A set is called open when none of its points is on the edge: around every one of them there is a ball, however small, that the set contains entirely.

A country is open in this sense if from anywhere inside it you can walk some distance in every direction without ever crossing the border.

A subset $U$ of a metric space is open when every point of it has some ball around it lying entirely inside $U$. The radius is allowed to depend on the point, and usually must. Being open is a property of the set together with the metric, not of the set on its own: the same collection of points can be open under one distance rule and not under another.

• The empty set is open: it has no point at which the condition could fail.
• The whole space is open: every ball lies inside it, whatever the radius.
• On the real line $(0,1)$ is open, and $[0,1]$ is not: no ball around 0 stays inside.
• The radius generally has to shrink near the edge, which is why it is chosen per point.

A set is open when every one of its points has room: around $x$ there is a ball, however small, still inside $U$. No point of an open set is on its own edge, which is why both rings here are dashed.

The name would be badly chosen if this failed. Take any point of the ball; the previous lesson already found a smaller ball around it lying inside. That is exactly what the definition of open asks for at that point, and the point was arbitrary.

Take any point of the ball. The condition has to be checked at each one, so nothing may be assumed about which.

The point is strictly nearer the centre than the radius, so the leftover room is positive.

This is the theorem of the previous lesson, proved there by the triangle inequality.

A ball inside was produced around an arbitrary point, which is the definition of open.

$[0,1]$ contains 0, but every $B(0,r)$ holds negative numbers

Those numbers are outside $[0,1]$, so no ball around 0 fits.

in $(0,1)$, the point $0.001$ needs $r\le 0.001$

While the point $0.5$ has room for $r=0.5$.

The requirement never comes up, because nothing is in the set.

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

So a single point contains a ball, and every set is open.