From Distances to Open Sets
A metric gives a way to measure how far apart points are. It also gives a local rule for deciding which subsets count as open: each point of an open set must have some positive-radius ball around it that stays inside the set. The radius may depend on the point. This local dependence is important: a set can contain points with very different amounts of room around them.
We begin with the metric and ball definitions, then use them to compute open sets in several examples. The focus is the collection of sets specified by the local rule. The separate task of verifying all the topology axioms for this collection comes next.
The quantifiers in this definition matter. For each point \(x\) in \(U\), a suitable positive radius must exist, but different points may require different radii. The empty set satisfies the condition because it has no points for which a radius must be found. The whole set \(X\) satisfies it because for any \(x\in X\) and any \(r>0\), \(B_d(x,r)\subseteq X\).
Metric Balls Are Open
Proof. We must show that each \(x\in B_d(a,R)\) has a positive-radius ball contained in \(B_d(a,R)\). Fix such an \(x\). By membership in the ball, \(d(a,x)<R\), so \(r=R-d(a,x)>0\). If \(y\in B_d(x,r)\), then \(d(x,y)<r\), and the triangle inequality gives \(d(a,y)\le d(a,x)+d(x,y)<d(a,x)+r=R\). Thus \(y\in B_d(a,R)\), so \(B_d(x,r)\subseteq B_d(a,R)\). This works for every \(x\) in the ball, proving that the ball is metric-open. \(\square\)
This proof uses the strict inequality in the definition of a ball. It leaves room between \(d(a,x)\) and \(R\); the positive difference is precisely the radius used around \(x\). In particular, the radius is not chosen once for the entire ball: it depends on \(x\).
Proof. Suppose first that \(U\in\mathcal{T}_d\). For each \(x\in U\), the definition provides a radius \(r_x>0\) such that \(B_d(x,r_x)\subseteq U\). Also \(x\in B_d(x,r_x)\), since \(d(x,x)=0<r_x\). Consequently, every point of \(U\) belongs to one of these balls, and every one of the balls is contained in \(U\). Therefore \(U=\bigcup_{x\in U}B_d(x,r_x)\). If \(U=\varnothing\), this is the empty union. Conversely, suppose \(U\) is a union of open balls. Each such ball belongs to \(\mathcal{T}_d\) by the theorem. More directly, if \(x\in U\), then \(x\in B_d(a,R)\) for some ball in the union. The theorem gives a positive-radius ball around \(x\) contained in \(B_d(a,R)\), and hence contained in \(U\). This is the defining condition for \(U\in\mathcal{T}_d\). If the union is empty, it is open by the vacuous condition. \(\square\)
The proposition gives a useful way to construct metric-open sets: collect balls, possibly with different centres and radii. It also gives a local way to test a proposed set: at each of its points, find a ball that remains inside it. The following examples use that test directly.
Worked Example: An Interval in the Absolute-Value Metric
Equip \(X=\mathbb{R}\) with the metric \(d(x,y)=|x-y|\). Show that \(U=(-1,2)\) is metric-open, and show that \(A=[0,1)\) is not.
Take any \(x\in U\). Then \(x+1>0\) and \(2-x>0\). Set \(r=\frac12\min\{x+1,2-x\}\), which is positive. If \(y\in B_d(x,r)\), then \(|y-x|<r\), so \(x-r<y<x+r\). Since \(r\le(x+1)/2\), we have \(x-r\ge x-(x+1)/2=(x-1)/2>-1\). Since \(r\le(2-x)/2\), we also have \(x+r\le x+(2-x)/2=(x+2)/2<2\). Thus \(-1<y<2\), so \(B_d(x,r)\subseteq U\). This holds for every \(x\in U\), proving \(U\in\mathcal{T}_d\).
For \(A=[0,1)\), consider its point \(0\). For every proposed radius \(r>0\), let \(y=-r/2\). Then \(d(0,y)=r/2<r\), so \(y\in B_d(0,r)\), but \(y<0\), so \(y\notin A\). No ball about \(0\) is contained in \(A\). The local condition therefore fails, and \(A\notin\mathcal{T}_d\).
Check: The failure occurs at a point that belongs to \(A\), namely \(0\). It is not enough for most points of a set to have suitable balls; the condition must hold at every point in the set.
Worked Example: A Metric on Three Points
Let \(X=\{a,b,c\}\), and define a symmetric distance function by \(d(a,b)=2\), \(d(b,c)=2\), and \(d(a,c)=3\), with \(d(x,x)=0\). Verify that this is a metric, then decide whether \(U=\{a,c\}\) is metric-open.
All distances between distinct points are positive, and the distance function is symmetric by its definition. For the triangle inequality, the only largest distance is \(d(a,c)=3\), and the route through \(b\) gives \(d(a,b)+d(b,c)=2+2=4\), so \(3\le4\). For the other two pairs, \(d(a,b)=2\le d(a,c)+d(c,b)=3+2=5\), and \(d(b,c)=2\le d(b,a)+d(a,c)=2+3=5\). Triangles with a repeated point satisfy the inequality because one side is zero or the inequality reduces to equality. Hence \(d\) is a metric.
At \(a\), the ball of radius \(1\) is \(B_d(a,1)=\{a\}\): the distances from \(a\) are \(d(a,a)=0\), \(d(a,b)=2\), and \(d(a,c)=3\), and only \(0<1\). At \(c\), the ball of radius \(1\) is \(B_d(c,1)=\{c\}\), since the distances are \(d(c,c)=0\), \(d(c,b)=2\), and \(d(c,a)=3\). Thus \(a\) and \(c\) each have a ball contained in \(U\), so \(U\in\mathcal{T}_d\).
Check: The point \(b\) is not in \(U\), so the definition requires no ball about \(b\) to lie in \(U\). The only points at which we need to check the condition are \(a\) and \(c\), and the radius-\(1\) balls calculated above work.
A Sequence of Points with a Limit
A metric can be restricted to a subset: if \(X\subseteq\mathbb{R}\), the formula \(d(x,y)=|x-y|\) for \(x,y\in X\) defines a metric on \(X\). The balls are then formed using only points of \(X\). This can make a ball look different from an interval in the whole real line: points outside \(X\) are not part of the ball.
Worked Example: Isolated Points in a Convergent Sequence
Let \(X=\{0\}\cup\{1/n:n\in\mathbb{N}\}\) with the restricted metric \(d(x,y)=|x-y|\), and set \(A=X\setminus\{0\}=\{1/n:n\in\mathbb{N}\}\). Show that \(A\) is metric-open but \(\{0\}\) is not.
First take the point \(1\in A\). Every other point of \(X\) is \(0\) or \(1/m\) for \(m\ge2\); its distance from \(1\) is at least \(1/2\). Thus \(B_d(1,1/4)=\{1\}\subseteq A\).
Now fix \(n\ge2\). If \(m>n\), then \[ \left|\frac1n-\frac1m\right|=\frac1n-\frac1m\ge\frac1n-\frac1{n+1}=\frac1{n(n+1)}. \] If \(m<n\), then \[ \left|\frac1n-\frac1m\right|=\frac1m-\frac1n\ge\frac1{n-1}-\frac1n=\frac1{n(n-1)}>\frac1{n(n+1)}. \] Also the distance from \(1/n\) to \(0\) is \(1/n>1/[n(n+1)]\). Therefore, with \(r=1/[2n(n+1)]\), every other point of \(X\) is farther from \(1/n\) than \(r\), while \(d(1/n,1/n)=0<r\). Hence \(B_d(1/n,r)=\{1/n\}\subseteq A\). Every point of \(A\) has a ball contained in \(A\), so \(A\in\mathcal{T}_d\).
To test \(\{0\}\), let \(r>0\) be any radius. Choose \(n\in\mathbb{N}\) with \(n>1/r\). Then \(1/n<r\), so \(1/n\in B_d(0,r)\), while \(1/n\notin\{0\}\). No ball about \(0\) is contained in \(\{0\}\), and therefore \(\{0\}\notin\mathcal{T}_d\).
For an additional ball calculation, \(B_d(0,1/3)\) contains \(0\), and it contains \(1/n\) exactly when \(1/n<1/3\), or equivalently \(n>3\). Thus \(B_d(0,1/3)=\{0\}\cup\{1/n:n\ge4\}\). The points \(1/2\) and \(1/3\) are excluded: their distances from \(0\) are \(1/2>1/3\) and \(1/3\not<1/3\), respectively.
A Common Pitfall: The Radius Inequality Is Strict
Common pitfall. When testing whether \(y\) belongs to \(B_d(x,r)\), the condition is \(d(x,y)<r\), not \(d(x,y)\le r\). For example, in the preceding space, \(d(0,1/3)=1/3\). The incorrect calculation “\(1/3\le1/3\), so \(1/3\in B_d(0,1/3)\)” uses the wrong membership test. The correction is \(1/3\not<1/3\), so \(1/3\notin B_d(0,1/3)\). Equality with the radius does not qualify.
There is a related mistake in testing openness: finding one ball around one point of a set does not establish that the set is open. The definition requires a suitable ball around every point in the set. Conversely, a point outside the set creates no extra condition. Keeping track of which points must be tested, and of the strict ball inequality, makes local calculations reliable.
The central pattern is: identify a point of the proposed set, choose a positive radius, and verify that every point at distance strictly less than that radius remains in the set. For a whole set, the radius may vary from point to point. When this test succeeds everywhere, the set is metric-open; equivalently, it can be written as a union of metric balls.
Check Your Understanding
For each question, use the ball definition and keep the strict inequality in view.
- In \(\mathbb{R}\) with \(d(x,y)=|x-y|\), find a positive radius around \(x\in(2,5)\) whose ball is contained in \((2,5)\).
- In the three-point metric example, compute \(B_d(b,2)\). Which points are excluded, and why?
- In \(X=\{0\}\cup\{1/n:n\in\mathbb{N}\}\) with the restricted absolute-value metric, determine whether \(B_d(0,1/2)\) contains \(1/2\).
- Explain why a union of open balls is metric-open, using the local definition.
- Give a set that is not metric-open in the sequence-space example, and identify the point where the local condition fails.