Why Every Open Ball Is Open

The Room Around a Point Inside a Ball

An open ball is described by a center and a radius, but openness asks for a local condition at every point of the set: each point must itself have a ball around it that remains inside the set. The center of the original ball is not the only point for which this condition must be checked. The key is to measure how much of the original radius remains after reaching the point under consideration.

Throughout, \((X,d)\) is a metric space. Thus \(d:X\times X\to[0,\infty)\) satisfies \(d(x,y)=0\) exactly when \(x=y\), is symmetric, and satisfies 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 is \(B(x,r)=\{z\in X:d(x,z)<r\}\). The metric topology consists of the empty set and those subsets \(U\subseteq X\) for which every \(y\in U\) has some \(\rho>0\) with \(B(y,\rho)\subseteq U\). This is the local-ball test for openness.

The point \(x\) is the center and \(r\) is the radius of the original ball. To check openness, take any point \(y\) in that ball. Membership means \(d(x,y)<r\), so the difference \(r-d(x,y)\) is positive. We use this difference to choose a smaller radius around \(y\).

Proof. If \(X\) is empty, there is no \(x\in X\), so the statement holds vacuously. Otherwise fix \(x\in X\) and \(r>0\), and take an arbitrary \(y\in B(x,r)\). By the definition of the ball, \(d(x,y)<r\), so \(\delta=r-d(x,y)>0\). Set \(\rho=\delta/2\), which is positive. If \(z\in B(y,\rho)\), then \(d(y,z)<\rho\). The triangle inequality gives \(d(x,z)\le d(x,y)+d(y,z)<d(x,y)+\rho\). Since \(\rho=(r-d(x,y))/2\), we have \(d(x,y)+\rho=(r+d(x,y))/2<r\), where the last inequality follows from \(d(x,y)<r\). Therefore \(d(x,z)<r\), so \(z\in B(x,r)\). We have shown \(B(y,\rho)\subseteq B(x,r)\). As this works for every \(y\in B(x,r)\), the local-ball test says that \(B(x,r)\) is open. \(\square\)

Choosing the Smaller Radius

The proof gives a particular valid choice at \(y\): \(\rho=(r-d(x,y))/2\). It is half the remaining distance allowance. The factor \(1/2\) is convenient because it leaves room in the strict inequality; the argument does not say that this is the only possible choice. The radius must be positive, which is why the point \(y\) must be in the open ball rather than merely at distance less than or equal to \(r\).

Here is the same estimate in a form useful for checking an example. If \(z\in B(y,\rho)\), then \(d(x,z)\le d(x,y)+d(y,z)<d(x,y)+\rho\). Thus it is enough to choose \(\rho>0\) so that \(d(x,y)+\rho\le r\). The half-margin choice satisfies even the stronger inequality \(d(x,y)+\rho<r\). Nothing in this reasoning depends on coordinates or on a special shape of a ball; it uses only the metric axioms, in particular the triangle inequality.

Relative Balls and Discrete Balls

The same proof applies when a metric is restricted to a subset. A ball in a subspace is a ball relative to that subspace: its points must belong to the subspace as well as satisfy the distance inequality. The triangle inequality still holds for the restricted metric, so the proof does not need to change.

A Ball Is the Union of Its Smaller Neighborhoods

This equality makes the local content of the theorem explicit: the original ball is covered by smaller balls, one centered at each of its points. Each smaller radius depends on its center \(y\), through the distance \(d(x,y)\).

Proof. Write \(V\) for the union on the right. If \(z\in V\), then for some \(y\in B(x,r)\) we have \(z\in B(y,(r-d(x,y))/2)\). The theorem's triangle-inequality estimate shows that this smaller ball is contained in \(B(x,r)\), so \(z\in B(x,r)\). Hence \(V\subseteq B(x,r)\). Conversely, take any \(z\in B(x,r)\). The term in the union corresponding to \(y=z\) has positive radius \((r-d(x,z))/2\), because \(d(x,z)<r\). Also \(d(z,z)=0<(r-d(x,z))/2\), so \(z\) belongs to that term. Thus \(z\in V\), giving \(B(x,r)\subseteq V\). The two inclusions prove the equality. \(\square\)

The radii in this union generally cannot be replaced by one fixed positive radius for every point. For example, in \(\mathbb{R}\), points \(y_n=1-1/n\), for \(n\ge2\), lie in \(B(0,1)\), and their remaining distances are \(1-|y_n|=1/n\). These distances become arbitrarily small as \(n\) increases. The proof needs a positive radius at each point, not a single radius that works uniformly throughout the ball.

Common Pitfall: Using the Outer Radius Again

A tempting but incorrect choice is to give a point \(y\) the same radius \(r\) as the outer ball. Take \(B(0,1)\subset\mathbb{R}\) and \(y=9/10\). This point belongs to the ball because \(|9/10|=9/10<1\). But \(B(y,1)\) is not contained in \(B(0,1)\): the point \(z=19/10\) satisfies \(|z-y|=|19/10-9/10|=1\), so this particular point is on the boundary and is not in \(B(y,1)\). Instead take \(z=3/2\): then \(|3/2-9/10|=3/5<1\), while \(|3/2|=3/2>1\). Thus \(z\in B(y,1)\) but \(z\notin B(0,1)\).

The correction is to use the remaining distance, not the full outer radius. Here the margin is \(1-9/10=1/10\), and the half-margin choice is \(\rho=1/20\). If \(|z-9/10|<1/20\), then \(|z|\le |z-9/10|+9/10<1/20+9/10=19/20<1\). So this smaller ball really is contained in \(B(0,1)\). The strict membership condition matters twice: it ensures the margin is positive, and points at exactly the chosen radius are not included in the smaller open ball.

The method is short, but its logic is worth retaining: start with an arbitrary point of the ball, compute the positive gap between its distance from the center and the outer radius, choose a positive fraction of that gap, and apply the triangle inequality. The same sequence of steps proves openness regardless of whether distances are measured on the real line, in a coordinate metric, or in a restricted metric space.