sequential criterion for boundary points in $\mathbb{R}^k$

Question

Let $E\subseteq \mathbb{R}^k$ and $x\in\mathbb{R}^k$. I need to show if $x$ is a boundary point of $E$ if and only if there is a sequence $\{p_n\}$ of elements in $E$ and a sequence of $\{q_n\}$ of limit points of $E$ such that $\lim p_n = x = \lim q_n.$

I know $x\in E$ is a boundary point iff $N(x,\epsilon)$ contains a point of $E$ as well as a point from $E^c$, now I can choose $\epsilon=1,1/2,1/3,\dots,1/n$ and construct a sequence $p_n\in E\ni p_n\to x$ but how can I choose the sequence from limit points?

Are you sure this is asked from you? If e.g. $E$ is finite then it has no limit points at all, so such sequence cannot be constructed. — drhab, Mar 04 '17 at 08:56
This is not true. What if $E$ has no limit point at all e.g. $E$ is a singleton? — Vim, Mar 04 '17 at 08:57

Emanuele Paolini · Accepted Answer · 2017-03-04T13:03:32.507

1

What is a limit point?

Every point $x\in E$ is the limit of the sequence $x_n=x\in E$. So, if you think that every point of $E$ is a limit point, just take $q_n=p_n$.

Otherwise if you define a limit point $x$ as a point which is the limit of points of $E$ different from $x$, then the statement is not true. In fact if $E$ as an isolated point $x_0$ then this point is in the boundary of $E$ but there is no sequence in $E\setminus\{x_0\}$ converging to $x_0$.

edited Mar 04 '17 at 13:03

answered Mar 04 '17 at 08:58

Emanuele Paolini

21,447

AFAIK the correct definition of the limit point is just the second one. The first one is usually called a limit, not limit point. – Vim Mar 04 '17 at 09:00
@Vim The first defines $x$ as a limit point of a sequence $(x_n)_n$ (not of set $E$). E.g. sequence $((-1)^n)_n$ has no limit (i.e. is not convergent) but has two limit points: $1$ and $-1$. – drhab Mar 04 '17 at 09:17
@drhab never mind I just think we should be very careful using the terminology "limit point". – Vim Mar 04 '17 at 09:19
@Vim Indeed. The terminology can be confusing if sets and sequences are not well discerned. – drhab Mar 04 '17 at 09:20
@drhab but ${-1, 1}$ has no limit points as a set. A limit point of a sequence is (in my book) usually called an accumulation point of the sequence and are just the subsequential limits. – Henno Brandsma Mar 04 '17 at 15:38
@HennoBrandsma I agree with that. But here for instance the cluster/accumulation points of a sequence get the name "limit point". Also have a look at the answer on this question. – drhab Mar 04 '17 at 16:28

score 1 · Answer 2 · answered Mar 04 '17 at 15:44

If $E \subseteq \mathbb{R}^n$ and $x$ is a boundary point of $E$, this means by definition for ever $r>0$, $B(x,r) \cap E \neq \emptyset$ and $B(x,r) \cap E^c \neq \emptyset$. Pick for each $n$ $p_n \in B(x, {1 \over n})$, and $q_n \in B(x, {1 \over n }) \cap E^c$. By virtue of being in these decreasing balls, we thus have a sequence $p_n$ from $E$ and a suquence of points $q_n$ from $\mathbb{R}^n \setminus E$, such that $\lim p_n = x = \lim q_n$.

If $A = \mathbb{Z}$, then all points of $\mathbb{Z}$ are boundary points but $A$ has no limit points. So then your statement becomes false.

sequential criterion for boundary points in $\mathbb{R}^k$

2 Answers2