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

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The forum discussion centers on understanding Theorem 3.55 from Karl R. Stromberg's "An Introduction to Classical Real Analysis," specifically regarding the implications of the statement that the interior of the closure of a set \( A_1 \) is empty. Users seek clarification on how this leads to the selection of points in open sets and the choice of radii for open balls. Key points include the contradiction arising from assuming \( V \setminus A_1^{-} \) is empty and the necessity of choosing \( r_1 < 1 \) to ensure the sequence remains Cauchy.

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The post concerns an aspect of Stromberg's proof of the Baire Category Theorem ... ...
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:

Stromberg - Theorem 3.55 ... Baire Category Theorem ... .png
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 exactly/rigorously why/how it is that ##A_1^{ - \ \circ } = \emptyset## means that we can choose ##x_1## in the open set ##V## \ ##A_1^{ - }## ... it seems very plausibly the case ... but ... 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 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
==========================================================================================================================

The definitions of nowhere dense, first and second category and residual are relevant ... so I am providing Stromberg's definitions ... as follows:
Stromberg -  Defn 3.53 ... Nowhere Dense ...First and Second Category ... .png


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:

Stromberg -  Defn 3.11  ... Terminology for Topological Spaces ... .png


Hope that helps ...

Peter
 
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On the first one:

Assume the opposite: that V\A1- is empty. Then A1- contains V, which is a nonempty open set and has interior points (since all points in an open set are interior). So A1- has interior points, but it is the closure of A1, and that contradicts the statement that A1 is nowhere dense. From this contradiction we conclude that V\A1- cannot be empty.
 
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On the second one, note that V\A1- is open because it is the intersection of open set V with the complement of A1-, which is open because A1- is closed. Since it is open, all its points are interior and in particular x1 is, so there exists some h such that the ball Bh(x1) is contained in V\A1- .

He requires r1<1 because a little further down he requires that rn < 1/n. That's how he makes the sequence Cauchy.
 
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Thanks so much for the help, Andrew

Just working through your posts now ...

Thanks again,

Peter
 
andrewkirk said:
On the second one, note that V\A1- is open because it is the intersection of open set V with the complement of A1-, which is open because A1- is closed. Since it is open, all its points are interior and in particular x1 is, so there exists some h such that the ball Bh(x1) is contained in V\A1- .

He requires r1<1 because a little further down he requires that rn < 1/n. That's how he makes the sequence Cauchy.
Thanks again for the help Andrew ...

... but ... just a clarification ...

... how exactly do we demonstrate rigorously .. given the ball ##B_h (x_1) \subset V## \ ##A_1^{ - }##

... we then have ##B_h (x_1)^{ - } \subset V## \ ##A_1^{ - }## ... ... ?Hoe you can help further ...

Peter
 
Find an h' such that the open ball of radius h' is in that set. Then set h=h'/2. The closure of that ball will be inside the open ball of radius h' (as the closure is the set of points distant no further than h'/2 from the point, and that is a subset of the bigger open ball, set of points distance less than h' from the point), and hence will be in the required set.

This may relate to a tangential question you were asking earlier - something about checking the closures of balls with double or half the radius.
 
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