Remarks on Limit Superrior .... B&S Page 82 .... Example ....

[Math Amateur]

I am reading "Introduction to Real Analysis" (Fourth Edition) b Robert G Bartle and Donald R Sherbert ...

I am focused on Chapter 3: Sequences and Series ...

I need help in fully understanding an example given by B&S in some introductory remarks on Limit Superior and Limit Inferior ...The first part of the introductory remarks read as follows:View attachment 7315

In the third paragraph of the above text, starting: "An extreme example ... ... " we read the following:

" ... ... Then it follows from the Density Theorem 2.4.8 that every number in \(\displaystyle [0, 1]\) is a subsequential limit of \(\displaystyle ( r_n )\). ... ... "Can someone please explain exactly how it follows from the Density Theorem 2.4.8 that every number in \(\displaystyle [0, 1]\) is a subsequential limit of \(\displaystyle ( r_n )\)?

Peter==================================================================================The above post refers to B&S's Density Theorem so I am proving the text of the teorem and its corollary ... as follows ... ...View attachment 7316

 

[Opalg]

Can someone please explain exactly how it follows from the Density Theorem 2.4.8 that every number in \(\displaystyle [0, 1]\) is a subsequential limit of \(\displaystyle ( r_n )\)?

Let $x$ be a number in $[0,1]$. One idea would be to use induction to construct a sequence $(s_n)$ of rational numbers such that $|x-s_n| <1/n$. Such a sequence certainly converges to $x$.

The induction starts by taking $s_1$ to be any rational number in the unit interval $[0,1]$, say $s_1 = 1/2$.

For the inductive step, suppose that $s_1,s_2,\ldots,s_n$ have already been chosen. By the Density Theorem, there exists a rational number in the interval $ I_{n+1} = (x - \frac1{n+1},x + \frac1{n+1})$. We want to choose $s_{n+1}$ to be such a number. But there are two possible snags to doing this.

First, the interval $ I_{n+1}$ may extend beyond the unit interval. We can get around that by shortening $ I_{n+1}$, replacing its lower limit by $0$ or its upper limit by $1$, as necessary.

Second, and more seriously, $ I_{n+1}$ may contain one or more of the numbers $s_1,s_2,\ldots,s_n$ that have already been chosen. But there are only finitely many of those numbers, so again we can can shorten $ I_{n+1}$ to exclude them all. The shortened version of $ I_{n+1}$ will still, by the Density Theorem, contain a rational number, which we can choose as $s_{n+1}$.

 

[Math Amateur]

Let $x$ be a number in $[0,1]$. One idea would be to use induction to construct a sequence $(s_n)$ of rational numbers such that $|x-s_n| <1/n$. Such a sequence certainly converges to $x$.

The induction starts by taking $s_1$ to be any rational number in the unit interval $[0,1]$, say $s_1 = 1/2$.

For the inductive step, suppose that $s_1,s_2,\ldots,s_n$ have already been chosen. By the Density Theorem, there exists a rational number in the interval $ I_{n+1} = (x - \frac1{n+1},x + \frac1{n+1})$. We want to choose $s_{n+1}$ to be such a number. But there are two possible snags to doing this.

First, the interval $ I_{n+1}$ may extend beyond the unit interval. We can get around that by shortening $ I_{n+1}$, replacing its lower limit by $0$ or its upper limit by $1$, as necessary.

Second, and more seriously, $ I_{n+1}$ may contain one or more of the numbers $s_1,s_2,\ldots,s_n$ that have already been chosen. But there are only finitely many of those numbers, so again we can can shorten $ I_{n+1}$ to exclude them all. The shortened version of $ I_{n+1}$ will still, by the Density Theorem, contain a rational number, which we can choose as $s_{n+1}$.

Thanks Opalg ...

Still reflecting on what you have written ...

Peter