Topological spaces · Definition

Accumulation point

topologytopological spacesaccumulation point

Before you start you need to know

It separates the points a set “surrounds” from those it merely contains, linking limits, closures and isolated points.

Definition

A point \(x\) of a topological space \(X\) is an accumulation point of a subset \(A\) if every neighborhood \(N\) of \(x\) contains a point of \(A\) other than \(x\); these points form the derived set \(A'\).

\[x\in A'\iff \forall N\text{ neighborhood of }x:\ (N\setminus\{x\})\cap A\neq\varnothing\]

The condition requires that \(x\) be approximated by points of \(A\) different from \(x\) itself. Hence a point of \(A\) may fail to be an accumulation point (if it is isolated: some neighborhood meets \(A\) only at \(x\)) and a point outside \(A\) may be one. In usual \(\mathbb{R}\), \((0,1)'=[0,1]\), \(\{1/n\}'=\{0\}\) and \(\mathbb{Z}'=\varnothing\).

With the derived set, the closure is described directly: \(\overline{A}=A\cup A'\). Consequently, \(A\) is closed if and only if \(A'\subseteq A\), i.e. it contains all its accumulation points. The points of \(A\setminus A'\) are the isolated points of \(A\).

The behavior depends on the topology. In the discrete one every derived set is empty, because \(\{x\}\) is a neighborhood of \(x\). In the trivial one, if \(A=\{a\}\) then \(A'=X\setminus\{a\}\), and if \(A\) has at least two points, \(A'=X\). Caution: the neighborhood definition is not equivalent to “some sequence in \(A\) converges to \(x\)” except in spaces with a good local base, such as metric spaces.

In metric spaces, \(x\in A'\) exactly when every open ball around \(x\) contains infinitely many points of \(A\); in general spaces this “infinitely many” property can fail (in the trivial space a single neighborhood is already all of \(X\)).

Think about it

Can a finite subset of usual \(\mathbb{R}\) have an accumulation point?

No: if \(A\) is finite and \(x\in\mathbb{R}\), a small enough interval around \(x\) excludes the other points of \(A\), so \(A'=\varnothing\).

If \(A\subseteq B\), how are \(A'\) and \(B'\) related?

\(A'\subseteq B'\): every neighborhood of \(x\) meeting \(A\) at a point other than \(x\) also meets \(B\) at that point.

Exercises 10

References

  • James R. Munkres, Topology, 2nd edition, Pearson, 2000. §17
  • Bert Mendelson, Introduction to Topology, 3rd edition, Dover Publications, 1990.
  • Stephen Willard, General Topology, Dover reprint, Dover Publications, 2004. §6