Is $ \lim_{n\to \infty}supx_n$ + 1 > sup{$x_{n_k}:k \geq n$} for $n \in \mathbb{N}$

Question

The theorem states:

Let $\left (x_n \right )_{n=1}^{\infty}$ be a bounded sequence in $\mathbb{R}$, and let S be the set of its accumulation points then maxS exists and equals $ \lim_{n\to \infty}$sup$x_n$.

Proof Set x:= $ \lim_{n\to \infty}$sup$x_n$. We claim that x belongs to S and is an upper bound for S. For n$\in\mathbb{N}$, set: $v_n:=$sup{$x_{k}: k\geq n$}.

We will construct $n_{1}<n_{2}<n_{3}<...$ in $\mathbb{N}$ s.t.: \begin{align} \lim_{k\to \infty}x_{n_{k}} = \lim_{n\to \infty}v_n = \lim_{n\to \infty}supx_n = x \end{align} by defining $n_{1}<n_{2}<n_{3}<...$ inductively s.t.: \begin{align} x - \frac{1}{k} < x_{n_{k}}<x+\frac{1}{k} \end{align} The proof then starts by saying, by definition of the $\lim_{n\to \infty}$sup$x_n$ there is $\overline{n_1} \in \mathbb{N}$ s.t.: \begin{align} x - 1 < v_n<x+1 \end{align} I understand why $x-1<v_n$ but how can we claim $v_n<x+1$? Is $v_n-\lim_{n\to \infty}supx_n <1$ always the case? Thank you in advance.

Answer 1

The $ \limsup $ of a sequence $(x_n)$ can be defined as

\begin{equation} \limsup x_n = \lim_{n \rightarrow \infty} sup {\{x_k : k \geq n\}} \end{equation}

Which is equivalent to

\begin{equation} \limsup x_n = \lim_{n \rightarrow \infty} v_n \end{equation}

The limit always exists when $(x_n)$ is bounded cause in this case $(v_n)$ will be a monotonic bounded sequence and thus converge. If you put $\epsilon = 1$ we can find $\overline{n_1}$ s.t

\begin{equation} |x - v_n| < 1 \quad \forall n\geq \overline{n_1}. \end{equation}

Answer 2

Denote $l=\limsup x_{n}$. Since $(x_{n})$ is bounded, $l\in\mathbb{R}$. We go to prove that following:

(1) For any $b>l$, the set $\{n\mid x_{n}\in(b,\infty)\}$ is a finite set.

(2) For any $a<l$, the set $\{n\mid x_{n}\in(a,\infty)\}$ is an infinite set.

Proof: (1): For each $n$, let $y_{n}=\sup\{x_{k}\mid k\geq n\}$. Note that $\infty>y_{1}\geq y_{2}\geq\ldots$ and $y_{n}\rightarrow l$. Since $\lim_{n}y_{n}<b$, there exists $N$ such that $y_{n}<b$ whenever $n\geq N$. In particular, for any $n\geq N$, we have $x_{n}\leq y_{n}<b$. Therefore $\{n\mid x_{n}\in(b,\infty)\}\subseteq\{1,2,\ldots,N-1\}$.

(2): Let $a<l$. For each $n$, $y_{n}\geq l>a$. $y_{1}>a$ implies that $a$ is not an upper bound for the set $\{x_{n}\mid n\geq1\}$, so there exists $n_{1}\geq1$ such that $x_{n_{1}}>a$. Note that $y_{n_{1}+1}\geq l>a$ implies that $a$ is not an upper bound for the set $\{x_{n}\mid n\geq n_{1}+1\}$, so there exists $n_{2}>n_{1}+1$ such that $x_{n_{2}}>a$. Repeat the process indefinitely (formally, we need the recursion theorem here) and we obtain $n_{1}<n_{2}<n_{3}<\ldots$ such that $x_{n_{k}}>a$ for $k=1,2,\ldots$. Therefore $\{n\mid x_{n}\in(a,\infty)\}$ is infinite.

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From Bolzano-Weierstrass Theorem, the set $S$ of accumulation point of the sequence $(x_{n})$ is non-empty. From $(1)$, it follows that $S\cap(l,\infty)=\emptyset$. For, if there exists $b\in S\cap(l,\infty)$, then $b\in S\Rightarrow$there exists a subsequence $(x_{n_{k}})$ such that $\lim_{k}x_{n_{k}}=b>\frac{b+l}{2}>l\Rightarrow$there exists $K$ such that $x_{n_{k}}>\frac{b+l}{2}$ whenever $k\geq K$. Then $\{n\mid x_{n}\in(b,\infty)\}\supseteq\{n_{K},n_{K+1},n_{K+2},\ldots\}$ contradicting to $(1)$.

From (1) and (2), it follows that $l$ is an accumulation point. For, let $\varepsilon>0$ be arbitrary. By (1) and (2), $\{n\mid x_{n}\in(l-\varepsilon,\infty)\}\setminus\{n\mid x_{n}\in(l+\varepsilon,\infty)\}=\{n\mid x_{n}\in(l-\varepsilon,l+\varepsilon]\}$ is an infinite set. Take $\varepsilon=1$, then we choose $n_{1}$ such that $x_{n_{1}}\in(l-1,l+1]$. Take $\varepsilon=\frac{1}{2}$, then we choose $n_{2}>n_{1}$ such that $x_{n_{2}}\in(l-\frac{1}{2},l+\frac{1}{2}]$. Continue the process indefinitely (formally, we need recursion theorem), we obtain a subsequence $(x_{n_{k}})$ such that $x_{n_{k}}\in(l-\frac{1}{k},l+\frac{1}{k}]$. This shows that $x_{n_{k}}\rightarrow l$ and hence $l\in S$.

Together, this shows that $l=\max S$.