Open balls are open

Statement

Let $(X, d)$ be a metric space, let $x \in X$ and let $r > 0$. Then the open ball $B(x, r)$ is an open set of $X$.

Remarks

Proof: let $y \in B(x, r)$ and let $\varepsilon = r - d(x, y) > 0$. If $d(y, z) < \varepsilon$, then $d(x, z) \le d(x, y) + d(y, z) < r$ by the triangle inequality. Thus $B(y, \varepsilon) \subseteq B(x, r)$. The open sets of $X$ are exactly the unions of open balls.

Explanations

L1 human

A point inside an open ball is at some distance from the edge of the ball. A smaller ball about that point, with radius equal to this distance, stays inside the big ball.