Why Finite Sets Have Many Relatively Open Sets
A finite set of points in Euclidean space may look sparse, but the important topological fact is more precise: each point can be separated from the other points of the set by a sufficiently small ambient open ball. Once each singleton is open in the subspace, every collection of points is open there as well. The ambient set need not itself be open in Euclidean space; the conclusion concerns the topology induced on the finite set.
We use \(\mathbb{R}^n\) with its Euclidean topology. For \(x\in\mathbb{R}^n\) and \(r>0\), write \(B(x,r)=\{y\in\mathbb{R}^n:\|y-x\|<r\}\) for the open ball of radius \(r\). Let \(F\subseteq\mathbb{R}^n\). The subspace topology on \(F\) consists of all sets \(F\cap U\), where \(U\) is open in \(\mathbb{R}^n\). A set open in this topology is called relatively open in \(F\).
For a finite set, the distances between distinct points have a positive minimum, provided there are at least two points. Half of that minimum is a convenient radius: the corresponding ball around any chosen point contains no other point of the finite set. The proof must treat the empty set and singleton separately, since there are then no distances between distinct points to minimize.
Isolating Each Point with an Ambient Ball
Proof. If \(F=\varnothing\), there is no \(x\in F\), so the assertion has no points to check. If \(F=\{x\}\), take \(r_x=1\); the ball contains \(x\), and its intersection with \(F\) is \(\{x\}\). Now suppose \(F\) has at least two points and fix \(x\in F\). The set of real numbers \(\{\|x-y\|:y\in F,\ y\ne x\}\) is finite and nonempty. Every member is positive, because \(\|x-y\|=0\) holds only when \(x=y\). Therefore this finite set has a positive minimum; call it \(d_x\). Set \(r_x=d_x/2\), which is positive. Since \(\|x-x\|=0<r_x\), the point \(x\) lies in \(B(x,r_x)\). For any other \(y\in F\), the definition of \(d_x\) gives \(\|x-y\|\ge d_x>d_x/2=r_x\), so \(y\notin B(x,r_x)\). Consequently \(B(x,r_x)\cap F=\{x\}\). \(\square\)
The theorem gives a ball in the ambient space, not just an open set described abstractly within \(F\). This is useful in practice: calculate distances from the chosen point to the other members of \(F\), take a positive radius smaller than all of them, and intersect the resulting ambient ball with \(F\).
Worked Example: Isolating Points in the Plane
Let \(F=\{p,q,s\}\subseteq\mathbb{R}^2\), where \(p=(0,0)\), \(q=(3,0)\), and \(s=(0,4)\). Find radii that isolate each point, then use them to exhibit a relatively open subset containing \(q\) and \(s\) but not \(p\).
From \(p\), the distances to the other points are \(\|p-q\|=\sqrt{(3-0)^2+(0-0)^2}=\sqrt{9}=3\) and \(\|p-s\|=\sqrt{(0-0)^2+(4-0)^2}=\sqrt{16}=4\). The least is \(3\), so take \(r_p=3/2\).
From \(q\), the distances are \(\|q-p\|=3\) and \(\|q-s\|=\sqrt{(3-0)^2+(0-4)^2}=\sqrt{9+16}=5\). The least is \(3\), so take \(r_q=3/2\). From \(s\), the distances are \(\|s-p\|=4\) and \(\|s-q\|=5\); the least is \(4\), so take \(r_s=2\).
Let \(A=\{q,s\}\) and \(U=B(q,3/2)\cup B(s,2)\), an open set in \(\mathbb{R}^2\). The ball around \(q\) contains \(q\), while the distance from \(q\) to \(p\) is \(3>3/2\) and the distance from \(q\) to \(s\) is \(5>3/2\). Thus \(B(q,3/2)\cap F=\{q\}\). The ball around \(s\) contains \(s\), while the distances from \(s\) to \(p\) and \(q\) are \(4>2\) and \(5>2\), respectively. Thus \(B(s,2)\cap F=\{s\}\). Therefore \(U\cap F=\{q,s\}=A\), which proves that \(A\) is relatively open in \(F\).
Check: the two balls include their centres because \(0<3/2\) and \(0<2\). The other points are excluded by the strict inequalities \(3>3/2\), \(5>3/2\), \(4>2\), and \(5>2\). Hence the intersection is exactly \(\{q,s\}\), not a larger subset of \(F\).
Every Subset Is Relatively Open
Proof. If \(F\) is empty, its only subset is \(\varnothing\), which is open in every topology. Otherwise, for each \(x\in F\), use the preceding theorem to choose \(r_x>0\) with \(B(x,r_x)\cap F=\{x\}\). Let \(A\subseteq F\) and form the ambient set \(U=\bigcup_{x\in A}B(x,r_x)\). Each ball in this union is open in \(\mathbb{R}^n\), so \(U\) is open as a union of open sets. If \(A=\varnothing\), the union is \(\varnothing\), which is also open. For nonempty \(A\), intersecting the union with \(F\) gives $$ F\cap U=F\cap\bigcup_{x\in A}B(x,r_x) =\bigcup_{x\in A}\bigl(F\cap B(x,r_x)\bigr) =\bigcup_{x\in A}\{x\}=A. $$ The same equality holds when \(A=\varnothing\), since both sides are empty. By the definition of the subspace topology, \(A\) is relatively open in \(F\). Since this works for every \(A\subseteq F\), the subspace is discrete. \(\square\)
This proof uses a separate ball for each point of \(A\). It does not require the union of those balls to equal \(A\) in \(\mathbb{R}^n\); the required equality is only after intersecting that union with \(F\). Points of the ambient space outside \(F\) may lie in \(U\) without affecting the relative-open set \(F\cap U\).
Worked Example: A Finite Subset of Three-Dimensional Space
Let \(F=\{O,P,Q,R\}\subseteq\mathbb{R}^3\), where \(O=(0,0,0)\), \(P=(2,0,0)\), \(Q=(0,3,0)\), and \(R=(0,0,6)\). Construct ambient balls whose union intersects \(F\) in \(A=\{O,Q\}\).
For \(O\), the distances to \(P,Q,R\) are \(2,3,6\), respectively, so the least distance is \(2\). Take the radius \(1\). For \(Q\), the distances to \(O,P,R\) are \(\|Q-O\|=3\), \(\|Q-P\|=\sqrt{(0-2)^2+(3-0)^2+(0-0)^2}=\sqrt{13}\), and \(\|Q-R\|=\sqrt{(0-0)^2+(3-0)^2+(0-6)^2}=\sqrt{45}\). Since \(13>9\), \(\sqrt{13}>3\); since \(45>9\), \(\sqrt{45}>3\). Thus the least distance from \(Q\) is \(3\), and we take radius \(3/2\).
Set \(U=B(O,1)\cup B(Q,3/2)\). The first ball contains \(O\), and the distances from \(O\) to \(P,Q,R\) are \(2>1\), \(3>1\), and \(6>1\), so \(B(O,1)\cap F=\{O\}\). The second ball contains \(Q\), and the distances from \(Q\) to \(O,P,R\) are \(3>3/2\), \(\sqrt{13}>3>3/2\), and \(\sqrt{45}>3>3/2\). Thus \(B(Q,3/2)\cap F=\{Q\}\). It follows that \(U\cap F=\{O,Q\}=A\), so \(A\) is relatively open.
Check: the relevant centre distances are \(0<1\) and \(0<3/2\), while every listed distance to another member of \(F\) exceeds the ball's radius. This verifies that neither ball contributes an unintended point of \(F\).
Proof. The theorem shows that every \(A\subseteq F\) is relatively open. Its complement within \(F\), namely \(F\setminus A\), is also a subset of \(F\), so it is relatively open by the same theorem. A set is relatively closed precisely when its complement in the space is relatively open. Hence \(A\) is relatively closed as well. This includes \(A=\varnothing\) and \(A=F\): their relative complements are \(F\) and \(\varnothing\), respectively, and both are relatively open. \(\square\)
Relative Openness Is Not Ambient Openness
A set can be open in a subspace without being open in the surrounding Euclidean space. The distinction is visible even for a finite subset of the real line. When we establish relative openness, the object we must express as \(F\cap U\) is the subset \(A\), and \(U\) must be open in the ambient space. There is no requirement that \(A\) itself be an ambient open set.
Worked Example: A Relatively Open Set That Is Not Open in the Line
Let \(F=\{-1,2,7\}\subseteq\mathbb{R}\) and \(A=\{-1,7\}\). Find an ambient open set \(U\) with \(F\cap U=A\), and verify that \(A\) is not open in \(\mathbb{R}\).
For \(-1\), the distances to the other members of \(F\) are \(|-1-2|=3\) and \(|-1-7|=8\). A radius of \(3/2\) is smaller than both. For \(7\), the distances are \(|7-2|=5\) and \(|7-(-1)|=8\). A radius of \(5/2\) is smaller than both. Take \(U=(-1-3/2,-1+3/2)\cup(7-5/2,7+5/2)=(-5/2,1/2)\cup(9/2,19/2)\). Each interval contains its chosen point, and the distance calculations show that the first contains no other point of \(F\) and the second contains no other point of \(F\). Hence \(F\cap U=\{-1,7\}=A\), so \(A\) is relatively open in \(F\).
To check ambient openness, suppose \(A\) were open in \(\mathbb{R}\). Then some interval \((-1-\varepsilon,-1+\varepsilon)\), with \(\varepsilon>0\), would be contained in \(A\). But the point \(-1-\varepsilon/2\) lies in that interval because \(0<\varepsilon/2<\varepsilon\), and it is less than \(-1\), so it is not in \(A=\{-1,7\}\). This contradicts the proposed containment. Therefore \(A\) is relatively open in \(F\) but not open in \(\mathbb{R}\).
Common pitfall. Choosing a radius without comparing it with the distances between points can fail to isolate a singleton. For \(F=\{0,1/4,3\}\subseteq\mathbb{R}\), the incorrect choice \(B(0,1)\) gives \(B(0,1)\cap F=\{0,1/4\}\), not \(\{0\}\), because \(|1/4-0|=1/4<1\). The correction is to choose a radius smaller than the nearest other point's distance. Here the nearest distance from \(0\) is \(1/4\), so radius \(1/8\) works: \(0<1/8\), \(1/4>1/8\), and \(3>1/8\), giving \(B(0,1/8)\cap F=\{0\}\). The general construction uses half the least distance, rather than a radius selected without checking the other points.
The finite hypothesis matters to this construction. The proof obtains a positive minimum from a finite nonempty collection of distances. For an infinite set, distances from a point to the other points need not have a positive minimum, so this argument does not establish that each singleton is relatively open. The conclusion here is specifically about finite subsets of Euclidean space; it is not a claim that every subspace of \(\mathbb{R}^n\) is discrete.
Check Your Understanding
Use the subspace definition and, where appropriate, verify your choices by comparing Euclidean distances.
- For \(F=\{0,2,5\}\subseteq\mathbb{R}\), find a positive radius isolating each point of \(F\).
- Let \(F=\{(0,0),(1,0),(0,2)\}\subseteq\mathbb{R}^2\). Give an ambient open set \(U\) such that \(F\cap U=\{(1,0),(0,2)\}\).
- Explain why the empty set and a singleton satisfy the point-isolation theorem, although there is no least distance to another point in either case.
- Prove directly from the definition of the subspace topology that every subset of a finite discrete subspace is relatively closed.
- Give a finite subset \(F\subseteq\mathbb{R}\) and a set \(A\subseteq F\) that is relatively open but not open in \(\mathbb{R}\), and justify both claims.