Once there is a distance, the natural thing to collect is every point closer to a chosen point than some fixed amount. That collection is called a ball, and it is the basic shape of the whole subject.

Everywhere within twenty minutes of a station is a ball around it. Its shape on a map depends entirely on how the twenty minutes are measured.

Every point closer to $x$ than $r$, and nothing else. The edge is drawn dashed because it is not included: a ball is the strict $d(x,y)<r$, and that strictness is what makes it behave like an open set rather than a closed disc.

For a point $x$ of a metric space and a real number $r>0$, the open ball of centre $x$ and radius $r$ is the set of points strictly closer to $x$ than $$. The radius must be positive: with the collection would be empty, since no point is at distance less than zero. The inequality is strict, which is what the word open is doing here.

• $x\in B(x,r)$ for every $r>0$, because $d(x,x)=0<r$.

Take a point of the ball. It is strictly nearer the centre than the radius, so there is some room left over, and $s=r-d(x,y)$ measures it. Any point within that much of the chosen point reaches the centre by a detour costing less than the radius, so it lies in the original ball. This one fact is what the next lesson is built on.

The point lies in the ball, so its distance to the centre is strictly less than the radius and the leftover room is positive.

Take any point of the smaller ball, so its distance to $y$ is less than $s$.

Reach $z$ from the centre by way of $y$, using the triangle inequality.

The second term is smaller than $s$, and $s$ was chosen to make the total exactly the radius.

The distance from the centre is strictly less than the radius, which is what belonging to the ball means. This holds for every such $z$.

on $\mathbb{R}$, $B(0,2)$ holds 1.9 but not 2

The distance from 0 to 2 is 2, which is not less than 2.

True for every positive radius, in every metric space.

$d(x,y)=1$ for $x\neq y$, so $B(x,1)=\{x\}$

No other point is strictly closer than 1, because every other point is at exactly 1.

$y\in B(x,r)$, $s=r-d(x,y)$

Then $B(y,s)$ fits inside $B(x,r)$, by the triangle inequality.

Swapping $<$ for $\le$ in the definition gives the closed ball, which adds the points at exactly the radius.