A Local Test for Closure
Closure is defined using all closed sets containing a given set. That definition describes closure globally, but a question such as “does this particular point belong to the closure?” can be answered by looking only at open neighborhoods of the point. The criterion in this tutorial makes that local test precise and applies it in several spaces where neighborhoods behave differently.
Let \((X,\mathcal{T})\) be a topological space. An open neighborhood of \(x\in X\) is an open set \(U\in\mathcal{T}\) with \(x\in U\). Recall that a set is closed if its complement is open.
The last observation matters: a point can belong to the closure simply because it already belongs to the set. The neighborhood test captures that possibility as well as the case where points of \(A\) occur nearby.
Proof. Suppose first that \(x\in\overline{A}\). Let \(U\) be any open neighborhood of \(x\). If \(U\cap A=\varnothing\), then \(A\subseteq X\setminus U\). Since \(U\) is open, \(X\setminus U\) is closed. The defining minimality of \(\overline{A}\) therefore gives \(\overline{A}\subseteq X\setminus U\). This contradicts \(x\in\overline{A}\) and \(x\in U\). Thus every such \(U\) intersects \(A\).
Conversely, suppose every open neighborhood of \(x\) intersects \(A\). If \(x\notin\overline{A}\), then \(U=X\setminus\overline{A}\) is open because \(\overline{A}\) is closed. It contains \(x\), and it is disjoint from \(A\) because \(A\subseteq\overline{A}\). This gives an open neighborhood of \(x\) that does not intersect \(A\), contrary to the supposition. Hence \(x\in\overline{A}\). \(\square\)
The criterion is a statement about every open neighborhood: one neighborhood that misses \(A\) is enough to rule out membership in \(\overline{A}\). To prove membership, by contrast, we must know that no open neighborhood misses \(A\).
Using the Criterion in the Standard Real Line
Worked Example: The Reciprocal Sequence
In \(\mathbb{R}\) with its standard topology, let \(A=\{1/n:n\in\mathbb{N}\}\). We show that \(\overline{A}=A\cup\{0\}\) by testing all real numbers.
The point \(0\). Let \(U\) be any open neighborhood of \(0\). Since the standard topology on \(\mathbb{R}\) has open intervals as a basis, there is an \(r>0\) such that \((-r,r)\subseteq U\). Choose \(n\in\mathbb{N}\) with \(n>1/r\). Then \(0<1/n<r\), so \(1/n\in U\cap A\). Every open neighborhood of \(0\) therefore intersects \(A\), and the criterion gives \(0\in\overline{A}\).
A point \(1/k\in A\). Every open neighborhood of \(1/k\) contains \(1/k\), and \(1/k\in A\). Thus each such neighborhood intersects \(A\), so \(1/k\in\overline{A}\) for every \(k\in\mathbb{N}\).
A point \(x<0\). The interval \((x-1,0)\) is an open neighborhood of \(x\) and contains no element of \(A\), since all elements \(1/n\) are positive. Thus \(x\notin\overline{A}\). If \(x>1\), the interval \((1,x+1)\) is an open neighborhood of \(x\) and also misses \(A\), whose elements are at most \(1\).
A point \(0<x<1\) not in \(A\). Then \(1/x\) is not an integer. Let \(k\) be the greatest integer less than \(1/x\). Since \(1/x>1\), \(k\geq 1\), and \(k<1/x<k+1\). Taking reciprocals gives \(1/(k+1)<x<1/k\). The open interval \((1/(k+1),1/k)\) contains \(x\) but no point of \(A\), so \(x\notin\overline{A}\). The cases above cover every real number not in \(A\cup\{0\}\).
Conclusion and check. The criterion gives \(\overline{A}=A\cup\{0\}\). In particular, \(0\) is included even though \(0\notin A\), while every \(1/k\) is included because it is in \(A\). For any real \(x\) outside this displayed set, the neighborhoods constructed above miss \(A\), confirming that no such \(x\) belongs to the closure.
This example separates two ways a point can enter the closure. The points \(1/k\) are already in \(A\); the point \(0\) is not, but every open neighborhood of it reaches \(A\). The same neighborhood test works even when the topology is not the standard one on the real line.
When Neighborhoods Are Large
Worked Example: The Finite-Complement Topology
Give \(X=\mathbb{R}\) the finite-complement topology: its open sets are \(\varnothing\) and all subsets whose complement in \(X\) is finite. This is a topology. An arbitrary union that is nonempty contains a nonempty open set, whose complement is finite; the complement of the union is a subset of that finite set. A finite intersection of nonempty open sets has finite complement, since its complement is a finite union of finite sets. Intersections or unions involving the empty set also satisfy the topology axioms.
First take \(A=\mathbb{N}=\{1,2,\dots\}\). Let \(x\in X\) and let \(U\) be an open neighborhood of \(x\). Since \(U\ne\varnothing\), its complement \(X\setminus U\) is finite. If \(U\cap\mathbb{N}=\varnothing\), then \(\mathbb{N}\subseteq X\setminus U\), which would make \(\mathbb{N}\) finite. This is impossible. Hence every open neighborhood of every \(x\in X\) intersects \(\mathbb{N}\), and the criterion gives \(\overline{\mathbb{N}}=X=\mathbb{R}\).
Now take \(B=\{2,4,6\}\). Each point of \(B\) belongs to \(\overline{B}\), because every open neighborhood of that point contains the point itself. If \(x\notin B\), the set \(U=X\setminus B\) is open: its complement \(B\) is finite. It contains \(x\) and has empty intersection with \(B\), so \(x\notin\overline{B}\). Therefore \(\overline{B}=B\).
Check. The same test gives different behavior for the two sets: every neighborhood meets the infinite set \(\mathbb{N}\), while each point outside \(B\) has a neighborhood that avoids the finite set \(B\). The conclusion depends on the topology, not just on the underlying set \(\mathbb{R}\).
A Small Space with Unequal Neighborhoods
Worked Example: The Sierpinski Space
Let \(X=\{a,b\}\) with topology \(\mathcal{T}=\{\varnothing,\{a\},X\}\). These sets form a topology: the empty set and \(X\) are included, and unions and finite intersections of sets in the list remain in the list. Compute the closures of \(\{a\}\) and \(\{b\}\) using the neighborhood criterion.
For \(A=\{a\}\), every open neighborhood of \(a\) is either \(\{a\}\) or \(X\); both intersect \(A\). The only open neighborhood of \(b\) is \(X\), which also intersects \(A\). Thus every point of \(X\) passes the criterion, and \(\overline{\{a\}}=X\).
For \(B=\{b\}\), every open neighborhood of \(b\) is \(X\), so it intersects \(B\). But \(\{a\}\) is an open neighborhood of \(a\) and \(\{a\}\cap B=\varnothing\). Thus \(b\in\overline{B}\) and \(a\notin\overline{B}\), giving \(\overline{\{b\}}=\{b\}\).
Check. The asymmetry follows directly from the listed open sets: \(b\) has no open neighborhood smaller than \(X\), whereas \(a\) has the neighborhood \(\{a\}\). That neighborhood meets \(\{a\}\) but avoids \(\{b\}\).
Closure Membership Is Not Limit-Point Membership
A limit point of \(A\) is a point \(x\) such that every open neighborhood of \(x\) contains a point of \(A\) different from \(x\). The open-neighborhood criterion for closure does not require the point found in a neighborhood to be different from \(x\). A point of \(A\) can therefore belong to \(\overline{A}\) without being a limit point of \(A\).
Worked Example: A Singleton in a Discrete Space
Let \(X=\{p,q\}\) have the discrete topology, in which every subset of \(X\) is open, and set \(A=\{p\}\). Every open neighborhood of \(p\) contains \(p\), so it intersects \(A\). The open neighborhood \(\{q\}\) of \(q\) misses \(A\). The criterion gives \(\overline{A}=\{p\}\).
However, \(\{p\}\) is an open neighborhood of \(p\) containing no point of \(A\) different from \(p\). Thus \(p\) is not a limit point of \(A\). The same point belongs to the closure and fails to be a limit point.
Common mistake. A tempting but incorrect test is to require every neighborhood of \(x\) to contain a point of \(A\) different from \(x\) before concluding \(x\in\overline{A}\). Applying that test here would wrongly exclude \(p\) from \(\overline{\{p\}}\), even though every open neighborhood of \(p\) contains \(p\in A\). The correction is to use the criterion as stated: for closure, the neighborhood must intersect \(A\); the extra “different from \(x\)” condition belongs to the definition of a limit point.
The practical procedure is short: fix a point, consider an arbitrary open neighborhood when proving it lies in the closure, and find one open neighborhood that misses the set when proving it does not. The topology determines which neighborhoods must be tested, so familiar geometric pictures are not a substitute for the definition.
Check Your Understanding
Use open neighborhoods to justify each conclusion.
- In the standard topology on \(\mathbb{R}\), determine whether \(0\) belongs to the closure of \(\{1/n:n\in\mathbb{N}\}\).
- In the finite-complement topology on \(\mathbb{R}\), find the closure of \(\mathbb{Z}\).
- In the Sierpinski space, which open neighborhood of \(a\) shows that \(a\notin\overline{\{b\}}\)?
- In a discrete space \(X\), find \(\overline{A}\) for an arbitrary subset \(A\subseteq X\).
- State the difference between the neighborhood tests for closure membership and being a limit point.