From a Metric to a Topology
A metric gives a way to measure distances, and those distances specify which subsets count as open. The key condition is local: every point of an open set must have some ball around it that remains inside the set. We will prove that the collection of sets satisfying this condition obeys all the topology axioms. The important radius arguments work even for arbitrary unions and for finite intersections of any positive size.
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 the triangle inequality \(d(x,z)\le d(x,y)+d(y,z)\) for all \(x,y,z\in X\). For \(x\in X\) and \(r>0\), the open ball of radius \(r\) about \(x\) is \(B_d(x,r)=\{y\in X:d(x,y)<r\}\).
The topology axioms require that the empty set and \(X\) belong to the collection; that any union of sets in the collection belongs to it; and that any finite intersection of sets in it belongs to it. For the empty intersection, the convention is that the result is \(X\). The proof below checks each requirement directly from the ball condition.
Why Open Balls Are Open
Proof. Let \(y\in B_d(x,r)\). Then \(d(x,y)<r\), so \(s=r-d(x,y)>0\). If \(z\in B_d(y,s)\), then \(d(y,z)<s\), and the triangle inequality gives \(d(x,z)\le d(x,y)+d(y,z)<d(x,y)+s=r\). Thus \(z\in B_d(x,r)\), so \(B_d(y,s)\subseteq B_d(x,r)\). Each point \(y\) of the original ball therefore has a ball around it contained in that ball. This is exactly the definition of metric-open. \(\square\)
Worked Example: Finding a Ball Inside Another Ball
Take \(X=\mathbb{R}\) with the metric \(d(x,y)=|x-y|\). Consider \(B_d(2,5)\) and the point \(y=4\). Since \(d(2,4)=|2-4|=2<5\), the point lies in the ball. The lemma’s radius at \(y\) is \(5-2=3\), so it gives \(B_d(4,3)\subseteq B_d(2,5)\).
For a direct check, if \(z\in B_d(4,3)\), then \(|z-4|<3\), and \(|z-2|\le |z-4|+|4-2|<3+2=5\). Hence \(z\in B_d(2,5)\). For instance, \(z=6\) satisfies \(|6-4|=2<3\) and \(|6-2|=4<5\), checking a specific point against both ball conditions.
Check: The computed radius is positive, \(3>0\), as required. The triangle-inequality bound is strict at the last step because \(|z-4|<3\); this is why the resulting distance is strictly less than \(5\).
The Topology Axioms
Proof. First, \(\varnothing\) is metric-open because it has no points for which the ball condition needs to be checked. The set \(X\) is metric-open because for any \(x\in X\), \(B_d(x,1)\subseteq X\). If \(X=\varnothing\), this statement also holds: there are no points to check.
Now let \(\{U_\alpha:\alpha\in J\}\) be any family of metric-open sets, where \(J\) is an index set. If \(J=\varnothing\), its union is \(\varnothing\), which was just shown to be metric-open. If \(J\ne\varnothing\), take any \(x\in\bigcup_{\alpha\in J}U_\alpha\). By the definition of union, \(x\in U_{\alpha_0}\) for some \(\alpha_0\in J\). Since \(U_{\alpha_0}\) is metric-open, there is an \(r>0\) such that \(B_d(x,r)\subseteq U_{\alpha_0}\). Because \(U_{\alpha_0}\subseteq\bigcup_{\alpha\in J}U_\alpha\), the same ball lies in the union. This works for every point of the union, so the union is metric-open.
Finally, consider a finite family \(U_1,\ldots,U_n\) of metric-open sets with \(n\ge1\), and put \(F=\bigcap_{i=1}^n U_i\). If \(F=\varnothing\), it is metric-open. Otherwise, take \(x\in F\). For each \(i\), \(x\in U_i\), so there is an \(r_i>0\) with \(B_d(x,r_i)\subseteq U_i\). Since there are finitely many positive numbers \(r_1,\ldots,r_n\), their minimum \(r=\min\{r_1,\ldots,r_n\}\) exists and is positive. For each \(i\), \(r\le r_i\), so \(B_d(x,r)\subseteq B_d(x,r_i)\subseteq U_i\). Therefore \(B_d(x,r)\subseteq F\). Every point of \(F\) has such a ball, proving that \(F\) is metric-open. If the family has no members, its intersection is \(X\), already shown to be metric-open. All topology axioms hold, so \(\mathcal{T}_d\) is a topology on \(X\). \(\square\)
Radius Choices in Examples
Worked Example: An Arbitrary Union in a Subspace of the Real Line
Let \(X=\{0\}\cup\{1/n:n\in\mathbb{N}\}\) with the metric \(d(x,y)=|x-y|\) restricted from \(\mathbb{R}\). For each \(n\in\mathbb{N}\), set \(U_n=B_d(1/n,1/(2n(n+1)))\). We check that this ball contains exactly the point \(1/n\).
The point \(1/n\) is in \(U_n\), since its distance from the center is \(0\), and \(0<1/(2n(n+1))\). The distance from \(1/n\) to \(0\) is \(1/n\), which is greater than \(1/(2n(n+1))\): multiplying by the positive number \(2n(n+1)\), this comparison becomes \(2(n+1)>1\).
If \(m>n\), then \(|1/n-1/m|\ge 1/n-1/(n+1)=1/(n(n+1))\). If \(1\le m<n\), a case that can occur only when \(n\ge2\), then \(|1/m-1/n|\ge 1/(n-1)-1/n=1/(n(n-1))>1/(n(n+1))\). In both cases the distance is at least \(1/(n(n+1))\), which is strictly greater than the ball radius \(1/(2n(n+1))\). Thus no other point of \(X\) belongs to \(U_n\), and \(U_n=\{1/n\}\).
Consequently, \(\bigcup_{n\in\mathbb{N}}U_n=\{1/n:n\in\mathbb{N}\}=X\setminus\{0\}\). This union is metric-open: each of its points \(1/n\) has the ball \(U_n\) contained in it. The point \(0\) is not in the union, so no neighborhood condition is required there. This explicit infinite union illustrates why the union proof chooses a ball from one set that contains the point, rather than trying to find a single radius that works for every member of the family.
Check: At \(n=2\), the radius is \(1/(2\cdot2\cdot3)=1/12\). The distances from \(1/2\) to \(1/3\) and to \(0\) are \(1/6\) and \(1/2\), respectively, both greater than \(1/12\). This agrees with \(U_2=\{1/2\}\).
Worked Example: A Finite Intersection of Two Balls
Again use \(X=\mathbb{R}\) with \(d(x,y)=|x-y|\). Let \(U_1=B_d(0,3)\), \(U_2=B_d(2,2)\), and consider \(x=1\). The point belongs to both: \(d(0,1)=1<3\), and \(d(2,1)=1<2\). The available radii from the definition are \(3-1=2\) for \(U_1\) and \(2-1=1\) for \(U_2\). Their minimum is \(1\), so the proof gives \(B_d(1,1)\subseteq U_1\cap U_2\).
To verify the inclusion, let \(z\in B_d(1,1)\), so \(|z-1|<1\). Then \(|z-0|\le|z-1|+|1-0|<1+1=2<3\), and \(|z-2|\le|z-1|+|1-2|<1+1=2\). Thus \(z\in U_1\) and \(z\in U_2\). The intersection is metric-open by the theorem, and the calculation shows the radius supplied at this particular point.
Check: The point \(z=1\) lies in the chosen ball because \(|1-1|=0<1\), and in the intersection because \(1<3\) and \(1<2\).
A Common Radius Mistake
For a finite intersection, each set may require a different radius at the same point. Choosing one of those radii without checking the others can fail.
Common pitfall. In the preceding example, one might choose the radius \(2\) available from \(U_1\) and claim \(B_d(1,2)\subseteq U_1\cap U_2\). This is wrong: \(z=-1/2\) satisfies \(|z-1|=3/2<2\), so it lies in \(B_d(1,2)\), but \(|z-2|=5/2>2\), so it does not lie in \(U_2\). The correction is to use the minimum of the radii available from every set containing the point. Here that minimum is \(\min\{2,1\}=1\). For any \(z\in B_d(1,1)\), the triangle inequality gives \(|z-2|\le|z-1|+1<1+1=2\), as required. The strict inequality matters: a point at distance exactly \(2\) from the center of \(U_2\) would not be in that open ball.
The two radius arguments explain the difference between the axioms. For a union, a point belongs to at least one member of the family, so that member supplies a suitable ball. For an intersection, a point must belong to every member, and the chosen ball must satisfy every radius constraint; for a finite family the smallest radius does so. This proves the topology axioms directly from the metric rather than assuming that the collection of metric-open sets already has topological structure.
Check Your Understanding
For each question, use the ball condition and keep strict inequalities for membership in open balls.
- Why is the empty set metric-open, even when \(X\) is nonempty?
- For a point in a union of metric-open sets, which member of the family supplies the ball required by the definition?
- Why does the finite-intersection proof require the minimum of finitely many positive radii to be positive?
- In \(\mathbb{R}\) with \(d(x,y)=|x-y|\), find a positive radius \(s\) such that \(B_d(1,s)\subseteq B_d(0,4)\cap B_d(3,3)\).
- Explain why the proof for arbitrary unions does not need a single radius that works for all sets in the family.