# Baire Category Theorem .... Stromberg, Theorem 3.55 .... ....

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

My questions are as follows:

Question 1

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

Question 2

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

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

*** EDIT ***

My thoughts on Question 2 ...

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

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

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

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

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

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 9167

Hope that helps ...

Peter

#### Attachments

• Stromberg - Theorem 3.55 ... Baire Category Theorem ... .png
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• Stromberg - Defn 3.53 ... Nowhere Dense ...First and Second Category ... .png
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• Stromberg - Defn 3.11 ... Terminology for Topological Spaces ... .png
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Last edited:

Gold Member
MHB
Question 1

Can someone explain and demonstrate why/how it is that $$\displaystyle A_1^{ - \ \circ } = \emptyset$$ means that we can choose $$\displaystyle x_1$$ in the open set $$\displaystyle V$$ \ $$\displaystyle A_1^{ - }$$ ... 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 $$\displaystyle 0 \lt r_1 \lt 1$$ such that $$\displaystyle B_{ r_1 } ( x_1 )^{ - } \subset V$$ \ $$\displaystyle A_1^{ - }$$ ... ?

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

*** EDIT ***

My thoughts on Question 2 ...

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

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

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

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

... and further why is Stromberg talking about $$\displaystyle r_1$$ between $$\displaystyle 0$$ and $$\displaystyle 1$$ ...?
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 $r_n<1/n$.

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MHB
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

Gold Member
MHB
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

Last edited:
Gold Member
MHB
... 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 $y\in B_{2r}(x).$

Gold Member
MHB
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 $$\displaystyle y$$ is a limit point of $$\displaystyle B_r(x)^-$$ ... by why is this true when $$\displaystyle y$$ is an interior point of $$\displaystyle B_r(x)^-$$ ... given we are dealing with a general metric space ...

Hope you can help further ...

Peter

Gold Member
MHB
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 $$\displaystyle y$$ is a limit point of $$\displaystyle B_r(x)^-$$ ... by why is this true when $$\displaystyle y$$ is an interior point of $$\displaystyle B_r(x)^-$$ ... given we are dealing with a general metric space ...
If $$\displaystyle y\in B_r(x)^-$$ is not a limit point, then $$\displaystyle y\in B_r(x)$$, which is obviously contained in $$\displaystyle B_{2r}(x)$$.

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

Thanks for all your help, Opalg ...

It is much appreciated ...

Peter