}}$) with finite complement is a subset of the set of all infinite subsets. The third assertion is now clear from the previous two propositions. QED

Corollary. If $\{A_{i}\}$ is a decreasing sequence of sets, then

$\liminf A_{i}=\limsup A_{i}=\lim A_{i}=\bigcap A_{i}.$ |

Similarly, if $\{A_{i}\}$ is an increasing sequence of sets, then

$\liminf A_{i}=\limsup A_{i}=\lim A_{i}=\bigcup A_{i}.$ |

Proof. We shall only show the case when we have a descending chain of sets, since the other case is completely analogous. Let $A_{1}\supseteq A_{2}\supseteq\ldots$ be a descending chain of sets. Set $A=\bigcap_{{i=1}}^{\infty}A_{i}$. We shall show that

$\limsup A_{i}=\liminf A_{i}=\lim A_{i}=A.$ |

First, by the definition of limit superior of a sequence of sets:

$\limsup A_{i}=\bigcap_{{n=1}}^{\infty}\bigcup_{{i=n}}^{\infty}A_{k}=\bigcap_{{% n=1}}^{\infty}A_{n}=A.$ |

Now, by Assertion 3 of the above Theorem, $\liminf A_{i}\subseteq\limsup A_{i}=A$, so we only need to show that $A\subseteq\liminf A_{i}$. But this is immediate from the definition of $A$, being the intersection of all $A_{i}$ with subscripts $i$ taking on all values of $\mathbb{Z}^{{+}}$. Its complement is the empty set, clearly finite. Having shown both the existence and equality of the limit superior and limit inferior of the $A_{i}$’s, we conclude that the limit of $A_{i}$’s exist as well and it is equal to $A$. QED

## Mathematics Subject Classification

03E20*no label found*28A05

*no label found*60A99

*no label found*

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