This is a question regarding exercise 27 in chapter 2 from baby Rudin:

How does one prove that P is perfect?

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- Nov 19th 2010, 07:08 AMjefferson_lcBaby Rudin - ch 2, ex 27
This is a question regarding exercise 27 in chapter 2 from baby Rudin:

How does one prove that P is perfect? - Nov 19th 2010, 07:28 AMTheEmptySet
- Nov 19th 2010, 11:52 AMHallsofIvy
That depends on two things: (1) What is the definition of P? (2) What is the definition of "perfect"?

- Nov 19th 2010, 12:25 PMChris L T521
Here's the problem from Rudin's "Principles of Mathematical Analysis":

Quote:

Originally Posted by**Chapter 2, Exercise 27**

- Nov 21st 2010, 12:24 PMDrexel28
It should be uncountable, otherwise it's trivially true that is perfect since it' empty!

**Lemma:**Let be a second countable metric (or topological) space with countable basis and . Then, if is an open cover for then admits a countable subcover.

**Proof:**For each there exists a neighborhood containing it. But, since is a basis we may find some basic open set for which . Evidently then is an open cover for , and since it's a subset of it must be countable. Thus, for each element of choosing some for which produces a countable subcover for as desired.

So, we now claim that , as described in the problem, is perfect. To sync our notations call (the derived set) the set of all limit points of . Thus, we aim to show that .

So, to first see that let . Then, for each neighborhood of we have that there is some . But, since is a neighborhood of we see that is uncountable. Thus, since was arbitrary it follows that .

Conversely, suppose that then, there exists a neighborhood of for which every element of has a neighborhood for which clearly then is an open cover for . Thus, by our lemma it must admit a countable subcover . Note though that (this last part since by construction is countable and the countable union of countable sets is countable). and thus and thus . Thus, .

Thus, putting these two together gives .

So, now for the second part of the problem let be the countable basis for (the set of all open balls with rational radii at points of ) and let . We claim that . To see this let , then for some , but evidently then is a neighborhood of for which and so . Conversely, let , then there is some neighborhood of for which , and since is a basis we man find some for which and so evidently and so and thus . It follows that as desired. Note though that as desired (this past part gotten since by construction for each and since we have that is countable).