Rudin's Theorem 2.27

  • Thread starter Bachelier
  • Start date
  • #1
376
0

Main Question or Discussion Point

Prove the closure of E in a Metric Space X is closed. (page 35)

Rudin states:

if p∈X and p∉E then p is neither a point of E nor a limit point of E..

Hence, p has a neighborhood which does not intersect E. (Great)

The compliment of the closure of E is therefore open. WHY? I don't see it...

BTW, I know there are different ways to proving this, but I want to understand the last line jump. Thanks.
 
Last edited:

Answers and Replies

  • #2
806
23
The compliment of E is therefore open. WHY? I don't see it...
What is the definition of open? What has Rudin just shown about an arbitrary point in X?
 
Last edited:
  • #3
22,097
3,283
What is your definition of "open"?
 
  • #4
376
0
What is the definition of open? What has Rudin just shown about an arbitrary point in E?
E is Open if every p in E is an interior point (meaning there exists a neighborhood of p that is in E)
The problem we should say that the complement of E is open, not the complement of the closure of E.
 
  • #5
376
0
I guess since the intersection of N(p) and E is empty then no point q of N(p) can be a limit point of E as this would mean every neighborhood of q will contain an infinite number of points in E. Hence the intersection of N(p) and "closure of E" is empty.

Is this correct?
 
Last edited:
  • #6
lavinia
Science Advisor
Gold Member
3,236
623
If any point not in E has an open neighborhood that does not intersect E then by definition the complement of E is open.
 
Last edited:
  • #7
376
0
my bad, I forgot to add the word: "closure" in the last line of the proof. I just re-read it.

This is what is confusing me:

FROM: Hence, p has a neighborhood which does not intersect E.

We get: The compliment of the closure of E is therefore open.
 
  • #8
We have just shown that p is in the complement of the closure of E (call it A). We also showed that p has a neighborhood that is entirely in A. Hence, p is an interior point of that set A. Hence A is open.
 
  • #9
Erland
Science Advisor
738
136
if p∈X and p∉E then p is neither a point of E nor a limit point of E..
This doesn't make sense. It must be ##p\notin \overline E##, where ##\overline E## is the closure of ##E##, instead of ##p\notin E##.
Then, everthing is clear.
 
Last edited:
  • #10
376
0
This doesn't make sence. It must be ##p\notin \overline E##, where ##\overline E## is the closure of ##E##, instead of ##p\notin E##.
Then, everthing is clear.
Since ##p\notin \overline E## then it is not a limit point, hence not every neighborhood of ##p## contains a point of ##E##.

Let ##N(p)## be the neighborhood with no common points with ##E##.

What about ##\overline E##? Is the ##\overline E \cap N(p)## an empty set because if it wasn't, then ##N(p)## will contain a limit point of ##E## and these will have neighborhoods that contain a point of ##E##?

"I understand everything about the proof, except for the part where we go from ##E## to ## \overline E## when we mention the complement. I want to make sure my reasoning is correct" Thanks.
 
Last edited:
  • #11
Erland
Science Advisor
738
136
Since ##p\notin \overline E## then it is not a limit point, hence not every neighborhood of ##p## contains a point of ##E##.

Let ##N(p)## be the neighborhood with no common points with ##E##.

What about ##\overline E##? Is the ##\overline E \cap N(p)## an empty set because if it wasn't, then ##N(p)## will contain a limit point of ##E## and these will have neighborhoods that contain a point of ##E##?

"I understand everything about the proof, except for the part where we go from ##E## to ## \overline E## when we mention the complement. I want to make sure my reasoning is correct" Thanks.
You are right, I, and Rudin it seems, were a little bit too quick here. But it is as you write. If ##N(p)## intersects ##\overline E## in a point ##q##, say, then ##N(p)## is also a neighborhood of ##q##, and it must contain an element of ##E##, which was not the case. Thus ##N(p)## is a neighborhood of ##p## which does not intersect ##\overline E##.
 
Last edited:
  • #12
376
0
You are right, I, and Rudin it seems, were a little bit to quick here. But it is as you write. If ##N(p)## intersects ##\overline E## in a point ##q##, say, then ##N(p)## is also a neighborhood of ##q##, and it must contain an element of ##E##, which was not the case. Thus ##N(p)## is a neighborhood of ##p## which does not intersect ##\overline E##.
Thank you. :)
 

Related Threads on Rudin's Theorem 2.27

Replies
10
Views
3K
  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
1
Views
3K
  • Last Post
Replies
5
Views
3K
  • Last Post
Replies
2
Views
3K
  • Last Post
Replies
5
Views
2K
  • Last Post
Replies
5
Views
4K
Replies
3
Views
1K
Replies
3
Views
2K
Replies
7
Views
3K
Top