When Does a Point Accumulate Near a Set?
The closure of a set records points whose every neighborhood meets the set. But that test does not distinguish between a neighborhood meeting the set only at the point itself and meeting it at other points as well. Limit points make this distinction. They capture genuine accumulation around a point, whether or not the point belongs to the set.
We use the neighborhood test for closure from Interior and Closure Topologically. The key adjustment is to remove the point being tested from the set first: a point is a limit point when every one of its neighborhoods meets what remains.
The definition does not require \(x\in E\). A point outside \(E\) can be a limit point if points of \(E\) occur in every neighborhood of it. Conversely, a point of \(E\) need not be a limit point: it may have a neighborhood that contains no other points of \(E\).
A Neighborhood-Base Test
Often a topology is described using a neighborhood base at each point rather than all neighborhoods. It is enough to test the members of that base.
Proof. If \(x\in E'\), then every neighborhood of \(x\), including every member of \(\mathcal{B}_x\), meets \(E\setminus\{x\}\). Conversely, suppose every member of \(\mathcal{B}_x\) meets \(E\setminus\{x\}\). Given any neighborhood \(N\) of \(x\), the base property supplies \(B\in\mathcal{B}_x\) with \(B\subseteq N\). Since \(B\) contains a point of \(E\setminus\{x\}\), so does \(N\). Thus every neighborhood of \(x\) meets \(E\setminus\{x\}\), and \(x\in E'\). \(\square\)
Worked Example: A Sequence of Points in the Real Line
Give \(\mathbb{R}\) its usual topology and take \(E=\{1/n:n\geq 1\}\). The point \(0\) is a limit point: every open interval around \(0\) contains \(1/n\) for all sufficiently large \(n\), and each \(1/n\) is different from \(0\).
No point \(1/k\) is a limit point of \(E\). For \(k=1\), the interval \((3/4,5/4)\) meets \(E\) only at \(1\). For \(k\geq 2\), the neighboring values \(1/(k-1)\) and \(1/(k+1)\) lie on either side of \(1/k\). The positive distances to them are $$ \frac{1}{k-1}-\frac{1}{k}=\frac{1}{k(k-1)} \qquad\text{and}\qquad \frac{1}{k}-\frac{1}{k+1}=\frac{1}{k(k+1)}. $$
An open interval centered at \(1/k\) with radius smaller than both distances contains no other point of \(E\). Finally, let \(x\neq 0\) and suppose \(x\notin E\). Choose a small interval around \(x\) that stays a positive distance from \(0\). Only finitely many points \(1/n\) can lie in that interval, because \(1/n\) tends to \(0\). Since none of those finitely many points equals \(x\), the interval can be narrowed to avoid them all. Thus \(x\notin E'\), and \(E'=\{0\}\).
Limit Points and Closure
The closure test gives a direct relationship between the closure and the derived set. If \(x\) is in the closure but not in \(E\), every neighborhood must meet \(E\) at a point other than \(x\). If \(x\) belongs to \(E\), it is already accounted for, whether or not it is a limit point.
Proof. Suppose \(x\in\overline{E}\). If \(x\in E\), then \(x\in E\cup E'\). If \(x\notin E\), the closure test says that every neighborhood of \(x\) intersects \(E\). Since \(x\notin E\), each such intersection contains a point different from \(x\). Hence \(x\in E'\), so again \(x\in E\cup E'\).
For the reverse inclusion, \(E\subseteq\overline{E}\) by the closure properties established earlier. If \(x\in E'\), every neighborhood of \(x\) meets \(E\setminus\{x\}\), and therefore meets \(E\). The closure test gives \(x\in\overline{E}\). This proves the identity.
A set is closed exactly when it equals its closure. By the identity just proved, \(\overline{E}=E\) exactly when \(E\cup E'=E\), which is equivalent to \(E'\subseteq E\). \(\square\)
Worked Example: Infinite Sets in the Cofinite Topology
Let \(X\) be infinite with the cofinite topology, and let \(E\subseteq X\). If \(E\) is infinite, then every neighborhood \(U\) of any \(x\in X\) has finite complement (unless \(U=X\), which also has empty complement). The set \(E\setminus\{x\}\) is infinite, so it cannot be contained in the finite set \(X\setminus U\). Thus \(U\) meets \(E\setminus\{x\}\), and every \(x\in X\) is a limit point: \(E'=X\).
If \(E\) is finite, then for each \(x\in X\), the set \(U=X\setminus(E\setminus\{x\})\) is open and contains \(x\). It is disjoint from \(E\setminus\{x\}\), so \(x\notin E'\). Therefore finite sets have empty derived set in this topology. These conclusions include \(E=\varnothing\), whose derived set is empty.
Isolated Points and the Importance of \(T_1\)
A point \(x\in E\) is called an isolated point of \(E\) if some neighborhood of \(x\) meets \(E\) only at \(x\). For points that belong to \(E\), this is exactly the failure to be a limit point: \(x\in E\) is isolated in \(E\) if and only if \(x\notin E'\). The qualification \(x\in E\) matters, since a point outside \(E\) cannot be an isolated point of \(E\) under this definition.
Worked Example: A Derived Set Need Not Be Closed
Let \(X\) be an indiscrete space with at least two points, and choose \(a\in X\). The only neighborhood of any point is \(X\). Take \(E=\{a\}\). At \(a\), no neighborhood contains a point of \(E\setminus\{a\}=\varnothing\), so \(a\notin E'\). At each \(x\neq a\), the neighborhood \(X\) contains \(a\in E\setminus\{x\}\), so \(x\in E'\). Therefore $$ E'=X\setminus\{a\}. $$
This is a nonempty proper subset of \(X\), and the only closed sets in an indiscrete space are \(\varnothing\) and \(X\). Hence \(E'\) is not closed. This example warns against assuming that the derived set must always be closed in an arbitrary topological space.
A useful positive result holds when points can be separated from one another in the sense of the \(T_1\) axiom: every singleton is closed. In that setting, the derived set is closed.
Proof. Take \(x\notin E'\). By the definition of limit point, there is an open neighborhood \(U\) of \(x\) such that \(U\cap(E\setminus\{x\})=\varnothing\). We show that \(U\) contains no point of \(E'\). The point \(x\) itself is not in \(E'\) by assumption. If \(y\in U\) and \(y\neq x\), the \(T_1\) property makes \(X\setminus\{x\}\) open. Thus \(U\cap(X\setminus\{x\})\) is an open neighborhood of \(y\). It contains no point of \(E\): it excludes \(x\), and the rest of \(U\) contains no point of \(E\) by the choice of \(U\). Therefore \(y\notin E'\). We have shown \(U\cap E'=\varnothing\). Every point outside \(E'\) consequently has an open neighborhood disjoint from \(E'\), so \(X\setminus E'\) is open and \(E'\) is closed. \(\square\)
The \(T_1\) hypothesis is doing real work: it lets us remove \(x\) from \(U\) while keeping an open neighborhood of \(y\). The indiscrete example shows that the conclusion can fail without this separation property. In particular, since every Hausdorff space is \(T_1\), the theorem applies in Hausdorff spaces as well.
Viewing a Set as a Subspace
Limit points can be calculated either in the ambient space or within a subspace, provided the set and the point being tested lie in that subspace. This follows from the form of subspace neighborhoods.
Proof. Every open neighborhood of \(x\) in \(A\) has the form \(A\cap U\), where \(U\) is open in \(X\) and contains \(x\). Since \(E\subseteq A\), a point of \((A\cap U)\cap(E\setminus\{x\})\) is exactly a point of \(U\cap(E\setminus\{x\})\). If \(x\) is a limit point in \(X\), every such \(U\) contains such a point, so every subspace neighborhood does too. Conversely, if every subspace neighborhood meets \(E\setminus\{x\}\), then for any open neighborhood \(U\) of \(x\) in \(X\), the subspace neighborhood \(A\cap U\) meets \(E\setminus\{x\}\). The point it contains also lies in \(U\). Hence \(x\) is a limit point in \(X\). \(\square\)
Why the Distinction Matters
The closure identity separates two ways a point can belong to \(\overline{E}\): it may already be in \(E\), or it may be a limit point of \(E\). In a \(T_1\) space, the theorem that \(E'\) is closed gives a further useful consequence. If \(E\) is closed, then \(E'\subseteq E\), so every limit point of \(E\) belongs to \(E\); the set can still contain isolated points that are not limit points.
Do not treat “belongs to the closure” and “is a limit point” as interchangeable. A singleton in a discrete space is closed and equals its closure, but it has no limit points: each point has an open neighborhood containing only itself. In contrast, every point is a limit point of an infinite subset in the cofinite topology. These examples show that accumulation is determined by the topology, not merely by the set’s size or by an assumed notion of distance.
Check Your Understanding
Use the neighborhood definition and the results above to answer the following questions.
- What must every neighborhood of \(x\) contain for \(x\) to be a limit point of \(E\)?
- Why is \(\overline{E}=E\cup E'\), and what does this imply about closed sets?
- In the cofinite topology on an infinite space, what is the derived set of an infinite subset? What about a finite subset?
- Why can the derived set fail to be closed in an indiscrete space?
- Where does the \(T_1\) property enter the proof that a derived set is closed?
- If \(E\subseteq A\) and \(x\in A\), does viewing \(E\) inside \(A\) change whether \(x\) is a limit point? Explain.