Can't understand a proof in Rudin

[kostas230]

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?

 

[verty]

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.

 

[kostas230]

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)

 

[verty]

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.