Here's Theorem 2.27 (a) in the book Principles of Mathematical Analysis by Walter Rudin, 3rd edition:

If $X$ is a metric space and $E \subset X$, then $\overline{E}$ is closed.

Now here's Rudin's proof:

If $p \in X$ and $p \not\in \overline{E}$ 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 $\overline{E}$ is therefore open. Hence $\overline{E}$ is closed.

Is the above proof good enough, especially at the level Rudin is intended for?

Now here's the proof I propose:

If $p \in X$ and $p \not\in \overline{E}$ then $p$ is neither a point of $E$ nor a limit point of $E$. Hence $p$ has a neighborhood which does not intersect $E$. Let $N_\epsilon (p)$ be this neighborhood.

Now we show that no point of $N_\epsilon (p)$ can be in $\overline{E}$. Let $q \in N_\epsilon (p)$. Then $d(q,p) < \epsilon$, where $d$ denotes the metric on $X$.

Let $\delta \colon= \epsilon - d(p,q)$. Then $0 < \delta < \epsilon$. Now if $a \in N_\delta (q)$, then $d(a,q) < \delta = \epsilon - d(q,p)$, which implies that $$d(a,p) \leq d(a,q) + d(q,p) < \epsilon,$$ and so $a \in N_\epsilon (p)$.

Thus we have shown that $N_\delta (q) \subset N_\epsilon (p)$. Since $ N_\epsilon (p) \cap E = \emptyset$, we have $N_\delta (q) \cap E = \emptyset$ as well. That is, the point $q$ has a neighborhood --- namely $N_\delta (q)$ --- which does not intersect $E$ at all. So $q \not\in \overline{E}$.

But $q$ was an arbitrary point in $N_\epsilon (p)$. So $N_\epsilon (p) \subset \left( \overline{E} \right)^c$.

But $p$ was an arbitrary point in $\left( \overline{E} \right)^c$. Thus, we can conclude that every point of $\left( \overline{E} \right)^c$ is an interior point. Hence $\left( \overline{E} \right)^c$ is open.

Now is my proof any better than Rudin's? Are there any extra advantages to be had from inclusion or exclusion of extra details?

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    $\begingroup$ You may be right that Rudin could have supplied more details, but you may just be getting to the point where texts will supply less details for you. Usually this is done to save space when the details are rather straightforward to supply, as you have done. It also helps for clarity. $\endgroup$ – Andrew Jan 29 '16 at 16:17
  • $\begingroup$ Always relevant: abstrusegoose.com/12 $\endgroup$ – PseudoNeo Jan 29 '16 at 16:30
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    $\begingroup$ This level of detail is excessive once you're at a sufficiently high level. You could simply add the following sentence to Rudin's proof: "this neighborhood $N$ cannot intersect $\bar{E}$ either, for each of its points is contained within an open ball lying entirely within $N$." I agree, however, that for beginners some indication such as this would have been helpful. $\endgroup$ – symplectomorphic Jan 29 '16 at 16:52
  • $\begingroup$ Once you have that neighborhood you're done. What did you find insufficient about Rudin's proof? $\endgroup$ – zhw. Jan 29 '16 at 17:44
  • $\begingroup$ I don't think we can look at Rudin's proof in the abstract. It has to be intelligible to someone who has only read up to 2.27. Nonetheless, I would add just a slight bit more: "The complement of the neighborhood is closed by 2.23. It contains $E$, and being closed it contains all its own limit points (2.18(d)); therefore the complement contains all the limit points of $E$, and therefore the complement contains $\overline{E}$." $\endgroup$ – ForgotALot Jan 29 '16 at 19:52

I would say that your proof is better. I would assume that he missed a detail, and left out a proof that if it holds for $E$ then it holds for $\overline{E}$. Certainly, if I were grading a course I would mark his proof as incomplete - even in a course not for first or second years. Especially since his book is a standard introductory text for first and second years, I think this oversight is problematic.

As a commenter notes, authors have a habit of increasing the details they omit as time goes on, out of a combination of laziness, a desire to save space, and a desire to write less, but at chapter 2 of an intro book rigor should be the standard.

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    $\begingroup$ I think it's wrong to assume Rudin confuses a set with its closure when another interpretation is available: he just omits the straightforward step of checking that if a neighborhood doesn't intersect E, it can't intersect the closure of E. I agree, though, that for beginners even a short indication that this extra step is required would be helpful. $\endgroup$ – symplectomorphic Jan 29 '16 at 16:35
  • $\begingroup$ @symplectomorphic true. Changed my wording. $\endgroup$ – Stella Biderman Jan 29 '16 at 16:36

I believe that the part of the proof that confuses you would be between the two sentences

Hence $p$ has a neighborhood which does not intersect $E$.


The complement of $\overline{E}$ is therefore open.

So for now, the problem is that it is not clear why such neighborhood cannot include an element of $E'$. And if it can, then it will not be an interior point of $E^c$, then $E^c$ will not be an open set since there exists a point $p$ that is not an interior point.

In another word, we now have $\exists r>0.\,\forall q\in N_r(p).\,q\notin E$, but in order to finish the proof, we need $\exists r>0.\,\forall q\in N_r(p).\,q\notin \overline{E}$.

Do you see the logic gap here? Well, perhaps for Professor Rudin and his students, this is too trivial to even bring up, but for us who are reading the textbook and studying mathematical analysis, formalizing the proof filling up this gap is undoubtedly beneficial.

The goal is to prove the following statement.

$$ \forall r>0.\, N_r(p)\cap E=\emptyset\implies N_r(p)\cap E'=\emptyset $$

Because once this is proven, it suffices to say, after the first sentence, that such neighborhood of $p$ will not intersect $E'$ thus $\overline{E}$, too.

Now carry out the proof by reductio ad absurdum.

First, we have the assumption $N_r(p)\cap E=\emptyset$, and we assume that $N_r(p)\cap E'\neq\emptyset$.

Now the latter statement basically means that

$$ \exists q\in N_r(p).\,\exists S>0.\,\forall s\in(0,S).\,\exists e\in E.\,e\in N_s(p) $$

Denote $\xi=r-d(p,q)$, thus $N_\xi(q)\subset N_r(p)$, then take a specific $S=\xi$, there will be a $e\in E$ for any $s\in(0,\xi)$ such that $e\in N_s(q)$.

Trivially $N_s(q)\subset N_\xi(q)\subset N_r(p)$ since $s\in(0,\xi)$.

Now, we have both $e\in N_r(p)$ and $e\in E$, which contradicts the premise of the proposition we wish to prove, which concludes the reductio ad absurdum.

This part confuses me at first, also, and yes, your proof also works.

I understand what you wish to say: Each point $q$ in $N_r(p)$ is not an element of $E$, then for each of them it will not be a limit point of $E$, an element of $E'$, thus not an element of $\overline{E}$, since $\exists \xi=r-d(p,q).\, \neg\exists\mu\in E.\,\mu\in N_\xi(q)$ (for this part you are saying the same thing as I), which concludes that $N_r(p)\in\overline{E}^c$.


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