Proof of Basic Properties of the Limit Inferior and Limit Superior of a Bounded Real Sequence

Throughout write $A_{k}=\{a_{m}:m\ge k\}$, $\beta_{k}=\inf A_{k}$ and $\gamma_{k}=\sup A_{k}$, so that $\ell:=\liminf_{n}a_{n}=\sup\{\beta_{k}:k\in\mathbb{N}\}$ and $L:=\limsup_{n}a_{n}=\inf\{\gamma_{k}:k\in\mathbb{N}\}$ by Limit Inferior of a Bounded Sequence of Real Numbers and Limit Superior of a Bounded Sequence of Real Numbers. Since $a_{k}\in A_{k}$ we have $\beta_{k}\le a_{k}\le\gamma_{k}$. Order manipulations in $\mathbb{R}$ use Elementary Order Arithmetic in an Ordered Field, and claim numbers for the order on $\mathbb{N}$ refer to Properties of the Order on the Natural Numbers.

Monotonicity of the tails. If $k\le k'$ then $A_{k'}\subseteq A_{k}$, since $m\ge k'$ and $k'\ge k$ give $m\ge k$ by transitivity (claim 1). Hence $\beta_{k}$ is a lower bound of $A_{k'}$ and $\gamma_{k}$ is an upper bound of $A_{k'}$, so $\beta_{k}\le\beta_{k'}$ and $\gamma_{k'}\le\gamma_{k}$ by the defining properties of the infimum and the supremum.

Claim 1. Let $k,k'\in\mathbb{N}$. By trichotomy (claim 3) one of $k,k'$ is greater than or equal to the other; call it $k''$, so $k\le k''$ and $k'\le k''$. Then

Thus $\gamma_{k'}$ is an upper bound of $\{\beta_{k}:k\in\mathbb{N}\}$, so $\ell\le\gamma_{k'}$ for every $k'$; that is, $\ell$ is a lower bound of $\{\gamma_{k'}:k'\in\mathbb{N}\}$, whence $\ell\le L$.

Claim 2. Assume $a_{n}\le b_{n}$ for every $n$, and write $B_{k}=\{b_{m}:m\ge k\}$ with $\beta'_{k}=\inf B_{k}$ and $\gamma'_{k}=\sup B_{k}$.

Fix $k$. For every $m\ge k$ we have $\beta_{k}\le a_{m}\le b_{m}$, so $\beta_{k}$ is a lower bound of $B_{k}$ and therefore $\beta_{k}\le\beta'_{k}\le\liminf_{n}b_{n}$, the last inequality because $\beta'_{k}$ belongs to the set whose supremum is $\liminf_{n}b_{n}$. Hence $\liminf_{n}b_{n}$ is an upper bound of $\{\beta_{k}:k\in\mathbb{N}\}$ and $\ell\le\liminf_{n}b_{n}$.

Similarly, for every $m\ge k$ we have $a_{m}\le b_{m}\le\gamma'_{k}$, so $\gamma'_{k}$ is an upper bound of $A_{k}$ and $L\le\gamma_{k}\le\gamma'_{k}$, the first inequality because $\gamma_{k}$ belongs to the set whose infimum is $L$. Hence $L$ is a lower bound of $\{\gamma'_{k}:k\in\mathbb{N}\}$ and $L\le\limsup_{n}b_{n}$.

For the two special cases, note that for the constant sequence with value $M$ every tail set is $\{M\}$, whose infimum and supremum are both $M$, so its limit inferior and limit superior are both $M$. If $a_{n}\le M$ for every $n$, comparison with that constant sequence gives $L\le M$; if $M\le a_{n}$ for every $n$, comparison in the other direction gives $M\le\ell$.

Claim 3. Let $\varepsilon>0$ be real. By claim 3 of Approximation Property of the Supremum and the Infimum in $\mathbb{R}$, applied to the set $\{\beta_{k}:k\in\mathbb{N}\}$, whose supremum is $\ell$, there is $k$ with $\ell-\varepsilon<\beta_{k}$. For every $m\ge k$ we have $\beta_{k}\le a_{m}$, hence $\ell-\varepsilon<a_{m}$; take $N=k$. Dually, by claim 4 of that lemma, applied to the set $\{\gamma_{k}:k\in\mathbb{N}\}$, whose infimum is $L$, there is $k'$ with $\gamma_{k'}<L+\varepsilon$, and for every $m\ge k'$ we have $a_{m}\le\gamma_{k'}<L+\varepsilon$; take $N'=k'$.

Claim 4. Let $\varepsilon>0$ be real.

Every tail contains an index with $a_{m}<\ell+\varepsilon$. Fix $k\in\mathbb{N}$. Since $\beta_{k}$ belongs to the set whose supremum is $\ell$, we have $\beta_{k}\le\ell$. By claim 4 of Approximation Property of the Supremum and the Infimum in $\mathbb{R}$, applied to $A_{k}$, whose infimum is $\beta_{k}$, there is an element of $A_{k}$, say $a_{m}$ with $m\ge k$, satisfying $a_{m}<\beta_{k}+\varepsilon\le\ell+\varepsilon$.

Let $P=\{n\in\mathbb{N}:a_{n}<\ell+\varepsilon\}$. Taking $k=1$ above shows $P\ne\emptyset$. Taking $k=n+1$ shows that for every $n\in P$ there is $n'\in P$ with $n+1\le n'$; then $n<n'$, because either $n'=n+1$ and $n<n+1$ by claim 6, or $n+1<n'$ and $n<n+1<n'$ gives $n<n'$ by transitivity (claim 1). So the relation on $P$ that links $n$ to $n'$ when $n'>n$ has the property that every element of $P$ is related to some element of $P$, and Axiom of Dependent Choice, started at an element of $P$, yields a sequence $(n_{j})_{j\in\mathbb{N}}$ in $P$ with $n_{j+1}>n_{j}$ for every $j$. Thus $n_{1}<n_{2}<\dots$ and $a_{n_{j}}<\ell+\varepsilon$ for every $j$.

Claim 5. Suppose first that $(a_{n})$ has limit $a$, and let $\varepsilon>0$ be real. There is $N$ with $|a_{m}-a|<\varepsilon$ for every $m\ge N$, that is, by claim 9 of Properties of the Absolute Value in an Ordered Field, $a-\varepsilon<a_{m}<a+\varepsilon$ for every $m\ge N$. Then $a-\varepsilon$ is a lower bound of $A_{N}$ and $a+\varepsilon$ is an upper bound of $A_{N}$, so

using claim 1 in the middle. Since this holds for every real $\varepsilon>0$, we get $a\le\ell$ and $L\le a$: if $\ell<a$, then $\varepsilon=(a-\ell)/2$ is positive and $a-\varepsilon=(a+\ell)/2>\ell$, contradicting $a-\varepsilon\le\ell$; and if $a<L$, then $\varepsilon=(L-a)/2$ is positive and $a+\varepsilon=(a+L)/2<L$, contradicting $L\le a+\varepsilon$. With $\ell\le L$ this gives $a\le\ell\le L\le a$, so $\ell=L=a$.

Conversely suppose $\ell=L=a$ and let $\varepsilon>0$ be real. By claim 3 there are $N$ and $N'$ with $a-\varepsilon<a_{m}$ for every $m\ge N$ and $a_{m}<a+\varepsilon$ for every $m\ge N'$. Let $N''$ be the greater of $N$ and $N'$ (claim 3 of the order lemma). For $m\ge N''$ both inequalities hold, so $|a_{m}-a|<\varepsilon$ by claim 9 of Properties of the Absolute Value in an Ordered Field. Hence $(a_{n})$ has limit $a$ in the sense of Limit of a Sequence of Real Numbers.