Can't understand a proof in Rudin

[kostas230]

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

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

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

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

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

 

[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 [itex]p[/itex] is a limit point of [itex]E'[/itex]. Then, for every neighborhood [itex]N_r (p)[/itex] there exists a limit point [itex]q[/itex] of [itex]E[/itex]. There exists a real number [itex]h>0[/itex] with [itex]d(p,q)=r-h[/itex]. Consider the neighborhood [itex]N_h (q)[/itex]

[itex]d(s,p) \leqslant d(p,q)+d(q,s) < r[/itex]

Hence, [itex]s\in N_r (p)[/itex]. Therefore, [itex]p[/itex] is a limit point of [itex]E[/itex].
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.

 

[blue_raver22]

I would approach this question by first understanding the definitions and concepts involved. In this case, we need to understand what a limit point is and how it relates to neighborhoods. A limit point of a set E is a point p for which every neighborhood of p contains a point of E other than p itself. In other words, p is a point where E "accumulates" or "approaches" in some sense.

Now, in order to prove that there exists a neighborhood of x that does not contain any limit points of E, we can use the fact that x is not a limit point of E. This means that there exists a neighborhood of x that does not intersect E. Since this neighborhood does not intersect E, it also does not contain any limit points of E. This is because if it did contain a limit point of E, then it would also intersect E.

Furthermore, since x is not a limit point of E, this means that there exists a positive distance between x and any point in E. This distance can be used to define a neighborhood of x that does not contain any limit points of E. For example, we can take a neighborhood with radius smaller than this distance. This neighborhood will only contain points that are a positive distance away from x, and therefore cannot be limit points of E.

In summary, we can prove that there exists a neighborhood that does not contain any limit points of E by using the fact that x is not a limit point of E and defining a neighborhood with a radius smaller than the distance between x and any point in E. This neighborhood will not intersect E and therefore will not contain any limit points of E.