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Hi! This is from Rudin's RCA about $\limsup$. I would like to check some moments.

$1)$ Since $b_k\to \beta$ as $k\to \infty$. For $\varepsilon>0$ $\exists N=N_{\varepsilon}>0$ such that for any $k\geqslant N_{\varepsilon}$ we have $|b_k-\beta|<\varepsilon$ $\Leftrightarrow$ $b_k\in (\beta-\varepsilon,\beta+\varepsilon)$. But by definition $b_k=\sup \limits_{m\geqslant k} a_m$.

Then for $\varepsilon_1=\dfrac{1}{2}\min\{\beta+\varepsilon-b_N, b_N-\beta+\varepsilon\}>0$ exists $a_{n_1}$ such that $a_{n_1}\in (\beta-\varepsilon,\beta+\varepsilon).$

Then making analogous reasoning for $b_n$ for $n>\max\{n_1,N\}$ we find $a_{n_2} $ such that $a_{n_2}\in (\beta-\varepsilon,\beta+\varepsilon)$.

Proceeding this procees we get increasing sequence $\{n_i\}_{i\geqslant 1}$ such that $a_{n_i}\to \beta$ as $i\to \infty$.

$2)$ Also $\beta$ is the largest number with this property. Suppose that exists some subsequence $\{a_{n_i}\}$ such that $a_{n_i}\to \gamma$ as $i\to \infty$ where $\gamma>\beta$.

Taking $\varepsilon=\frac{\gamma-\beta}{2}>0$ then exists $N$ such that $\forall n_i>N$ we have $|a_{n_i}-\gamma|<\varepsilon$ $\Leftrightarrow$ $a_{n_i}>\frac{\beta+\gamma}{2}.$ But since $\beta=\inf \limits_{m\geqslant 1}b_m$ then for $\varepsilon_1=\frac{\gamma-\beta}{2}$ exists $b_k$ such that $b_k<\beta+\varepsilon_1=\frac{\beta+\gamma}{2}$. In other words, $\sup \{a_k, a_{k+1},\dots\}<\frac{\beta+\gamma}{2}$. And we get contradiction.

Are my reasoning right? Can anyone check them please.

Any comment/remark will be appreciated!

  • $\begingroup$ Note that $\beta$ may be $\infty$ or $-\infty$. $\endgroup$ – Bungo May 19 '16 at 21:14
  • $\begingroup$ @Bungo, I posted an answer for cases $\pm \infty$. Can you take a look? $\endgroup$ – ZFR May 22 '16 at 6:54

Case $\beta=\infty$. Then by definition $\inf \{b_1,b_2,b_3,\dots\}=\infty$ $\Rightarrow$$b_k=\infty$ for any $k\in \mathbb{N}$ $\Rightarrow$ $\sup \{a_k, a_{k+1}, a_{k+2},\dots\}=\infty$. Then property $1)$ is obvious. For $2)$ there is $a_{n_1}>1$, $a_{n_2}>2, \dots, a_{n_i}>i,\dots$ Hence $a_{n_i}\to \infty$ as $i\to \infty$.

Case $\beta=-\infty$.

$1^o$. Let $\exists j\in \mathbb{N}$ such that $b_j=-\infty$. Then $\sup \{a_j, a_{j+1}, a_{j+2},\dots\}=-\infty$ $\Rightarrow$ $a_j=a_{j+1}=\dots=-\infty$. In this case $\beta=-\infty$ is the largest number with this property (obvious).

$2^o$. Suppose that for any $j$ we have $b_j>-\infty$. Since sequence $\{b_1, b_2, \dots\}$ is not bounded below then exists subsequence $\{b_{n_i}\}$ such that $b_{n_i}<-i$. And we can find subsequence $\{a_{n_{i_1}}, a_{n_{i_2}},a_{n_{i_3}},\dots\}$ such that $a_{n_{i_j}}\to -\infty$ as $j\to \infty$. Also it's easy to prove that $\beta=-\infty$ is the largest number with this property.


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