Rudin's Theorem 2.27


by Bachelier
Tags: rudin, theorem
Bachelier
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#1
Jan2-13, 09:42 PM
P: 376
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.
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Number Nine
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#2
Jan2-13, 09:46 PM
P: 771
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?
micromass
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#3
Jan2-13, 09:47 PM
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What is your definition of "open"?

Bachelier
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#4
Jan2-13, 09:49 PM
P: 376

Rudin's Theorem 2.27


Quote Quote by Number Nine View Post
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.
Bachelier
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#5
Jan2-13, 10:08 PM
P: 376
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?
lavinia
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#6
Jan2-13, 11:31 PM
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If any point not in E has an open neighborhood that does not intersect E then by definition the complement of E is open.
Bachelier
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#7
Jan2-13, 11:48 PM
P: 376
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.
Useful nucleus
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#8
Jan3-13, 06:34 AM
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P: 240
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.
Erland
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#9
Jan3-13, 09:59 AM
P: 302
Quote Quote by Bachelier View Post
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.
Bachelier
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#10
Jan3-13, 03:28 PM
P: 376
Quote Quote by Erland View Post
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.
Erland
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#11
Jan3-13, 04:11 PM
P: 302
Quote Quote by Bachelier View Post
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##.
Bachelier
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#12
Jan3-13, 05:06 PM
P: 376
Quote Quote by Erland View Post
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. :)


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