Rudin's Theorem 2.27

Bachelier

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.

Number Nine

What is the definition of open? What has Rudin just shown about an arbitrary point in X?

micromass

What is your definition of "open"?

Bachelier

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

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

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

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

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

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

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

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

Thank you. :)