MHB Closed and Bounded Intervals are Compact .... Sohrab, Propostion 4.1.9 .... ....

[Math Amateur]

Closed and Bounded Intervals are Compact ... Sohrab, Proposition 4.1.9 ... ...

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

I am focused on Chapter 4: Topology of [FONT=MathJax_AMS]R and Continuity ... ...

I need help in order to fully understand the proof of Proposition 4.1.9 ...Proposition 4.1.9 and its proof read as follows:
View attachment 9091
My questions are as follows:Question 1

In the above proof by Sohrab we read the following:

" ... ... Now, if $$c \lt b$$, then we can pick $$d \in ( c, c + \epsilon )$$ such that $$c \lt d \lt b$$ ... ... "My question is as follows:

How (... rigorously speaking ... ) do we know such a $$d$$ exists ...

In other words, what is the rigorous justification that if $$c \lt b$$, the we can pick $$d \in ( c, c + \epsilon )$$ such that $$c \lt d \lt b$$ ... ...?

Question 2

In the above proof by Sohrab we read the following:

" ... ... Now, if $$c \lt b$$, then we can pick $$d \in ( c, c + \epsilon )$$ such that $$c \lt d \lt b$$, and it follows that $$[a,d ]$$ can also be covered by a finite subcover. i.e. $$d \in S$$ ... ... "Can someone please explain why/how it follows that $$[a,d ]$$ can also be covered by a finite subcover. i.e. $$d \in S$$ ... ... ?

Help will be appreciated ...

Peter

 

[castor28]

Hi Peter,

For question 1, since $c<b$ and $\epsilon>0$, you can take $d$ between $c$ and $\min(c+\epsilon,b)$.

For question 2, since $c<d<c+\epsilon$, and $(c-\epsilon,c+\epsilon)\subset O_\lambda$, $d\in O_\lambda$, and you can add $O_\lambda$ to the finite cover of $[a,c)$ to get a finite cover of $[a,d]$.

 

[castor28]

Hi Peter,

My answer of question 2 is not quite correct. In fact, we do not necessarily have a finite cover of the interval $[a,c)$ (which is not compact).

We do not know if $c\in S$. What we do know is that, for any $x$ with $a<x<c$, we have a finite cover $U_x$ of $[a,x]$. We can take $x$ between $c-\epsilon$ and $c$, which gives the inequalities:
$$
a<c-\epsilon<x<c<d<c+\epsilon
$$
Now, $U_x$ covers $[a,x]$ and $O_\lambda$ covers $(c-\epsilon,c+\epsilon)$ (which contains $d$); therefore $U_x\cup O_\lambda$ covers $[a,d]$; this means that $d\in S$, which contradicts $c=\sup S$.

 

[Math Amateur]

Hi Peter,

My answer of question 2 is not quite correct. In fact, we do not necessarily have a finite cover of the interval $[a,c)$ (which is not compact).

We do not know if $c\in S$. What we do know is that, for any $x$ with $a<x<c$, we have a finite cover $U_x$ of $[a,x]$. We can take $x$ between $c-\epsilon$ and $c$, which gives the inequalities:
$$
a<c-\epsilon<x<c<d<c+\epsilon
$$
Now, $U_x$ covers $[a,x]$ and $O_\lambda$ covers $(c-\epsilon,c+\epsilon)$ (which contains $d$); therefore $U_x\cup O_\lambda$ covers $[a,d]$; this means that $d\in S$, which contradicts $c=\sup S$.

Thanks for the guidance and assistance ... and the correction ...

Most helpful ...

Peter