A neighbourhood of a point is any set roomy enough around it to hold a whole open set containing it. Being open then turns out to be exactly the property of being a neighbourhood of every one of your own points. In a metric space the open set can always be taken to be a ball, which is how chapter 1 would have said it.

A region counts as your surroundings if it holds everything within some distance of you. It need not be tidy at its far edge, only roomy where you are.

A set $N$ is a neighbourhood of a point $x$ when some open set $U$ satisfies $x\in U\subseteq N$, and $x$ is then called an interior point of $N$. The interior of $N$, written $\operatorname{int}N$, is the set of all its interior points. A neighbourhood need not be open: it may be perfectly roomy at $x$ and ragged elsewhere. In a metric space this says exactly what chapter 1 said with balls, since the open sets are unions of balls.

• $\operatorname{int}N\subseteq N$: an interior point lies in a ball inside $N$, so it lies in $N$.
• $[0,1]$ is a neighbourhood of but not of 0, and its interior is .

$N$ is a neighbourhood of $x$ because it has room for a whole ball around it. $N$ itself need not be open — only roomy at $x$ — and a set that is a neighbourhood of every one of its own points turns out to be exactly an open set.

One direction is a rewriting: openness asks for a ball inside $U$ around each of its points, and having such a ball is what makes a point interior, so every point of an open set is interior and the two sets agree. The other direction runs backwards through the same sentence: if every point of $U$ is interior then every point has a ball inside, which is the definition of open.

Assume the set is open and take any of its points.

Openness supplies the ball, and having one is precisely what makes the point interior.

Every point of $U$ is interior, and interior points always lie in $U$, so the two sets are equal.

Now assume the equality instead, and take any point of $U$.

The point is interior, so some ball around it lies inside $U$, and that is the definition of open.

$[0,1]$ is a neighbourhood of $0.5$

Because $B(0.5,0.4)\subseteq[0,1]$.

$[0,1]$ at the point $0.5$

Roomy at $0.5$, but not open, because 0 has no room.

Both endpoints are lost, because no ball around either stays inside.

The interval is open, and nothing is lost by taking the interior.

In a topological space the definition reads with an open set in place of a ball.