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Accumulation point: A point in a metric space where any open nbhd contains a point other than the originally considered point itself

The question I am going to ask occurred to me when I read this answer by Peter Petrov where a stricter definition of Accumulation point was used:

Accumulation point: A point in a set where every neighborhood of which contains infinitely many points of the set.

In $\mathbb{R}$ it is true that both definitions of Accumulation point are equivalent.

What is the characteristic of $\mathbb{R}$ as a metric space/ topology which causes having one point in open set imply infinitely many?

tryst with freedom
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2 Answers2

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It's important to observe that in general we can consider accumulation points not only of the whole spaces but also of their subsets. Therefore let $A\subset X$ be any subset.

Consider two statements:

  • (i) $x$ is an accumulation point of $A$ of the first type
  • (ii) $x$ is an accumulation point of $A$ of the second type.

Of course (ii) always implies (i). Now we'll show that in $T_1$ spaces (i) implies (ii).

Assume that $X\in T_1$. Let $x\in X$ satisfy (i). Assume that $x$ doesn't satisfy (ii). Then there exists an open neighbourhood $U$ of $x$ such that $B:=A\cap U\setminus\{x\}$ is finite, hence closed. Then the set $U\setminus B$ is an open neighbourhood of $x$ having no points from $A$ but $x$. A contradiction.

Now consider the space $(\Bbb R,\mathcal T)$ where $\mathcal T=\{\emptyset,\Bbb R\}\cup\{(a,\infty):a\in \Bbb R\}$. Let $A=\{1\}$ and $x=0$. Then each nbhood of $x$ has only one point from $A$, so $x$ satisfies (i) but not (ii).

Final remark It's important to distinguish the notion of accumulation point of a set and of the sequence. For example $1$ is an acc. point of the sequence $((-1)^n)_{n=1}^\infty$ but not of the set $\{(-1)^n:n\in\Bbb N\}$. Definition The point $x$ is an acc. point of the sequence $((-1)^n)_{n=1}^\infty$ iff for any open nbhood $U$ of $x$ the set of indices $\{n\in\Bbb N:x_n\in U\}$ is infinte.

Off topic: To understand why we think differently in this case (for example we consider points in $U$ and not in $U\setminus\{x\}$ and count indices rather than elements) I like thinking of the sequence $(x_n)_n$ as a multiset $\{x_n:n\in\Bbb N\}$ and in this situation the sequence $(-1)^n$ can be treated as a set $\{-1,1\}$ but with infinitely many numbers $-1,1$. Then each neighbourhood of $1$ has infinitely many elements from this set minus $\{1\}$ (one element $1$ is disposed).

Mateo
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0

Let is consider a topological space $(X, \tau) $ and $A\subset X$ .

$\textbf{Definition}$: $x\in X$ is said to be an accumulation point of $A$ if $\forall U\in\tau$ with $x\in U$ satisfy $A\cap U\setminus \{x\}\neq \emptyset$

In other words every open sets containing $x$ contains $\color{red}{a \space point}$ of $A$ other than $x$.

Consider two examples:

  1. Consider $(0,1)\subset(\Bbb{R}, \tau_{std}) $.Then any open interval $0\in (-\delta, \delta) $ contains infinitely many points of $(-\delta, \delta) $

  2. Let $X=\{a,b\}$ and $\tau=\{\emptyset,\{a\},\{a,b\},\{a,b,c\}\}$

Let $A=\{a,b\}$

Then $c$ is a limit point of $A$ as any open set containing $c$ $[$infact there is only one such open set , $\{a, b, c\}]$ contains a point other than $c$ i.e $\{a, b, c\}\cap A\setminus \{c\}=\{a, b\}$

Note: $b$ is also limit point of $A$ but $a$ is not a limit point of $A$ (verify)

In example $1$ , any open set containing the limit point contains infinitely many elements whereas in example $2$ , we have seen that open set containing the limit point is finite. Why?

$\textbf{Definition}$ : A topological space $(X, \tau) $ is said to a $T_1$-space if $\forall x\neq y\in X$ , there exists open sets $U\ni x$ and $y\ni V$ such that $y\notin U$ and $x\notin V$.

The following conditions are equivalent :

  1. $(X, \tau) $ is $T_1$ -space.

  2. $\forall x\in X ,\{x\}$ is closed.

  3. $A\subset X$ is closed whenever $A$ is finite.

$\textbf{Examples}$ : $(\Bbb{R}, \tau_{std}) $, any metric space, $(X, \tau_{\text{cofinte}}) $, $(X, \tau_{\text{cocountable}}) $ etc.

$\textbf{Non Examples}$: $(X, \tau_{\text{indiscrete}}) $, $(X=\{a, b\}, \tau=\{\emptyset,\{a\},\{a,b\}\}) $ , $(X=\{a,b\},\tau=\{\emptyset,\{a\},\{a,b\},\{a,b,c\}\})$ etc.

$\textbf{Theorem}$: Let $(X, \tau) $ be a $T_1$-space and $A\subset X$ . Then $x\in X$ is a limit point of $A$ . Then any open set containing $x$ contains an infinite subset of $A$ .

Proof: Suppose there is an open set $U$ containing $x$ such that $A\cap U\setminus \{x\}$ is finite. Then $X\setminus (U\setminus \{x\})$ is an open set containing $x$ which doesn't contain any point of $A$ . Hence it contradict that $x$ is a limit point of $A$.

Some interesting questions related to topology.

Sourav Ghosh
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