Rudin's Theorem 2.27

1. Jan 2, 2013

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.

Last edited: Jan 2, 2013
2. Jan 2, 2013

Number Nine

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

Last edited: Jan 2, 2013
3. Jan 2, 2013

micromass

What is your definition of "open"?

4. Jan 2, 2013

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.

5. Jan 2, 2013

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?

Last edited: Jan 2, 2013
6. Jan 2, 2013

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.

Last edited: Jan 3, 2013
7. Jan 2, 2013

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.

8. Jan 3, 2013

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.

9. Jan 3, 2013

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.

Last edited: Jan 3, 2013
10. Jan 3, 2013

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.

Last edited: Jan 3, 2013
11. Jan 3, 2013

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 $\overline E$.

Last edited: Jan 3, 2013
12. Jan 3, 2013

Bachelier

Thank you. :)