Openness Is a Pointwise Condition
In a metric space, distance gives a concrete way to test whether points of a set have room around them. The key is that the room is measured locally: each point may use its own radius. This pointwise test applies in many metrics, not just the usual distance on the real line.
Let \((X,d)\) be a metric space. For \(x\in X\) and \(r>0\), write \(B_d(x,r)=\{y\in X:d(x,y)<r\}\) for the open ball of radius \(r\) centred at \(x\). The inequality defining the ball is strict: a point at distance exactly \(r\) is not in the ball.
To prove a set is open, start with an arbitrary point \(x\) in it and produce a positive radius whose entire ball stays in the set. To prove it is not open, it is enough to find one point \(x\) in the set for which every positive-radius ball contains a point outside the set. These are different tasks: an open set needs a suitable ball at each of its points, while a non-open set fails at at least one point.
Intervals in the Usual Metric
On \(\mathbb{R}\), use the usual metric \(d(x,y)=|x-y|\), so \(B_d(x,r)=(x-r,x+r)\). The endpoints of an interval matter because a ball around an included endpoint extends beyond it.
Proof. For \((a,b)\), take any \(x\in(a,b)\). Both \(x-a\) and \(b-x\) are positive. Set \(r=\min\{x-a,b-x\}/2\), which is positive. If \(y\in B_d(x,r)\), then \(|y-x|<r\), so \(y>x-r\geq a\) and \(y<x+r\leq b\). Hence \(B_d(x,r)\subseteq(a,b)\). Since this works for each \(x\in(a,b)\), that interval is open.
For \([a,b]\), the point \(a\) belongs to the set. For any \(r>0\), let \(y=a-r/2\). Then \(|y-a|=r/2<r\), so \(y\in B_d(a,r)\), but \(y<a\), so \(y\notin[a,b]\). Thus no positive-radius ball around \(a\) is contained in \([a,b]\), and the interval is not open. \(\square\)
Worked Example: Testing Two Real Intervals
In \(\mathbb{R}\) with the usual metric, test \(U=(-2,3)\) and \(F=[-2,3]\).
For \(U\). Take \(x\in(-2,3)\) and choose \(r=\min\{x+2,3-x\}/2\). Both quantities inside the minimum are positive, so \(r>0\). If \(y\in B_d(x,r)\), then \(x-r<y<x+r\). From \(r\leq(x+2)/2\), we have \(x-r\geq(x-2)/2>-2\), since \(x>-2\). From \(r\leq(3-x)/2\), we have \(x+r\leq(x+3)/2<3\), since \(x<3\). Therefore \(y\in(-2,3)\), and \(U\) is open.
For instance, at \(x=0\), the two distances to the endpoints are \(0-(-2)=2\) and \(3-0=3\). The formula gives \(r=\min\{2,3\}/2=1\), and \(B_d(0,1)=(-1,1)\subseteq(-2,3)\).
For \(F\). The endpoint \(-2\) belongs to \(F\). Given any \(r>0\), the point \(y=-2-r/2\) has distance \(r/2<r\) from \(-2\), but is less than \(-2\), so it is outside \(F\). Thus \(F\) is not open.
Check: at the chosen interior point \(0\), the ball has endpoints \(-1\) and \(1\), both strictly between \(-2\) and \(3\). At the included endpoint \(-2\), every positive radius admits the explicitly constructed point \(-2-r/2\) outside the interval.
Local Balls in a Subspace
A subset can be viewed as a metric space in its own right. If \(A\subseteq X\), its subspace metric is the restriction of \(d\) to pairs of points in \(A\). A ball in \(A\) consists only of those points of \(A\) that lie in the corresponding ball of \(X\). Thus a ball in the subspace need not contain all the points of an ambient ball.
Proof. For any \(x\in A\) and \(r>0\), the ball in the restricted metric is \(B_{d|_A}(x,r)=\{y\in A:d(x,y)<r\}=B_d(x,r)\cap A\). By the definition of openness applied to the metric space \(A\), \(V\) is open there exactly when every \(x\in V\) has a positive-radius restricted ball contained in \(V\). Substituting the displayed identity gives precisely the stated condition, in both directions. \(\square\)
Worked Example: Openness Relative to an Interval
Let \(X=[0,2]\) with the metric inherited from \(\mathbb{R}\), and let \(V=[0,1)\). Test whether \(V\) is open in \(X\), and compare with its status in \(\mathbb{R}\).
Take any \(x\in V\), so \(0\leq x<1\), and set \(r=(1-x)/2\). This radius is positive. If \(y\in B_d(x,r)\cap X\), then \(y<x+r=(x+1)/2<1\). Also \(y\in X=[0,2]\), so \(y\geq0\). Therefore \(y\in[0,1)=V\), proving that \(V\) is open in \(X\).
For a numerical check at \(x=0.8\), \(r=(1-0.8)/2=0.1\). The ambient ball is \((0.7,0.9)\), and its intersection with \(X\) is still contained in \([0,1)\). At \(x=0\), the formula gives \(r=(1-0)/2=0.5\); the ball in \(X\) is \([0,0.5)\), again contained in \(V\).
In the ambient space \(\mathbb{R}\), \(V\) is not open: it contains \(0\), and for every \(r>0\), the point \(-r/2\) lies in \(B_d(0,r)\) but not in \(V\). The subspace test succeeds at \(0\) because points below \(0\) are not in \(X\) at all.
Check: the same set has different openness answers in the two metric spaces. In the subspace, the test uses \(B_d(x,r)\cap X\); in \(\mathbb{R}\), it uses the entire ball.
Examples in Other Metrics
For metrics other than the usual distance, the definition stays the same; only the meaning of a ball changes. In particular, to use a coordinate formula for a metric, one must show that the chosen radius controls every coordinate sufficiently to keep the point in the set.
Worked Example: A Rectangle in the Taxicab Metric
On \(\mathbb{R}^2\), use the taxicab metric \(d_1((x,y),(u,v))=|x-u|+|y-v|\). Show that \(U=(0,2)\times(-1,1)\) is open in this metric.
Take any \((x,y)\in U\). The four numbers \(x\), \(2-x\), \(y+1\), and \(1-y\) are positive. Choose \(r=\min\{x,2-x,y+1,1-y\}/2>0\). If \((u,v)\in B_{d_1}((x,y),r)\), then \(|u-x|+|v-y|<r\). Both absolute values are nonnegative, so each is less than \(r\). Therefore \(u>x-r\geq x/2>0\) and \(u<x+r\leq(x+2)/2<2\). Also \(v>y-r\geq(y-1)/2>-1\) and \(v<y+r\leq(y+1)/2<1\). Thus \((u,v)\in U\). This gives a suitable ball at every point of \(U\).
At the specific point \((1/2,0)\), the four margins are \(1/2\), \(3/2\), \(1\), and \(1\). Their minimum is \(1/2\), so the prescribed radius is \(r=1/4\). If \(d_1((u,v),(1/2,0))<1/4\), then \(|u-1/2|<1/4\) and \(|v|<1/4\), giving \(1/4<u<3/4\) and \(-1/4<v<1/4\). These inequalities place the point in \(U\).
Check: the taxicab distance of any such point is less than \(1/4\), so each coordinate change is less than \(1/4\); the resulting coordinate bounds lie strictly inside the rectangle.
Proof. Let \((X,d)\) be finite and \(V\subseteq X\). If \(V=\varnothing\), it is open by the definition. If \(X\) has one point, every subset is either empty or all of \(X\); for the latter, any radius such as \(r=1\) works. Now suppose \(X\) has at least two points. The collection of distances \(d(x,y)\) for distinct \(x,y\in X\) is finite and consists of positive numbers, by the metric axioms. It therefore has a positive minimum \(\delta\). For each \(x\in V\), use radius \(r=\delta\). If \(y\in B_d(x,\delta)\), then \(d(x,y)<\delta\). No point \(y\ne x\) can satisfy this, because every distance between distinct points is at least \(\delta\). Hence \(B_d(x,\delta)=\{x\}\subseteq V\). This supplies a ball at each point of \(V\), proving the claim. \(\square\)
Worked Example: A Set in a Finite Graph Metric
Consider the graph with vertices \(v_1,v_2,v_3\) and edges joining \(v_1\) to \(v_2\) and \(v_2\) to \(v_3\), each of length \(1\). Use the graph metric, in which distance is the length of a shortest path. Test \(A=\{v_1,v_3\}\).
The distances between distinct vertices are \(d(v_1,v_2)=1\), \(d(v_2,v_3)=1\), and \(d(v_1,v_3)=2\). At \(v_1\), choose \(r=1\). The ball \(B_d(v_1,1)\) includes vertices at distance strictly less than \(1\); \(v_1\) has distance \(0\), while \(v_2\) and \(v_3\) have distances \(1\) and \(2\). Thus \(B_d(v_1,1)=\{v_1\}\subseteq A\). At \(v_3\), the distances to \(v_1,v_2,v_3\) are \(2,1,0\), so \(B_d(v_3,1)=\{v_3\}\subseteq A\). The condition holds at both points of \(A\), so \(A\) is open.
Check: the radius is exactly \(1\), and the strict ball condition excludes the other vertices at distance exactly \(1\). If the ball were defined using distance at most \(1\), these calculations would give different sets.
A Common Mistake: Demanding One Radius
Openness does not require a single radius to work uniformly at every point. For example, no fixed positive radius works for all points of \((0,1)\): given \(r>0\), choose \(x=\min\{r/2,1/2\}\). Then \(x\in(0,1)\), \(x\leq r/2\), and \(y=x-r/2\) satisfies \(|y-x|=r/2<r\) but \(y\leq0\). Thus \(B_d(x,r)\) is not contained in \((0,1)\).
Wrong conclusion: “No one radius works everywhere, so \((0,1)\) is not open.” This applies a stronger condition than the definition. The required radius can vary with \(x\): \(r_x=\min\{x,1-x\}/2\) is positive for every \(x\in(0,1)\). If \(|y-x|<r_x\), then \(y>x-r_x\geq x/2>0\) and \(y<x+r_x\leq(x+1)/2<1\). Hence the ball at \(x\) stays inside \((0,1)\), as required.
The local-ball criterion is useful because it turns openness into a point-by-point task. In coordinate metrics, distances can bound each coordinate change; in a subspace, only points of that subspace need to be considered; and in a finite metric space, a smallest positive distance gives a ball isolating each point.
Check Your Understanding
For each question, apply the pointwise ball condition and pay attention to which metric space is being used.
- In the usual metric on \(\mathbb{R}\), is \((1,4]\) open? Identify a point where the local-ball test succeeds or fails.
- In \(X=[-1,1]\) with the restricted usual metric, is \(V=[-1,0)\) open in \(X\)? Give a radius formula or show why no suitable radius exists.
- For the taxicab metric on \(\mathbb{R}^2\), find a positive radius that works at \((1,1)\) for the rectangle \((0,3)\times(0,2)\).
- Why does the strict inequality in the definition of a ball matter in the graph example with radius \(1\)?
- Explain why the absence of one radius working at every point does not by itself show that a set is not open.