Limits and accumulation points

Question

• Why is the definition of a limit point in topology the same as saying $x$ is a limit point if it is in the closure of $A\backslash \{x\}$?

---

Related: Measurable functions

---

In Euclidean space §

Definition: Limit point of a set

Let $D\subset {\mathbb{R}}^n$. We say that $a\in D$ is a limit point of $D$ if there exists a sequence $\{x_n\}\in D\backslash \{a\}$ such that $x_n\to a$.

Definition: Limit point of a function

Given a function $f:D\to {\mathbb{R}}^m$, the limit of $f(x)$ as $x\to a$ exists and is $c\in {\mathbb{R}}^m$ if for all $ϵ>0$, there exists $\delta >0$ such that

$$0<\Vert x-a\Vert <\delta ⟹\Vert f(x)-c\Vert <ϵ.$$

In this case, we write ${\lim }_{x\to a}f(x)=c$.

Limits for combinations of functions are defined as follows:

• Sums of functions: Given $f,g:D\to {\mathbb{R}}^m$, if $\text{lim}(f)=\vec{c}$ and $\text{lim}(g)=\vec{d}$ exist, then $\text{lim}(f+g)=\vec{c}+\vec{d}$ exists.
• Products of functions: given $f,g:D\to \mathbb{R}$, if $\text{lim}(f)=c$ and $\text{lim}(g)$ exists, then $\text{lim}(fg)=cd$ exists.
• Composite functions: Given $f:D\subset {\mathbb{R}}^n\to {\mathbb{R}}^m$ and $g:D^{\prime }\subset {\mathbb{R}}^m\to {\mathbb{R}}^{\ell }$ so that the composite function $g\circ f:D\to {\mathbb{R}}^{\ell }$ is defined, if $\text{lim}f(\vec{x})=\vec{c}$ as $\vec{x}\to {\vec{x}}_0$ exists and the limit for $g(\vec{y})$ as $\vec{y}\to \vec{c}$ exists, then the limit $g(f(\vec{x}))$ as $\vec{x}\to {\vec{x}}_0$ exists, and is equal to the limit $g(\vec{y})$ as $\vec{y}\to \vec{c}$.
• Composites with continuous functions: If $g$ is continuous, then ${\text{lim}}_{\vec{x}\to {\vec{x}}_0}g(f(\vec{x}))=g({\text{lim}}_{\vec{x}\to x_0}f(\vec{x}))$.

We have the following facts about limits:

• The limit on an open set does not change with a larger domain. Given an open set $U\subset D\subset {\mathbb{R}}^m$, the function $f:D\to {\mathbb{R}}^m$, and an element ${\vec{x}}_0\in U$, then the limit as $\vec{x}\to {\vec{x}}_0$ for some $\vec{x}\in U$ is the same as the limit for some $\vec{x}\in D$.

---

In general metric spaces §

Definition: Accumulation point

Given any set $E\subset X$, a point $x\in X$ is an accumulation point if for all $ϵ>0$, there exists $y\in E$ such that $y\ne x$ and the distance metric has $d(x,y)<ϵ$. That is, for all $ϵ>0$,

$$E\cap [B(x,ϵ)\backslash \{x\}]\ne \varnothing .$$

Proposition: Sequence characterization of limit points

Given a metric space $(X,d)$ and a subset $A\subseteq X$, if $x\in X$ is a limit point of $A$, then there exists a sequence $(x_n{)}_{n\in \mathbb{N}}$ in $A$ which converges to $x$.

Proposition: Sequence characterization of closed sets

Let $(X,d)$ be a metric space and $A\subseteq X$. Then $A$ is closed if and only if every convergent sequence in $A$ converges to some $a\in A$.

Link to original

---

In topological spaces §

Definition: Limit point in topological space

Given a subset $A$ of a topological space $X$, a point $x\in X$ is a limit point if every neighborhood of $x$ intersects $A$ at some point other than $x$ itself; that is, for all neighborhoods $U\subseteq X$ with $x\in U$, we have

$$A\cap (U\backslash \{x\})\ne \varnothing .$$

Equivalently, $x$ is a limit point if it is in the closure of $A\backslash \{x\}$. We write $A’$ to denote the set of all limit points of $A$ in $X$.

Transclude of (Theorem)-The-closure-of-a-set-is-the-union-of-the-set-with-its-limit-points#^f956e9

---

Review §

Definitions §

Topology

• Limit point in topological space

Exercises §

Topology

• Prove $\bar{A}=A\cup A’$ for a subset of topological space $A\subseteq X$.

Honors Mathematics B

• Use the definitions of limits and continuity, respectively, to show that a function $f:D\to {\mathbb{R}}^m$ is continuous at $x_0\in D$ if and only if ${\lim }_{x\to x_0}f(x)$ exists and is equal to $f(x_0)$. ⭐