MHB Baire Category Theorem .... Stromberg, Theorem 3.55 .... ....

[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 3.55 on page 110 ... ... Theorem 3.55 and its proof read as follows:
View attachment 9165
At the start of the second paragraph of the above proof by Stromberg we read the following:

" ... ...Since $$A_1^{ - \ \circ } = \emptyset$$, we can choose $$x_1$$ in the open set $$V$$ \ $$A_1^{ - }$$ and then we can choose $$0 \lt r_1 \lt 1$$ such that $$B_{ r_1 } ( x_1 )^{ - } \subset V$$ \ $$A_1^{ - }$$ [ check that $$B_r (x)^{ - } \subset B_{ 2r } (x) $$ ] ... ...My questions are as follows:Question 1

Can someone explain and demonstrate why/how it is that $$A_1^{ - \ \circ } = \emptyset$$ means that we can choose $$x_1$$ in the open set $$V$$ \ $$A_1^{ - }$$ ... how are we (rigorously) sure this is true ... ?
Question 2

How/why can we choose $$0 \lt r_1 \lt 1$$ such that $$B_{ r_1 } ( x_1 )^{ - } \subset V$$ \ $$A_1^{ - }$$ ... ?

... and why are we checking that $$B_r (x)^{ - } \subset B_{ 2r } (x)$$ ... ... ?
*** EDIT ***

My thoughts on Question 2 ...

Since $$V$$ \ $$A_1^{ - }$$ is open ... $$\exists \ r_1$$ such that $$B_{ r_1 } ( x_1 ) \subset V$$ \ $$A_1^{ - }$$ ...

... BUT ... how do we formally and rigorously show that ...

... we can choose an $$r_1$$ such that the closure of $$B_{ r_1 } ( x_1 )$$ is a subset of $$V$$ \ $$A_1^{ - }$$ ...

... (intuitively I think we just choose $$r_1$$ somewhat smaller yet ...)

... and further why is Stromberg talking about $$r_1$$ between $$0$$ and $$1$$ ...?
Help will be much appreciated ...

Peter
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The definitions of nowhere dense, first and second category and residual are relevant ... so I am providing Stromberg's definitions ... as follows:
View attachment 9166

Stromberg's terminology and notation associated with the basic notions of topological spaces are relevant to the above post ... so I am providing the text of the same ... as follows:

View attachment 9167Hope that helps ...

Peter

 

[Opalg]

Question 1

Can someone explain and demonstrate why/how it is that $$A_1^{ - \ \circ } = \emptyset$$ means that we can choose $$x_1$$ in the open set $$V$$ \ $$A_1^{ - }$$ ... how are we (rigorously) sure this is true ... ?

If $V\setminus A_1^-$ is empty then $V\subseteq A_1^{-\mathrm o}$, contradicting the fact that $A_1^{-\mathrm o}$ is empty. Therefore $V\setminus A_1^-$ is not empty and so it contains some point, which we can choose as $x_1$.

Question 2

How/why can we choose $$0 \lt r_1 \lt 1$$ such that $$B_{ r_1 } ( x_1 )^{ - } \subset V$$ \ $$A_1^{ - }$$ ... ?

... and why are we checking that $$B_r (x)^{ - } \subset B_{ 2r } (x)$$ ... ... ?
*** EDIT ***

My thoughts on Question 2 ...

Since $$V$$ \ $$A_1^{ - }$$ is open ... $$\exists \ r_1$$ such that $$B_{ r_1 } ( x_1 ) \subset V$$ \ $$A_1^{ - }$$ ...

... BUT ... how do we formally and rigorously show that ...

... we can choose an $$r_1$$ such that the closure of $$B_{ r_1 } ( x_1 )$$ is a subset of $$V$$ \ $$A_1^{ - }$$ ...

... (intuitively I think we just choose $$r_1$$ somewhat smaller yet ...)

... and further why is Stromberg talking about $$r_1$$ between $$0$$ and $$1$$ ...?

Your thoughts on Question 2 are quite correct. If you can choose a number, say $s$ such that $B_{ s } ( x_1 ) \subset V\setminus A_1^{ - }$, then let $r_1 = \frac12s$. Then $s=2r_1$, so that $B_{ 2r_1 } ( x_1 ) \subset V\setminus A_1^{ - }$ and hence the closure of $B_{ r_1 } ( x_1 )$ is contained in $V\setminus A_1^{ - }$.

Given $r_1$ with that property, any smaller (positive) value of $r_1$ will have the same property. So you can always assume that $0<r_1<1$. Stromberg's reason for wanting that is that this is the first step of an inductive construction in which he wants to ensure that $r_n\to0$ as $n\to\infty$. The easiest way to do that is to require that $r_n<1/n$.

 

[Math Amateur]

If $V\setminus A_1^-$ is empty then $V\subseteq A_1^{-\mathrm o}$, contradicting the fact that $A_1^{-\mathrm o}$ is empty. Therefore $V\setminus A_1^-$ is not empty and so it contains some point, which we can choose as $x_1$.
Your thoughts on Question 2 are quite correct. If you can choose a number, say $s$ such that $B_{ s } ( x_1 ) \subset V\setminus A_1^{ - }$, then let $r_1 = \frac12s$. Then $s=2r_1$, so that $B_{ 2r_1 } ( x_1 ) \subset V\setminus A_1^{ - }$ and hence the closure of $B_{ r_1 } ( x_1 )$ is contained in $V\setminus A_1^{ - }$.

Given $r_1$ with that property, any smaller (positive) value of $r_1$ will have the same property. So you can always assume that $0<r_1<1$. Stromberg's reason for wanting that is that this is the first step of an inductive construction in which he wants to ensure that $r_n\to0$ as $n\to\infty$. The easiest way to do that is to require that $r_n<1/n$.

Thanks Opalg ...

... very much appreciate your help...

Peter

 

[Math Amateur]

Thanks Opalg ...

... very much appreciate your help...

Peter

Hi Opalg ...

Thanks again for your help ...

Just a further point of clarification ...

You write:

" ... ... Then $s=2r_1$, so that $B_{ 2r_1 } ( x_1 ) \subset V\setminus A_1^{ - }$ and hence the closure of $B_{ r_1 } ( x_1 )$ is contained in $V\setminus A_1^{ - }$. ... ... "So ... as I understand it, you are arguing that because $B_{ 2r_1 } ( x_1 ) \subset V\setminus A_1^{ - }$ ...

... ... that therefore ... the closure of $B_{ r_1 } ( x_1 )$ is contained in $V\setminus A_1^{ - }$. ...

... that is that $B_{ r_1 } ( x_1 )^{ - } \subset V\setminus A_1^{ - } $ ... ...

But why ... rigorously ... is this true ... (... it certainly seems plausible ... but rigorously ...? )
Hope that you can help further ...

Peter

 

[Opalg]

... as I understand it, you are arguing that because $B_{ 2r_1 } ( x_1 ) \subset V\setminus A_1^{ - }$ ...

... ... that therefore ... the closure of $B_{ r_1 } ( x_1 )$ is contained in $V\setminus A_1^{ - }$. ...

... that is that $B_{ r_1 } ( x_1 )^{ - } \subset V\setminus A_1^{ - } $ ... ...

But why ... rigorously ... is this true ... (... it certainly seems plausible ... but rigorously ...? )

The result that you need is that if $x$ is an element in a metric space $(X,d)$, and $r>0$, then $B_r(x)^- \subseteq B_{2r}(x)$.

To see that, notice that if $y\in B_r(x)^-$ then there must be a point $z\in B_r(x)$ with $d(z,y)<r$. By the triangle inequality, $d(y,x) \leqslant d(y,z) + d(z,x) < r + r = 2r$. Therefore $y\in B_{2r}(x).$

 

[Math Amateur]

The result that you need is that if $x$ is an element in a metric space $(X,d)$, and $r>0$, then $B_r(x)^- \subseteq B_{2r}(x)$.

To see that, notice that if $y\in B_r(x)^-$ then there must be a point $z\in B_r(x)$ with $d(z,y)<r$. By the triangle inequality, $d(y,x) \leqslant d(y,z) + d(z,x) < r + r = 2r$. Therefore $y\in B_{2r}(x).$

Thanks for the help Opalg ...

But ... just a point of clarification ...

You write:

" ... ... notice that if $y\in B_r(x)^-$ then there must be a point $z\in B_r(x)$ with $d(z,y)<r$ ... ... "I can see what you write must be true (by definition) when $$y$$ is a limit point of $$B_r(x)^-$$ ... by why is this true when $$y$$ is an interior point of $$B_r(x)^-$$ ... given we are dealing with a general metric space ...
Hope you can help further ...

Peter

 

[Opalg]

You write:

" ... ... notice that if $y\in B_r(x)^-$ then there must be a point $z\in B_r(x)$ with $d(z,y)<r$ ... ... "I can see what you write must be true (by definition) when $$y$$ is a limit point of $$B_r(x)^-$$ ... by why is this true when $$y$$ is an interior point of $$B_r(x)^-$$ ... given we are dealing with a general metric space ...

If $$y\in B_r(x)^-$$ is not a limit point, then $$y\in B_r(x)$$, which is obviously contained in $$B_{2r}(x)$$.

 

[Math Amateur]

If $$y\in B_r(x)^-$$ is not a limit point, then $$y\in B_r(x)$$, which is obviously contained in $$B_{2r}(x)$$.

Thanks for all your help, Opalg ...

It is much appreciated ...

Peter