Baire Category Theorem proof in Stromberg 3.55

  • Level: Graduate 
  • Thread starter Thread starter Math Amateur
  • Start date Start date
  • Tags Tags
    Theorem
Join the discussion
Registration is free. Start your own thread to ask a follow-up.
5 replies · 4K views · last reply
Math Amateur
Gold Member
MHB
Messages
3,920
Reaction score
48
TL;DR
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:

Printed page showing Theorem 3.55, the Baire Category Theorem, and its proof, with the proof using open sets V and A_n, a Cauchy sequence x_n with radii r_n, and a closing box marking the end of the proof.
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:
Printed definitions from a textbook, item (3.53): nowhere dense sets, first and second category sets, and residual sets in a topological space X, with the closure condition A-bar interior equals the empty set.


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:

Textbook page of definitions from topology: neighborhood, limit point, isolated point, interior, closed set, closure, perfect set, dense set and boundary, with the closure formula D bar = D union limit points of D.


Hope that helps ...

Peter
 
Last edited:
Physics news on Phys.org
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.
 
  • Like
Likes   Reactions: Math Amateur
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.
 
  • Like
Likes   Reactions: Math Amateur
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.
 
  • Like
Likes   Reactions: Math Amateur