From Metric Openness to a Topology
A metric tells us how far apart points are. Its open balls provide a local way to describe which subsets count as open. The important structural question is whether those subsets obey the rules required of a topology. The answer follows from the local definition: a radius may depend on the point, and for a finite intersection we can choose the smallest of finitely many available radii.
Let \((X,d)\) be a metric space. Thus \(d:X\times X\to[0,\infty)\) satisfies \(d(x,y)=0\) if and only if \(x=y\), \(d(x,y)=d(y,x)\), and \(d(x,z)\le d(x,y)+d(y,z)\) for all \(x,y,z\in X\). For \(x\in X\) and \(r>0\), its open ball is \(B_d(x,r)=\{y\in X:d(x,y)<r\}\).
We will verify those rules for the collection \(\mathcal{T}_d\) of all metric-open subsets of \(X\). The verification uses only the definition of metric openness and the metric axioms. In particular, it does not require one radius to work at every point of an open set.
The Empty Set and the Whole Space
Proof. For \(\varnothing\), the definition asks us to check every point \(x\in\varnothing\). There are no such points, so there is no condition to fail; hence \(\varnothing\) is metric-open. For \(X\), take any \(x\in X\). The radius \(r=1\) is positive, and \(B_d(x,1)\subseteq X\) because every ball is a subset of the space in which it is defined. Thus every point of \(X\) has a ball contained in \(X\), so \(X\) is metric-open. If \(X=\varnothing\), both sets are the same set and the first argument already applies. \(\square\)
This argument also accounts for the empty space, whose metric axioms hold vacuously. In topology, the empty union is \(\varnothing\), and the intersection of an empty finite family of subsets of \(X\) is defined to be \(X\). These conventions match the two sets just checked.
Arbitrary Unions
Proof. Let \((U_i)_{i\in J}\) be a family of metric-open subsets, indexed by a set \(J\), and put \(U=\bigcup_{i\in J}U_i\). If \(J=\varnothing\), then \(U=\varnothing\), which is metric-open by Proposition 1. Otherwise, take any \(x\in U\). By the definition of union, there is an index \(i\in J\) with \(x\in U_i\). Since \(U_i\) is metric-open, there is an \(r>0\) such that \(B_d(x,r)\subseteq U_i\). As \(U_i\subseteq U\), this gives \(B_d(x,r)\subseteq U\). Each point of \(U\) therefore has a ball contained in \(U\), proving that \(U\) is metric-open. \(\square\)
The choice of a set \(U_i\) may vary from point to point. The proof does not require one set in the family, or one radius, to work for all points of the union.
Worked Example: An Increasing Union in the Real Line
Use the standard metric \(d(x,y)=|x-y|\) on \(\mathbb{R}\). For each integer \(n\ge2\), let \(U_n=(-1+1/n,\,1-1/n)\). We compute their union.
Every \(U_n\) is contained in \((-1,1)\): its left endpoint is greater than \(-1\), and its right endpoint is less than \(1\). Hence \(\bigcup_{n\ge2}U_n\subseteq(-1,1)\).
Conversely, take \(x\in(-1,1)\), so \(1-|x|>0\). Choose an integer \(n\ge2\) with \(n>1/(1-|x|)\). Then \(1/n<1-|x|\). This implies \(-1+1/n<x<1-1/n\), so \(x\in U_n\). Therefore \(x\) belongs to the union, and \(\bigcup_{n\ge2}U_n=(-1,1)\).
Check: at any \(x\in(-1,1)\), the radius \(r=\min(x+1,1-x)\) is positive. If \(|y-x|<r\), then \(y>x-r\ge-1\) and \(y<x+r\le1\). The strict inequality in \(|y-x|<r\) gives \(-1<y<1\). Thus each point has a ball contained in the union, as the arbitrary-union result predicts.
Finite Intersections
Proof. For the empty family, the intersection is \(X\), which is metric-open by Proposition 1. Now take a nonempty finite family \(U_1,\ldots,U_m\) of metric-open sets, where \(m\) is a positive integer, and write \(U=\bigcap_{j=1}^{m}U_j\). If \(U=\varnothing\), it is open by Proposition 1. Otherwise, fix \(x\in U\). Then \(x\in U_j\) for every \(j\), so for each \(j\) there is a radius \(r_j>0\) such that \(B_d(x,r_j)\subseteq U_j\). The finite list \(r_1,\ldots,r_m\) has a minimum \(r=\min\{r_1,\ldots,r_m\}>0\). If \(y\in B_d(x,r)\), then \(d(x,y)<r\le r_j\) for every \(j\), so \(y\in B_d(x,r_j)\subseteq U_j\) for every \(j\). Hence \(y\in U\), and \(B_d(x,r)\subseteq U\). This works at every \(x\in U\), so \(U\) is metric-open. \(\square\)
Finiteness matters in this argument: it ensures that the smallest of the chosen radii is still positive. For an infinite family, a positive radius common to all the sets need not exist. The topology axiom asks for finite intersections, not arbitrary intersections.
Worked Example: A Finite Intersection in the Real Line
Again use \(\mathbb{R}\) with its standard metric. Let \(U=(-1,1)\) and \(V=(0,2)\). Their intersection is \(W=(0,1)\). For any \(x\in W\), both \(x\) and \(1-x\) are positive. Set \(r=\min(x,1-x)>0\).
If \(y\in B_d(x,r)\), then \(|y-x|<r\le x\), so \(y>x-r\ge0\), with strictness giving \(y>0\). Also \(|y-x|<r\le1-x\), so \(y<x+r\le1\), with strictness giving \(y<1\). Thus \(B_d(x,r)\subseteq W\), and \(W\) is metric-open.
Check: \(U\cap V=(-1,1)\cap(0,2)=(0,1)\): membership requires both \(0<x<2\) and \(-1<x<1\), which together are exactly \(0<x<1\). This agrees with the local-ball calculation.
Worked Example: The Discrete Metric
Let \(X=\{a,b,c\}\) and define the discrete metric by \(d(x,y)=0\) when \(x=y\), and \(d(x,y)=1\) when \(x\ne y\). Consider \(A=\{a,c\}\) and \(B=\{b,c\}\). For any \(x\in X\), the ball \(B_d(x,1/2)\) is \(\{x\}\): the distance from \(x\) to itself is \(0<1/2\), while the distance from \(x\) to each other point is \(1\), which is not less than \(1/2\).
Thus \(B_d(a,1/2)=\{a\}\subseteq A\), and \(B_d(c,1/2)=\{c\}\subseteq A\), proving \(A\) is open. The same calculation gives \(B_d(b,1/2)=\{b\}\subseteq B\) and \(B_d(c,1/2)=\{c\}\subseteq B\), so \(B\) is open. Their union is \(A\cup B=\{a,b,c\}=X\), and their intersection is \(A\cap B=\{c\}\). The set \(\{c\}\) is open because \(B_d(c,1/2)=\{c\}\).
Check: the union contains all three points, and the only point common to \(\{a,c\}\) and \(\{b,c\}\) is \(c\). The computed union and intersection therefore agree with the set operations, and both are open.
The Topology Axioms and a Common Mistake
Proof. Proposition 1 shows that \(\varnothing,X\in\mathcal{T}_d\). Proposition 2 shows that the union of any family of members of \(\mathcal{T}_d\) belongs to \(\mathcal{T}_d\), including the empty family. Proposition 3 shows that the intersection of every finite family of members belongs to \(\mathcal{T}_d\), including the empty finite family. These are precisely the topology axioms, so \(\mathcal{T}_d\) is a topology on \(X\). \(\square\)
Common pitfall. One might claim that \(W=(0,1)\) in the preceding real-line example is open because the same radius \(1\) works at every point. That calculation is false: at \(x=0.9\), the ball \(B_d(0.9,1)\) contains \(-0.09\), since \(|-0.09-0.9|=0.99<1\), but \(-0.09\notin W\). The correction is to choose a radius depending on the point. At \(x=0.9\), \(r=\min(0.9,1-0.9)=0.1\), and \(|y-0.9|<0.1\) implies \(0.8<y<1\), so \(B_d(0.9,0.1)\subseteq W\). Openness requires a suitable radius at each point, not one fixed radius for the entire set.
The topology axioms explain why metric spaces naturally provide topological spaces: local metric neighborhoods combine consistently under arbitrary unions and finite intersections. The finite-intersection proof also shows the key limit of the argument. A finite collection of positive radii has a positive minimum; an infinite collection need not. That is why the topology axioms distinguish arbitrary unions from finite intersections.
Check Your Understanding
Use the local-ball definition and the topology axioms to answer the following questions.
- Why is the empty set metric-open even though it contains no point at which to choose a ball?
- For a family of metric-open sets indexed by the empty set, what is its union, and why is it open?
- Explain where finiteness is used when proving that an intersection of open sets is open.
- In the standard metric on \(\mathbb{R}\), find a radius around \(x=0.2\) whose ball lies in \((0,1)\).
- Why does the proof of the topology axioms not require showing that every open ball is itself open?