Two Local Tests, with Different Requirements
To test whether \(x\) lies in the closure of a set \(A\), we ask whether every open neighborhood of \(x\) meets \(A\). To test whether \(x\) is an accumulation point of \(A\), we ask for more: every such neighborhood must contain a point of \(A\) other than \(x\). The distinction matters when \(x\) belongs to \(A\). A point can be in the closure simply because it is in the set, even if some neighborhood contains no other points of the set.
Throughout, \((X,\mathcal{T})\) is a topological space, \(A\subseteq X\), and a neighborhood means an open set containing the point in question. In \(\mathbb{R}\), unless stated otherwise, we use the standard topology, whose basic open sets are intervals \((a,b)\). In particular, every open neighborhood of \(x\in\mathbb{R}\) contains an interval \((x-\varepsilon,x+\varepsilon)\) for some \(\varepsilon>0\).
In the usual topology on \(\mathbb{R}\), the definition has a numerical form: \(x\) is an accumulation point of \(A\) exactly when, for every \(\varepsilon>0\), there is \(a\in A\) such that \(0<|a-x|<\varepsilon\). The strict inequality \(0<|a-x|\) excludes \(x\) itself; the other strict inequality requires the chosen point to lie inside the radius-\(\varepsilon\) interval.
Proof. Suppose first that \(x\in A\). Every open neighborhood of \(x\) contains \(x\), so it meets \(A\). By the neighborhood test for closure, \(x\in\overline{A}\). If instead \(x\) is an accumulation point of \(A\), every open neighborhood of \(x\) contains a point of \(A\) different from \(x\), and therefore meets \(A\). Again the closure test gives \(x\in\overline{A}\). Conversely, suppose \(x\in\overline{A}\) and \(x\notin A\). Every open neighborhood of \(x\) meets \(A\), by the closure test. Any point in that intersection differs from \(x\), since \(x\notin A\). Thus every open neighborhood contains a point of \(A\) different from \(x\), so \(x\) is an accumulation point. This proves both directions, including when \(A=\varnothing\): neither membership in \(A\) nor accumulation is possible, and its closure is empty. \(\square\)
Finite Sets in the Usual Real Line
A finite subset of \(\mathbb{R}\) has no accumulation points. Indeed, fix \(x\in\mathbb{R}\) and a finite set \(A\). If \(x\in A\) and \(A\setminus\{x\}\ne\varnothing\), the finitely many distances \(|a-x|\), for \(a\in A\setminus\{x\}\), are all positive. Their minimum is positive, so an interval of radius less than that minimum contains no other point of \(A\). If \(x\notin A\) and \(A\ne\varnothing\), the finitely many distances \(|a-x|\), for \(a\in A\), are positive and have a positive minimum; an interval of radius less than that minimum misses \(A\). If \(A\) is empty or \(A=\{x\}\), an interval around \(x\) contains no point of \(A\) other than possibly \(x\). In every case some neighborhood fails the accumulation-point test. This argument uses the usual real-line topology and the fact that a nonempty finite collection of positive real numbers has a positive minimum.
Worked Example: A Three-Point Set
Let \(A=\{-1,2,5\}\subseteq\mathbb{R}\). We determine its accumulation points and closure.
For the three points of \(A\), the intervals \((-2,0)\), \((1,3)\), and \((4,6)\) contain \(-1\), \(2\), and \(5\), respectively, and no other point of \(A\). In particular, for \((1,3)\) the ordering \(-1<1<2<3<5\) verifies that it contains \(2\) and excludes \(-1\) and \(5\). Thus none of the three points is an accumulation point.
If \(x\notin A\), each of the three distances \(|x+1|\), \(|x-2|\), and \(|x-5|\) is positive. Let \(r\) be half the smallest of these distances. Then \(r>0\), and the interval \((x-r,x+r)\) contains none of \(-1,2,5\). So \(x\) is not an accumulation point and is not in \(\overline{A}\).
Every point of \(A\) is in \(\overline{A}\), since every neighborhood of a point of \(A\) meets \(A\). The calculation therefore gives accumulation-point set \(\varnothing\) and closure \(\{-1,2,5\}\). Check: the three displayed intervals verify directly that no member of \(A\) is an accumulation point, while each member belongs to the closure by the neighborhood test.
The finite-set calculation should not be transferred to every topology without checking its hypotheses. For example, give \(X=\{a,b\}\) the topology \(\mathcal{T}=\{\varnothing,\{a\},X\}\). The only open neighborhood of \(b\) is \(X\), and it contains \(a\in\{a\}\) with \(a\ne b\). Hence \(b\) is an accumulation point of the finite set \(\{a\}\) in this topology. In the usual real line, by contrast, finite sets have no accumulation points. The topology determines which neighborhoods are available.
Accumulation Points of an Open Interval
Worked Example: The Interval \((2,4)\)
Let \(A=(2,4)\subseteq\mathbb{R}\). We show that its accumulation points are exactly \([2,4]\), and then determine its closure.
First take \(x\in[2,4]\) and any \(\varepsilon>0\). If \(x=2\), put \(t=\min(\varepsilon,2)/2\) and choose \(a=2+t\). Then \(t>0\), \(t\le 1\), so \(2<a\le3<4\), and \(0<|a-2|=t<\varepsilon\). If \(x=4\), use the same \(t\) and choose \(a=4-t\); then \(3\le a<4\), and \(0<|a-4|=t<\varepsilon\). If \(2<x<4\), let \(t=\min(\varepsilon/2,(x-2)/2,(4-x)/2)\) and choose \(a=x+t\). Each quantity in the minimum is positive, so \(t>0\). Also \(t\le(x-2)/2\) gives \(a\le(x+2)/2<x\) only if the sign were negative; here \(a=x+t\), so the needed upper bound follows instead from \(t\le(4-x)/2\): \(a\le(x+4)/2<4\). And \(a>x>2\), while \(t\le\varepsilon/2<\varepsilon\). Thus in every case there is \(a\in A\) with \(0<|a-x|<\varepsilon\), proving that each \(x\in[2,4]\) is an accumulation point.
Now suppose \(x<2\). The interval with centre \(x\) and radius \((2-x)/2\) lies entirely to the left of \(2\), so it misses \(A\). If \(x>4\), the interval with centre \(x\) and radius \((x-4)/2\) lies entirely to the right of \(4\), so it also misses \(A\). Neither kind of \(x\) is an accumulation point or a closure point. The accumulation-point set and closure are therefore both \([2,4]\). Check: the endpoints \(2\) and \(4\) are approached from inside \((2,4)\), while each point outside \([2,4]\) has a neighborhood disjoint from \((2,4)\).
In the interior of the interval, there is room on either side of a point; at an endpoint, points of the interval approach from only one side. The definition requires points in every neighborhood, not points on both sides. This is why the endpoints are accumulation points even though they are not members of \((2,4)\).
A Sequence and an Isolated Member
Worked Example: Reciprocal Integers and One Extra Point
Let \(A=\{1/n:n\in\mathbb{N}\}\cup\{3\}\subseteq\mathbb{R}\), where \(\mathbb{N}=\{1,2,\ldots\}\). We claim that the only accumulation point is \(0\).
For \(0\), take any \(\varepsilon>0\). Choose \(n\in\mathbb{N}\) with \(n>1/\varepsilon\). Then \(0<1/n<\varepsilon\), and \(1/n\in A\), so \(0\) is an accumulation point. For each \(m\in\mathbb{N}\), the interval centred at \(1/m\) with radius \(r_m=1/(2m(m+1))\) contains no other reciprocal \(1/n\). In fact, the nearest smaller reciprocal is \(1/(m+1)\), at distance \(1/(m(m+1))=2r_m\); when \(m>1\), the nearest larger one is \(1/(m-1)\), at distance \(1/(m(m-1))>2r_m\). For \(m=1\) there is no larger reciprocal in the set. These intervals also exclude \(3\), since \(r_m\le1/4\) and \(3-1/m\ge2\). Hence no \(1/m\) is an accumulation point. The interval \((2,4)\) meets \(A\) only at \(3\), so \(3\) is not an accumulation point either.
It remains to exclude every \(x\ne0\) that is not already covered. Choose \(N\ge2\) so large that \(1/N<|x|/2\). For \(n\ge N\), the triangle inequality gives \(|x-1/n|\ge |x|-1/n>|x|/2\). The remaining points of \(A\) are the finite set \(\{1,1/2,\ldots,1/(N-1),3\}\). Since \(x\) is not one of these points, its distance from each is positive. Take \(r>0\) smaller than \(|x|/2\) and smaller than half of each of these finitely many distances. The interval \((x-r,x+r)\) then misses the tail \(1/n\) for \(n\ge N\) and every point in the finite set. So \(x\) is not an accumulation point. The closure is \(A\cup\{0\}\), by the closure-and-accumulation theorem.
Check: \(0\notin A\) but is an accumulation point, while \(3\in A\) is not an accumulation point. The point \(3\) nevertheless belongs to the closure, because every neighborhood of \(3\) contains \(3\) itself.
A Common Mistake
A common error is to identify the closure with the accumulation-point set. For the concrete set \(\{3\}\subseteq\mathbb{R}\), that would give the wrong calculation \(\overline{\{3\}}=\varnothing\): the set has no accumulation points, since \((2,4)\) contains no point of \(\{3\}\) other than \(3\). But every neighborhood of \(3\) meets \(\{3\}\), so the closure test gives \(\overline{\{3\}}=\{3\}\). The correct relationship is that the closure consists of the points of the set together with its accumulation points. A point outside the set must be an accumulation point to lie in the closure; a point already in the set does not.
Check Your Understanding
For each question, distinguish carefully between membership in the set, accumulation-point status, and membership in the closure.
- In the usual topology on \(\mathbb{R}\), find the accumulation points and closure of \(\mathbb{Z}\).
- Let \(A=[0,1)\subseteq\mathbb{R}\). Which of its points are accumulation points, and what is \(\overline{A}\)?
- Give a finite subset of \(\mathbb{R}\) and a point in its closure that is not an accumulation point.
- In the topology \(\{\varnothing,\{a\},\{a,b\}\}\) on \(\{a,b\}\), determine whether \(b\) is an accumulation point of \(\{a\}\).
- Explain why the condition \(0<|a-x|<\varepsilon\), rather than just \(|a-x|<\varepsilon\), appears in the real-line test for accumulation points.