Accumulation point

Relation between accumulation point of a sequence and accumulation point of a set

Every sequence ${\displaystyle x_{\bullet }={\left(x_n\right)}_{n=1}^{\mathrm{\infty }}}$ in ${\displaystyle X}$ is by definition just a map ${\displaystyle x_{\bullet }:\mathbb{N}\to X}$ so that its image ${\displaystyle \mathrm{Im}{x}_{\bullet }:=\left\{x_n:n\in \mathbb{N}\right\}}$ can be defined in the usual way.

• If there exists an element ${\displaystyle x\in X}$ that occurs infinitely many times in the sequence, ${\displaystyle x}$ is an accumulation point of the sequence. But ${\displaystyle x}$ need not be an accumulation point of the corresponding set ${\displaystyle \mathrm{Im}{x}_{\bullet }.}$ For example, if the sequence is the constant sequence with value ${\displaystyle x,}$ we have ${\displaystyle \mathrm{Im}{x}_{\bullet }=\{x\}}$ and ${\displaystyle x}$ is an isolated point of ${\displaystyle \mathrm{Im}{x}_{\bullet }}$ and not an accumulation point of ${\displaystyle \mathrm{Im}{x}_{\bullet }.}$

• If no element occurs infinitely many times in the sequence, for example if all the elements are distinct, any accumulation point of the sequence is an ${\displaystyle \omega }$-accumulation point of the associated set ${\displaystyle \mathrm{Im}{x}_{\bullet }.}$

Conversely, given a countable infinite set ${\displaystyle A\subseteq X}$ in ${\displaystyle X,}$ we can enumerate all the elements of ${\displaystyle A}$ in many ways, even with repeats, and thus associate with it many sequences ${\displaystyle x_{\bullet }}$ that will satisfy ${\displaystyle A=\mathrm{Im}{x}_{\bullet }.}$

• Any ${\displaystyle \omega }$-accumulation point of ${\displaystyle A}$ is an accumulation point of any of the corresponding sequences (because any neighborhood of the point will contain infinitely many elements of ${\displaystyle A}$ and hence also infinitely many terms in any associated sequence).

• A point ${\displaystyle x\in X}$ that is not an ${\displaystyle \omega }$-accumulation point of ${\displaystyle A}$ cannot be an accumulation point of any of the associated sequences without infinite repeats (because ${\displaystyle x}$ has a neighborhood that contains only finitely many (possibly even none) points of ${\displaystyle A}$ and that neighborhood can only contain finitely many terms of such sequences).

Properties

Every limit of a non-constant sequence is an accumulation point of the sequence. And by definition, every limit point is an adherent point.

The closure ${\displaystyle \mathrm{cl}(S)}$ of a set ${\displaystyle S}$ is a disjoint union of its limit points ${\displaystyle L(S)}$ and isolated points ${\displaystyle I(S)}$; that is, $${\displaystyle \mathrm{cl}(S)=L(S)\cup I(S)\text{and}L(S)\cap I(S)=\mathrm{\varnothing }.}$$

A point ${\displaystyle x\in X}$ is a limit point of ${\displaystyle S\subseteq X}$ if and only if it is in the closure of ${\displaystyle S\setminus \{x\}.}$

Proof

We use the fact that a point is in the closure of a set if and only if every neighborhood of the point meets the set. Now, ${\displaystyle x}$ is a limit point of ${\displaystyle S,}$ if and only if every neighborhood of ${\displaystyle x}$ contains a point of ${\displaystyle S}$ other than ${\displaystyle x,}$ if and only if every neighborhood of ${\displaystyle x}$ contains a point of ${\displaystyle S\setminus \{x\},}$ if and only if ${\displaystyle x}$ is in the closure of ${\displaystyle S\setminus \{x\}.}$

If we use ${\displaystyle L(S)}$ to denote the set of limit points of ${\displaystyle S,}$ then we have the following characterization of the closure of ${\displaystyle S}$: The closure of ${\displaystyle S}$ is equal to the union of ${\displaystyle S}$ and ${\displaystyle L(S).}$ This fact is sometimes taken as the definition of closure.

Proof

("Left subset") Suppose ${\displaystyle x}$ is in the closure of ${\displaystyle S.}$ If ${\displaystyle x}$ is in ${\displaystyle S,}$ we are done. If ${\displaystyle x}$ is not in ${\displaystyle S,}$ then every neighbourhood of ${\displaystyle x}$ contains a point of ${\displaystyle S,}$ and this point cannot be ${\displaystyle x.}$ In other words, ${\displaystyle x}$ is a limit point of ${\displaystyle S}$ and ${\displaystyle x}$ is in ${\displaystyle L(S).}$

("Right subset") If ${\displaystyle x}$ is in ${\displaystyle S,}$ then every neighbourhood of ${\displaystyle x}$ clearly meets ${\displaystyle S,}$ so ${\displaystyle x}$ is in the closure of ${\displaystyle S.}$ If ${\displaystyle x}$ is in ${\displaystyle L(S),}$ then every neighbourhood of ${\displaystyle x}$ contains a point of ${\displaystyle S}$ (other than ${\displaystyle x}$), so ${\displaystyle x}$ is again in the closure of ${\displaystyle S.}$ This completes the proof.

A corollary of this result gives us a characterisation of closed sets: A set ${\displaystyle S}$ is closed if and only if it contains all of its limit points.

Proof

Proof 1: ${\displaystyle S}$ is closed if and only if ${\displaystyle S}$ is equal to its closure if and only if ${\displaystyle S=S\cup L(S)}$ if and only if ${\displaystyle L(S)}$ is contained in ${\displaystyle S.}$

Proof 2: Let ${\displaystyle S}$ be a closed set and ${\displaystyle x}$ a limit point of ${\displaystyle S.}$ If ${\displaystyle x}$ is not in ${\displaystyle S,}$ then the complement to ${\displaystyle S}$ comprises an open neighbourhood of ${\displaystyle x.}$ Since ${\displaystyle x}$ is a limit point of ${\displaystyle S,}$ any open neighbourhood of ${\displaystyle x}$ should have a non-trivial intersection with ${\displaystyle S.}$ However, a set can not have a non-trivial intersection with its complement. Conversely, assume ${\displaystyle S}$ contains all its limit points. We shall show that the complement of ${\displaystyle S}$ is an open set. Let ${\displaystyle x}$ be a point in the complement of ${\displaystyle S.}$ By assumption, ${\displaystyle x}$ is not a limit point, and hence there exists an open neighbourhood ${\displaystyle U}$ of ${\displaystyle x}$ that does not intersect ${\displaystyle S,}$ and so ${\displaystyle U}$ lies entirely in the complement of ${\displaystyle S.}$ Since this argument holds for arbitrary ${\displaystyle x}$ in the complement of ${\displaystyle S,}$ the complement of ${\displaystyle S}$ can be expressed as a union of open neighbourhoods of the points in the complement of ${\displaystyle S.}$ Hence the complement of ${\displaystyle S}$ is open.

No isolated point is a limit point of any set.

Proof

If ${\displaystyle x}$ is an isolated point, then ${\displaystyle \{x\}}$ is a neighbourhood of ${\displaystyle x}$ that contains no points other than ${\displaystyle x.}$

A space ${\displaystyle X}$ is discrete if and only if no subset of ${\displaystyle X}$ has a limit point.

Proof

If ${\displaystyle X}$ is discrete, then every point is isolated and cannot be a limit point of any set. Conversely, if ${\displaystyle X}$ is not discrete, then there is a singleton ${\displaystyle \{x\}}$ that is not open. Hence, every open neighbourhood of ${\displaystyle \{x\}}$ contains a point ${\displaystyle y\ne x,}$ and so ${\displaystyle x}$ is a limit point of ${\displaystyle X.}$

If a space ${\displaystyle X}$ has the trivial topology and ${\displaystyle S}$ is a subset of ${\displaystyle X}$ with more than one element, then all elements of ${\displaystyle X}$ are limit points of ${\displaystyle S.}$ If ${\displaystyle S}$ is a singleton, then every point of ${\displaystyle X\setminus S}$ is a limit point of ${\displaystyle S.}$

Proof

As long as ${\displaystyle S\setminus \{x\}}$ is nonempty, its closure will be ${\displaystyle X.}$ It is only empty when ${\displaystyle S}$ is empty or ${\displaystyle x}$ is the unique element of ${\displaystyle S.}$