Isolation Is Relative to a Subset
A point can be isolated within a subset even when it is not isolated in the whole space. The distinction is important: neighborhoods are open in the ambient space, but we test them by how they intersect the subset. This gives a local criterion for isolated points and a point-by-point way to recognize discrete subspaces.
Let \((X,\mathcal{T})\) be a topological space and let \(A\subseteq X\). An open neighborhood of \(x\in X\) is an open set \(U\in\mathcal{T}\) containing \(x\). Recall that \(x\) is a limit point of \(A\) if every open neighborhood of \(x\) contains a point of \(A\) different from \(x\).
The condition \(x\in A\) is part of the definition: a point outside \(A\) cannot be an isolated point of \(A\). The neighborhood \(U\) need not be the singleton \(\{x\}\). It only needs to contain no other points of \(A\). In particular, isolation is a property of the pair \(x\in A\), not just of the ambient space \(X\).
The Punctured-Neighborhood Test
Proof. Suppose first that \(x\) is isolated in \(A\). By definition, there is an open neighborhood \(U\) with \(U\cap A=\{x\}\). Removing \(x\) from \(A\) therefore leaves no point in \(U\), so \(U\cap(A\setminus\{x\})=\varnothing\). Conversely, suppose an open neighborhood \(U\) of \(x\) satisfies \(U\cap(A\setminus\{x\})=\varnothing\). Since \(x\in U\) and \(x\in A\), we have \(x\in U\cap A\). The assumed emptiness says that no other point of \(A\) lies in \(U\). Thus \(U\cap A=\{x\}\), and \(x\) is isolated in \(A\). \(\square\)
This criterion is the direct opposite of the neighborhood test for a limit point. For \(x\in A\), a limit point has other points of \(A\) in every open neighborhood, whereas an isolated point has at least one open neighborhood containing no other points of \(A\).
Proof. By definition, \(x\in A'\) means that every open neighborhood of \(x\) contains a point of \(A\) different from \(x\). Its negation is that some open neighborhood contains no such point. Since \(x\in A\), the punctured-neighborhood test says exactly that this is the condition for \(x\) to be isolated in \(A\). \(\square\)
Discrete Subspaces and Their Neighborhoods
The subspace topology on \(A\subseteq X\) consists of the sets \(A\cap U\), where \(U\) is open in \(X\). A topology on \(A\) is called discrete when every subset of \(A\) is open; equivalently, every singleton in \(A\) is open. The next result connects this definition with isolation in the ambient space.
Proof. Suppose first that \(A\) is discrete. For each \(x\in A\), the singleton \(\{x\}\) is open in the subspace topology. By the definition of that topology, there is an open set \(U\in\mathcal{T}\) such that \(A\cap U=\{x\}\). Since \(x\in A\cap U\), this \(U\) is an open neighborhood of \(x\) in \(X\), and \(x\) is isolated in \(A\).
Conversely, suppose every \(x\in A\) is isolated in \(A\). For each \(x\in A\), choose an open set \(U_x\) in \(X\) with \(A\cap U_x=\{x\}\). Then \(\{x\}\) is open in the subspace topology. Every subset \(B\subseteq A\) is the union of its singletons, \(B=\bigcup_{x\in B}\{x\}\), so it is open in the subspace topology. If \(A=\varnothing\), its only subset is \(\varnothing\), which is open, and both sides of the equivalence hold vacuously. Hence \(A\) is discrete. \(\square\)
The theorem says that every point of a discrete subspace has its own isolating neighborhood. It does not require one neighborhood to work for all points at once. Nor does it say that \(A\) is open in \(X\): discreteness concerns the topology inherited by \(A\).
Worked Example: The Set of Reciprocals
Give \(A=\{1/n:n\in\mathbb{N}\}\subseteq\mathbb{R}\) the subspace topology inherited from the standard topology on \(\mathbb{R}\). Determine which points of \(A\) are isolated, and examine the point \(0\).
Fix \(n\ge2\). The closest member of \(A\) below \(1/n\) is \(1/(n+1)\), at distance \[ \frac1n-\frac1{n+1}=\frac{1}{n(n+1)}. \] The closest member above \(1/n\) is \(1/(n-1)\), at distance \[ \frac1{n-1}-\frac1n=\frac{1}{n(n-1)}. \] The second distance is larger because \(n-1<n+1\). Choose the open interval centered at \(1/n\) with radius \(1/[2n(n+1)]\). Every other member of \(A\) is at distance at least \(1/[n(n+1)]\), twice this radius. Thus the interval intersects \(A\) only at \(1/n\), so \(1/n\) is isolated in \(A\).
For \(n=1\), every other member of \(A\) is at most \(1/2\), so its distance from \(1\) is at least \(1/2\). The interval \((3/4,5/4)\), of radius \(1/4\) around \(1\), therefore intersects \(A\) only at \(1\). This proves that every point of \(A\) is isolated and that \(A\) is a discrete subspace.
The point \(0\) is not in \(A\), so it is not an isolated point of \(A\). It is, however, a limit point: every open neighborhood of \(0\) contains an interval \((-\varepsilon,\varepsilon)\) for some \(\varepsilon>0\), and the Archimedean property gives an \(n\in\mathbb{N}\) with \(n>1/\varepsilon\). Then \(0<1/n<\varepsilon\), so the neighborhood contains a point of \(A\).
Check: at \(n=3\), the radius specified above is \(1/[2\cdot3\cdot4]=1/24\), giving the interval \((7/24,9/24)\) around \(1/3=8/24\). The adjacent terms are \(1/4=6/24\) and \(1/2=12/24\), both outside the interval; all other terms lie still farther away. This verifies the isolating-neighborhood calculation in a specific case.
Worked Examples in Other Topologies
Worked Example: A Dense Subset of the Real Line
Determine the isolated points of \(A=\mathbb{Q}\) in \(\mathbb{R}\) with its standard topology.
Take any \(q\in\mathbb{Q}\), and let \(U\) be any open neighborhood of \(q\). There is an \(\varepsilon>0\) such that \((q-\varepsilon,q+\varepsilon)\subseteq U\). Choose \(m\in\mathbb{N}\) with \(m>1/\varepsilon\). Then \(0<1/m<\varepsilon\), and \(q+1/m\) is rational, differs from \(q\), and lies in \(U\). Thus every open neighborhood of \(q\) contains another point of \(\mathbb{Q}\). No rational is isolated in \(\mathbb{Q}\), so this subspace is not discrete.
Check: the argument applies to each \(q\in\mathbb{Q}\), not just to a particular rational: adding the rational number \(1/m\) keeps the point rational, and the choice \(m>1/\varepsilon\) places it inside the given neighborhood.
Worked Example: Finite and Infinite Sets in the Finite-Complement Topology
Let \(X=\mathbb{R}\) with the finite-complement topology: its open sets are \(\varnothing\) and all subsets whose complements in \(X\) are finite. This space is not Hausdorff. Indeed, any two nonempty open sets have finite complements, so their intersection has finite complement and is nonempty; hence no two distinct points can have disjoint open neighborhoods. Determine which subsets of \(X\) are discrete.
First let \(A\subseteq X\) be finite and nonempty, and fix \(x\in A\). The set \(U=X\setminus(A\setminus\{x\})\) is open because \(A\setminus\{x\}\) is finite. It contains \(x\), and direct intersection gives \(U\cap A=\{x\}\). Thus every point of \(A\) is isolated, and \(A\) is discrete. The empty set is also discrete by definition.
Now let \(A\) be infinite and take any \(x\in A\). Every open neighborhood \(U\) of \(x\) has finite complement \(F=X\setminus U\). The set \(A\setminus\{x\}\) is infinite, so it cannot be contained in the finite set \(F\). Therefore \(U\cap(A\setminus\{x\})\ne\varnothing\). No open neighborhood isolates \(x\) in \(A\); since this holds for every \(x\in A\), \(A\) has no isolated points and is not discrete.
Check: if \(A=\{2,5\}\) and \(x=2\), the isolating neighborhood is \(U=X\setminus\{5\}\), which is open and satisfies \(U\cap A=\{2\}\). If \(A=\mathbb{N}\), removing the finite complement of any neighborhood from \(\mathbb{N}\setminus\{x\}\) leaves points behind, so no point is isolated.
Worked Example: The Sierpinski Space
Let \(X=\{a,b\}\) have topology \(\mathcal{T}=\{\varnothing,\{a\},X\}\). For \(A=X\), determine which points are isolated.
The open set \(\{a\}\) is a neighborhood of \(a\), and \(\{a\}\cap A=\{a\}\). Thus \(a\) is isolated in \(A\). The only open set containing \(b\) is \(X\). Its intersection with \(A\) is \(X=\{a,b\}\), not \(\{b\}\), so \(b\) is not isolated. Consequently \(A=X\) is not a discrete subspace.
This example also shows why isolation is relative. The singleton \(\{b\}\) is not open in \(X\), but the subspace \(\{b\}\) is discrete: its only subsets are \(\varnothing\) and \(\{b\}\), and both are open in its subspace topology. A singleton subspace is always discrete, whether or not its singleton is open in the ambient space.
A Common Mistake: Requiring an Open Singleton
A frequent error is to say that \(x\) is isolated in \(A\) only when \(\{x\}\) is open in \(X\). That condition is sufficient, because the open neighborhood \(\{x\}\) intersects \(A\) in \(\{x\}\), but it is not necessary. The relevant condition is that some ambient open neighborhood intersects \(A\) in a singleton.
For a concrete correction, take \(A=\{1/n:n\in\mathbb{N}\}\) in the standard real line. The interval \((7/24,9/24)\) is open in \(\mathbb{R}\) and intersects \(A\) in \(\{1/3\}\), so \(1/3\) is isolated in \(A\). But \(\{1/3\}\) is not open in \(\mathbb{R}\): every open interval containing \(1/3\) contains real numbers other than \(1/3\). The mistake confuses an open singleton in the subspace topology with an open singleton in the ambient space.
Isolation is useful whenever the local structure of a subset matters: it separates points that stand alone within the subset from points approached by other members. The examples show that the same subset can behave differently under different topologies, and that even a discrete subspace can have points of the ambient space lying arbitrarily close to it.
Check Your Understanding
For each question, use the definition with open neighborhoods in the stated ambient topology.
- In the standard topology on \(\mathbb{R}\), determine whether \(0\) is isolated in \(A=\{0\}\cup\{1/n:n\in\mathbb{N}\}\).
- In the standard topology on \(\mathbb{R}\), is \(\mathbb{Z}\) a discrete subspace? Give an isolating neighborhood for an arbitrary integer.
- In the finite-complement topology on \(\mathbb{R}\), explain why every finite subset is discrete, including the empty set.
- For the Sierpinski topology \(\{\varnothing,\{a\},\{a,b\}\}\), determine the isolated points of the subset \(\{b\}\).
- State why an open singleton in the subspace topology on \(A\) need not be open in \(X\).