MHB Closed Sets - Sohrab Exercise 2.2.4 - Part 2

[Math Amateur]

I am reading Houshang H. Sohrab's book: "Basic Real Analysis" (Second Edition).

I am focused on Chapter 2: Sequences and Series of Real Numbers ... ...

I need help with a part of Exercise 2.2.4 Part (2) ... ...

Exercise 2.2.4 Part (2) reads as follows:https://www.physicsforums.com/attachments/7183In the above text from Sohrab we read the following:

" ... ...Using the infinite collection $$[ \frac{1}{n}, 1 - \frac{1}{n} ], \ n \in \mathbb{N}$$, show the latter statement is false if $$\Lambda$$ is infinite ... ... "
I am unable to make a meaningful start on this problem ... can someone help me with the exercise ...

Peter
Note: For $$n = 1$$, the interval is [1,0] which does make much sense ... I guess we take that case as equal to $$\emptyset$$ ...

*** EDIT ***

After some reflection I am beginning to believe that $$\bigcup_{ n = 1}^{ \infty } I_n = (0,1)$$ where $$I_n = [ \frac{1}{n}, 1 - \frac{1}{n} ]$$ ... but ... sadly ... I cannot prove this intuition is correct ...

Note that Sohrab doesn't define limits or convergence until after setting this exercise ...=========================================================================================

The above exercise relies on the definition of open sets and related concepts and so to provide readers with a knowledge of Sohrab's definitions and notation I am provided the following text ...https://www.physicsforums.com/attachments/7184

 

[Evgeny.Makarov]

For $$n = 1$$, the interval is [1,0] which does make much sense ... I guess we take that case as equal to $$\emptyset$$

I think this is reasonable.

After some reflection I am beginning to believe that $$\bigcup_{ n = 1}^{ \infty } I_n = (0,1)$$ where $$I_n = [ \frac{1}{n}, 1 - \frac{1}{n} ]$$ ... but ... sadly ... I cannot prove this intuition is correct

This is correct and is proved by definition of union. Consider the cases $x\le 0$, $0<x<1$ and $x\ge 1$.

Note that the claims about the infinite intersection and finite union of closed sets, as well as the counterexample about the infinite union of closed sets, can be obtained from the corresponding claims about open sets by taking the complement.

 

[Math Amateur]

I think this is reasonable.

This is correct and is proved by definition of union. Consider the cases $x\le 0$, $0<x<1$ and $x\ge 1$.

Note that the claims about the infinite intersection and finite union of closed sets, as well as the counterexample about the infinite union of closed sets, can be obtained from the corresponding claims about open sets by taking the complement.

Thanks Evgeny ... but I am still perplexed as to how to proceed ...

Can you help further ...

Peter

 

[Evgeny.Makarov]

We want to show that
\[
\bigcup_{n=1}^\infty I_n=(0,1)\qquad(*)
\]
where $I_n=\left[\frac{1}{n},1-\frac{1}{n}\right]$. If $x\le0$, then $x\notin I_n$ for all $n$, so $x\notin\bigcup I_n$. Also, $x\notin(0,1)$. Similar for $x\ge1$. If $0<x<1$, then consider $n\in\mathbb{N}$ such that $n>\max\left(\frac{1}{x},\frac{1}{1-x}\right)$. Then $n>1/x$ and $n>1/(1-x)$, so $x>1/n$ and $1-x>1/n$, which means that $x\in[1/n,1-1/n]=I_n$. Thus $x\in\bigcup I_n$, and obviously $x\in (0,1)$. So in all three cases $x$ belongs or does not belong to the left- and right-hand sides of (*) simultaneously.