# Prove that boundary points are limit points.

Given the following definition, can someone help me figure out how to prove all boundary points are limits points?

Limit point: Let $A$ be a subset of a metric space $(X,d)$. A point $x\in X$ is called a limit point of $A$ if for every $r>0, B_r(x)\cap A\setminus\{x\}$ is not equal to the empty set.

A point $x$ in a subset $A$ of a metric space $(X,d)$ is an interior point of $A$ if there exists $r>0$ such that $B_r(x)\subseteq A$. The set of interior points of $A$ is denoted $\mathrm {int}(A)$. The exterior $\mathrm {ext}(A)$ of $A$ is defined to be the interior of the complement of $A$. The boundary of $A$ is the set of points which are neither in $\mathrm {int}(A)$ or in $\mathrm {ext}(A)$.

I am trying to figure out how to prove:

$$y \text{ is a limit point of }A \iff y \text{ is a boundary point of }A.$$

Thanks so much.

• Interior points of $A$ can be (often are) limit points of $A$, but aren't boundary points. Isolated points of $A$ can be (often are) boundary points, but aren't limit points. Commented Feb 18, 2017 at 15:38

Neither implication between limit points and boundary points holds: In $A = [0,1]$, usual topology, all points of $[0,1]$ are limit points of $A$, but only 2 are boundary points, namely $0$ and $1$. And in $A= \mathbb{Z} \subset \mathbb{R}$, also in the usual topology, all points of $A$ are boundary points but none of them is a limit point as $B(n, 1) \cap A = \{n\}$ for all $n \in \mathbb{Z}$.

This result is not true.

Take $A=\{1,2\}\subset \mathbb R$.

Then $1$ is not a limit point since for $r<1$ $$B_r(1)\cap A\setminus \{1\}=\emptyset.$$

But it is a boundary point since the exterior of $A$ is $\mathbb R\setminus\{1,2\}$ and the interior of $A$ is empty.

I think you just have an implication:

$$y\text{ is a limit point of }A\implies y\text { is a boundary point of } A.$$

• The complement of A is R\{1,2}. And according to my definitions, the exterior ext(A) of A is the interior of the complement. So it should be the interior of R\{1,2}. And that is int(R\{1,2}).
– user369792
Commented Feb 18, 2017 at 15:29
• Can you give a counter-example, please?
– user369792
Commented Feb 18, 2017 at 15:33
• I am sorry can you please explain your notation a little bit?
– user369792
Commented Feb 18, 2017 at 15:36
• @ButterMath Yes, but the interior of $\mathbb R\setminus\{1,2\}$ is exactly $\mathbb R\setminus\{1,2\}$. Commented Feb 18, 2017 at 15:40
• @E.Joseph Thanks so much. I got it now.
– user369792
Commented Feb 18, 2017 at 15:49