MHB Connectedness and Intervals in R .... Stromberg, Theorem 3.47 .... ....

[Math Amateur]

I am reading Karl R. Stromberg's book: "An Introduction to Classical Real Analysis". ... ...

I am focused on Chapter 3: Limits and Continuity ... ...

I need help in order to fully understand the proof of Theorem 3.47 on page 107 ... ... Theorem 3.47 and its proof read as follows:View attachment 9153
In the second paragraph of the above proof by Stromberg we read the following:

" ... ... Since $$U$$ is open we can choose $$c' \gt c$$ such that $$[ c, c' ] \subset U \cap [a, b]$$ ... ... "
My question is as follows:

Can someone please demonstrate rigorously why/how ...

$$U$$ is open $$\Longrightarrow$$ we can choose $$c' \gt c$$ such that $$[ c, c' ] \subset U \cap [a, b] $$ ... ...
Indeed I can see that ...

$$U$$ is open $$\Longrightarrow \exists$$ an open ball $$B_r(c) = \ ] c - r, c + r [ \ \subset U$$ ... ...but how do we conclude from here that

$$U$$ is open $$\Longrightarrow$$ we can choose $$c' \gt c$$ such that $$[ c, c' ] \subset U \cap [a, b]$$ ... ...*** EDIT ***

It may be that the solution is to choose $$s \lt r$$ so that $$[ c, c + s] \subset U$$ where $$c' = c + s$$ ... but how do we ensure this interval also belongs to $$[a, b]$$ ... ... ?

Help will be appreciated ... ...

Peter
=======================================================================================Stromberg uses slightly unusual notation for open intervals in $$\mathbb{R}$$ and $$\mathbb{R}^{\#} = \mathbb{R} \cup \{ \infty , - \infty \}$$ so I am providing access to Stromberg's definition of intervals in $$\mathbb{R}^{ \#} $$ ... as follows:

View attachment 9152

Hope that helps ...

Peter

 

[HallsofIvy]

I am reading Karl R. Stromberg's book: "An Introduction to Classical Real Analysis". ... ...

I am focused on Chapter 3: Limits and Continuity ... ...

I need help in order to fully understand the proof of Theorem 3.47 on page 107 ... ... Theorem 3.47 and its proof read as follows:
In the second paragraph of the above proof by Stromberg we read the following:

" ... ... Since $$U$$ is open we can choose $$c' \gt c$$ such that $$[ c, c' ] \subset U \cap [a, b]$$ ... ... "
My question is as follows:

Can someone please demonstrate rigorously why/how ...

$$U$$ is open $$\Longrightarrow$$ we can choose $$c' \gt c$$ such that $$[ c, c' ] \subset U \cap [a, b] $$ ... ...
Indeed I can see that ...

$$U$$ is open $$\Longrightarrow \exists$$ an open ball $$B_r(c) = \ ] c - r, c + r [ \ \subset U$$ ... ...but how do we conclude from here that

$$U$$ is open $$\Longrightarrow$$ we can choose $$c' \gt c$$ such that $$[ c, c' ] \subset U \cap [a, b]$$ ... …

Take c' to be the smaller of c+ r/2 and (c+ b)/2. Then c< c'< c+ r so is in U and c' is half way between c and b so c' is in [a b].

*** EDIT ***

It may be that the solution is to choose $$s \lt r$$ so that $$[ c, c + s] \subset U$$ where $$c' = c + s$$ ... but how do we ensure this interval also belongs to $$[a, b]$$ ... ... ?

Help will be appreciated ... ...

Peter
=======================================================================================Stromberg uses slightly unusual notation for open intervals in $$\mathbb{R}$$ and $$\mathbb{R}^{\#} = \mathbb{R} \cup \{ \infty , - \infty \}$$ so I am providing access to Stromberg's definition of intervals in $$\mathbb{R}^{ \#} $$ ... as follows:
Hope that helps ...

Peter

 

[soloenergy]

Hi Peter,

Thank you for reaching out for help with understanding Theorem 3.47 in Stromberg's book. I am also currently studying this chapter and can offer some clarification on the proof.

First, let's define some notation to make things clearer. Let U be an open subset of [a,b], and let c \in U. Since U is open, there exists an open interval B_r(c) = ]c-r, c+r[ \subset U. Now, we can choose s < r such that [c, c+s] \subset B_r(c). This is possible because the interval B_r(c) is open, so we can "shrink" it by choosing a smaller radius s.

Next, we need to show that [c, c+s] \subset [a,b]. Since c \in U \subset [a,b], we know that c \in [a,b]. And since c+s < c+r, we have c+s < b, which means that c+s \in [a,b]. Similarly, since c \in U \subset [a,b], we have c \in [a,b] and since c < c+s, we have c > a, which means that c \in [a,b]. Therefore, [c,c+s] \subset [a,b].

Now, we can choose c' = c+s, and we have shown that c' > c and [c,c'] \subset [a,b]. Also, since [c,c'] \subset B_r(c) \subset U, we have [c,c'] \subset U \cap [a,b], as desired.

I hope this helps clarify the proof for you. Let me know if you have any other questions or need further explanation.