Neighborhood of a point, the derived set, general topology and real analysis

Question

Required to prove: If Nδ(p) is a neighborhood of p that does not intersect E, show that it cannot intersect E′.

where E is a subset of R and where E' is the set made by all the accumulation points of E is called the derivative set and it is indicated with E′

Any help in this proof would be appreciated.

I have something on my mind and it's to prove this using the contrapositive. If anyone could help me supply the proof either in that way or any other way please. Thank you!

Here is my attempt: Suppose Nδ(p)∩E′. So there is some x∈Nδ(p), such that x is a limit point of E. Let ϵ<δ−d(x,p). Since x is a limit point of E, there is some e∈E such that e∈Nϵ(x). so if x is a limit point of E, then there are elements of E that are arbitrarily close to it, right? And since Nδ(p) is open, there is a smaller neighborhood Nϵ(x) such that Nϵ(x)⊂Nδ(p). since Nϵ(x)⊂Nδ(p) and e∈Nϵ(x) so e must ∈ Nδ(p). We also have that e ∈ E, and thus e is a common element of E and Nδ(p) and therefore E intersects Nδ(p). We conclude that if E intersects Nδ(p), then Nδ(p) also intersects E'. Thus proving, by the contrapositive that if If Nδ(p) is a neighborhood of p that does not intersect E, then it cannot intersect E′. Is this correct? Please can you give me any comments?

Answer 1

Yep contrapositive is a good way to go.

Just use the definition of a limit point! Assume $x\in E'\cap N_\delta(p)$. $x$ is a limit point of $E$, and $N_\delta(p)$ is an open neighborhood of $x$. Therefore...

Edit: Your proof is correct! (don't forget $N_\delta(p)\cap E'\neq\emptyset$ at the beginning) We can make it even simpler: the thing is, as you said, "then there are elements of E that are arbitrarily close to it, right", so the thing is you don't really need to take a smaller neighborhood by picking $\varepsilon<\delta-d(x,p)$. You can do simply as follows:

(By the way one thing I'm not sure of. Some people define a neighborhood of a point as an open set containing the point, while others define it as a set containing an open subset that contains the point. I'll assume you work with the definition that a neighborhood is open itself. The following still holds with the other definition)

Assume $N_\delta(p)\cap E'\neq\emptyset$. So there is $x\in N_\delta(p)$ such that $x$ is a limit point of $p$.

Recall the definition of a limit point:

$x$ is a limit point of $E$ if for every neighborhood $U$ of $x$, there exists $e\in E$ such that $e\in U$ and $e\neq x$ (in other words $(U\backslash\{x\})\cap E\neq\emptyset$).

The definition above works for any neighborhood of $x$. In particular, it works for $N_\delta(p)$, because it's an open set that contains $x$, hence a neighborhood of $x$. Therefore there exists $e\in E$ such that $e\in N_\delta(p)$ and $e\neq x$. This shows that $N_\delta(p)\cap E\neq\emptyset$.

Answer 2

So $E$ is a subset of some metric space, and $N_\delta(p)$ is a (ball) neighbourhood of $p$ that misses $E$ then $N_\delta(p)$ misses $E'$:

Suppose $q \in N_\delta(p)$. Then there is some $\delta' >0$ such that $N_{\delta'}(q) \subseteq N_\delta(p)$ (classic: take $\delta'=\delta-d(p,q)$ and apply the triangle inequality, or use the standard fact that the metric balls are open). It follows that $N_{\delta'}(q)$ misses $E$ (it is a subset of $N_\delta(p)$ that also misses $E$) and so $q \notin E'$. $q$ was arbitrary so QED.