A question about Baby Rudin Theorem 2.27 (a)

Theorem 2.27: If $X$ is a metric space and $E \subset X$, then $\bar E$ (the closure of $E$) is closed.

The proof says: If $p \in X$ and $p \not \in \bar E$ then $p$ is neither a point of $E$ nor a limit point of $E$. Hence $p$ has a neighborhood which does not intersect $E$. The complement of $\bar E$ is therefore open. Hence $\bar E$ is closed.

I'm particularly questioning about ''Hence $p$ has a neighborhood which does not intersect $E$. The complement of $\bar E$ is therefore open.'' Should we also prove that the neighborhood of $p$ also does not intersect $E'$ (the set of all limit points of $E$)?

Here's what I tried to prove, by contrapositive: ''For any $p \in {\bar E}^c$, if $N_r(p) \cap E' \ne \emptyset$ then $N_r(p) \cap E \ne \emptyset$''.

Proof: For any $p \in {\bar E}^c$, if $N_r(p) \cap E' \ne \emptyset$, then take $q \in N_r(p) \cap E'$, $\exists N_h(q)$ s.t. $N_h(q) \subset N_r(p)$. Since $q \in E'$ is a limit point, $N_h(q) \cap E \ne \emptyset$, and hence $N_r(p) \cap E \ne \emptyset$.

I'm not quite sure whether this is necessary. Or is there anything I missed from Rudin's proof?

$p$ was chosen randomly, so could be any point not in $E$, not in $E'$, as any such $p$ was explicitly ruled out as a limit point of $E$, hence the existence of a neighborhood of $p$ will not intersect $E'$.

Your proof is fine. You explained nicely why Rudin's claim follows.

Rudin is notorious for leaving some of the "links" between "stepping stones" (steps in his proofs) "to the reader". You've just filled in some of those unwritten details, and you are correct in those details. It's always a good idea to do so as you read any text, when anything is not immediately apparent to you while reading through a proof. In particular, that can often be the case when reading Rudin. When you revisit the proofs, then, that "extra work" will pay off, as you'll have made the connections, and will then be able to reread Rudin's proofs and follow them without so much effort.

Sometimes filling in the details may amount to simply "unpacking" the definition(s) of the terms being used in a theorem (what it means to be a limit point, e.g.). You'll find that's a sound way to learn the definitions inside-out, and to review theorems when they are used in subsequent proofs. (In all honesty, I personally "wrote, expanded upon, rewrote, extended, rewrote again" virtually all of Baby Rudin when I first encountered the text.

• amWhy, thanks a lot for the explanation. But I'm afraid I'm still a bit confused. I know it is stated explicitly that $p$ cannot be in $E$ and $E'$, so there is a neighborhood not intersecting with $E$ is obvious, since otherwise $p$ would be in $E'$, but how this implies that there is a neighborhood of $p$ which does not intersect with $E'$? Could you please explain a bit more? – David Tan Jun 17 '13 at 2:36
• Precisely for the reason you give. My point was simply that it follows for the reason you give, but Rudin is "leaving the details to the reader"...to apply the definitions of limit point, $E'$, neighborhoods, etc, to fill in the details as to why the claim is true. Reading and Understanding Rudin takes a lot of work. One aims to work through the text (and the "exercises" that requires), as well as aiming to complete the explicit exercises. – Namaste Jun 17 '13 at 2:48
• Don't let others fool you into thinking they never had to go through this...When you make it through Rudin, and refer back, the proofs will all seem more apparent, so for some folks here, they "are able" dismiss as "obvious" certain proofs only because they've been through such a process at one time or another, usually many times, and with many texts. And what once took them great pains to comprehend now comes much more easily. – Namaste Jun 17 '13 at 2:51
• You may find these supplementary notes (extensive) that expand tremendously on Baby Rudin useful. They are available to download in chapters, as pdf, and authored by a Professor to supplement Baby Rudin. It also includes many exercises. – Namaste Jun 17 '13 at 2:59
• I think now I got your point, simply because Rudin thought this is "obvious", but a good exercise for me. Actually I was just now looking at problem 6 of chapter 1, which states that $E'$ is closed. So if there is a neighborhood of $p$ that intersect with $E'$, then $p$ would be the limit point of $E'$, and should be in $E'$, which contradicts the assumption that $p$ is not in $E'$. Thanks a lot amWhy, for your great comments! – David Tan Jun 17 '13 at 3:00

To amplify slightly on @ncmathsadist's terse comment: It is a matter of negating a definition. If $p$ is not a limit point, then some neighborhood of $p$ must fail to intersect $E$. But then every point $q$ in this neighborhood has that same neighborhood that fails to intersect $E$, so $q\notin \bar E$.

P.S. Rudin is an ambitious choice for self-study. I continue to [complain] that he draws not a single picture, presumably as a matter of pride. Draw lots of pictures!

• Good point. I'm also self studying Rudin and I didn't stop at this proof precisely because of what you say. I must agree that drawing pictures is helping tremendously! (+1) – user70962 Jun 18 '13 at 19:30

Rudin's proof is spot on. He chose an arbitrary point and showed the existence of the desired neighborhood.