Baire Category Theorem: Question About Countable Dense Open Sets

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Discussion Overview

The discussion revolves around the Baire Category Theorem, specifically the properties of countable dense open sets in a complete metric space. Participants explore the implications of the theorem and seek clarification on the conditions under which the closure of a ball can be contained within these sets.

Discussion Character

  • Technical explanation
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • One participant questions the rigorous proof of why the closure of a ball can be contained in each of the dense open sets.
  • Another participant points out a typo regarding the nature of the intersection, clarifying that it is dense, not open.
  • A later reply asserts that if a point lies in an open set, then there exists a positive distance from the outside of that set, allowing for a ball of radius half that distance to have its closure contained entirely within the set.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the rigorous proof of the closure containment, and there are differing views on the clarity of the initial question posed.

Contextual Notes

The discussion includes assumptions about the properties of open sets in metric spaces and the implications of the Baire Category Theorem, but these assumptions are not fully explored or resolved.

mr.tea
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Hi,
I have a (probably stupid) question about the Baire Category Theorem. I am looking at the statement that says that in a complete metric space, the intersection of countable many dense open sets is dense in the metric space.
Assume that we have the countable collection of dense open sets ## \{U_n\} ## in a complete metric space ##X##, and let ##x \in X, \epsilon>0##. Since ##U_1## is dense in ##X##, there is ##y_1\in U_1## with ##d(x,y_1)<\epsilon##. Also, as ##U_1## is open, there is ##r_1>0## with ##B(y_1;r_1)\subset U_1##. Then, we can arrange ##r_1<1## such that ##\overline{B(y_1;r_1)} \subset U_1\cap B(x;\epsilon) ##.
Now my question is why we can arrange that the closure will be contained in each of them? I think intuitively it sounds correct, but I didn't succeed to prove it rigorously. Can you please help me here?
 
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there is a typo in your statement, the intersection is dense, not open.
 
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mathwonk said:
there is a typo in your statement, the intersection is dense, not open.
Sorry, fixed it.
 
i guess you mean by "each of them" the two open sets in your discussion. this is trivial. if a point lies in an open set in a metric space then it has some positive distance from the outside of that set, and hence the ball of radius 1/2 that distance has closure entirely contained in that set.
 
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