Baire Category Theorem

  • #1
102
12
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?
 
Last edited:

Answers and Replies

  • #2
mathwonk
Science Advisor
Homework Helper
2020 Award
11,100
1,302
there is a typo in your statement, the intersection is dense, not open.
 
  • Like
Likes mr.tea
  • #3
102
12
there is a typo in your statement, the intersection is dense, not open.
Sorry, fixed it.
 
  • #4
mathwonk
Science Advisor
Homework Helper
2020 Award
11,100
1,302
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.
 
  • Like
Likes mr.tea

Related Threads on Baire Category Theorem

Replies
2
Views
1K
Replies
4
Views
3K
Replies
5
Views
1K
Replies
4
Views
4K
Replies
1
Views
735
Replies
6
Views
3K
Replies
5
Views
441
  • Last Post
Replies
5
Views
932
  • Last Post
Replies
11
Views
11K
Top