# Can't understand a proof in Rudin

I can't understand a proof in the Theorem 2.27, part (a).

If $X$ is a metric space, $E$ a subspace of X, and $E'$ the set of all limit points of $E$, we denote by $\bar{E}$ the set: $\bar{E}=E \cup E'$

We need to prove that $\bar{E}$ is a closed set. Rudin's proof is this:

If $x\in X$ and $x \notin \bar{E}$ then $p$ is neither a point of $E$ nor a limit point. Hence, $p$ has a neighborhood which does not intersect $E$. Therefore, the complement of $\bar{E}$ is closed.

My question is this: How do we prove that there exists a neighborhood that does not contain any limit points of $E$?

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verty
Homework Helper
If every open ball at x contains a limit point of E, x would be a limit point of those limit points. Can you prove that it is contradictory for an open ball at x to be disjoint from E?

Edit: I changed "neighborhood" to "open ball", I think neighborhood is wrong although Rudin may define it differently.

I found a more straightforward proof. Suppose $p$ is a limit point of $E'$. Then, for every neighborhood $N_r (p)$ there exists a limit point $q$ of $E$. There exists a real number $h>0$ with $d(p,q)=r-h$. Consider the neighborhood $N_h (q)$

$d(s,p) \leqslant d(p,q)+d(q,s) < r$

Hence, $s\in N_r (p)$. Therefore, $p$ is a limit point of $E$.
Did I get it right? xD
(BTW, I should mention that I'm a physics undergrad; no training in rigorous math except a linear algebra self study with Axler. So, I'm a little slow in this. xD)

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verty
Homework Helper
That works. It is more technical than what I was alluding to above but it's fine.

You should probably mention that s is a point in E.

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