Any subset of a metric space is a metric space in its own right: keep the same distance rule and simply stop asking about points outside. What changes is not the distances but the balls — and with them which sets count as open, because a ball of the subset is only the part of the old ball that survived.

Measure distances in a town with the same tape measure you used for the county. Nothing about the tape changes when you stop walking outside the town boundary, but a circle drawn around a house at the edge is now a circle with a piece missing.

The space here is $A=[0,1]$, with the metric of the line. The ball $B_A(1,\tfrac12)$ is everything in $A$ within $\tfrac12$ of $1$, which is $(\tfrac12,1]$ — the trace on $A$ of the interval $(\tfrac12,\tfrac32)$. It contains its right-hand end, and in $A$ it is open, although as a subset of $\mathbb{R}$ it is not.

Let $(X,d)$ be a metric space and $A\subseteq X$ any subset. Restricting $d$ to pairs of points of $A$ gives a metric on $A$, called the induced metric, and is a subspace of . The three conditions are inherited unchanged, because each of them is a statement about particular points and those points are still there.

• The three conditions on a metric pass to every subset, so no checking is needed: a subset of a metric space is a metric space.
• A ball of the subspace is the trace of a ball of the whole space, which the theorem below proves. Every question about $A$ can therefore be answered by intersecting with $A$.
• Openness is not inherited in the same way: $(\tfrac12,1]$ is open in , because it is the ball , and it is not open in . Open is always open in a stated space, and the space has to be named.

Both sides are sets of points of A within r of a, so the proof is a matter of reading each side. A point of the left-hand side lies in A and satisfies the distance condition, so it lies in the ball of X and in A. A point of the right-hand side lies in A and satisfies the same condition, because the induced metric gives it the same distance. Nothing deeper happens: the induced metric was defined to make this true.

Unfold the ball of the subspace: its points come from A and satisfy the distance condition there.

The induced metric is the old one restricted, so the distance is unchanged.

The same two conditions describe the trace on A of the ball of X.

The two sets have the same members, so they are equal.

A subset becomes a metric space by keeping the distance rule and forgetting the rest of the space. No condition needs rechecking.

Openness is always relative to a space. A set can be open in a subspace and not in the whole space.

• A subset with the restricted metric is a metric space, the subspace.
• $B_A(a,r)=B_X(a,r)\cap A$: the balls are traces.

Great work! You now know how to:

• form the induced metric on a subset;
• compute balls in a subspace as traces of balls;
• give a set open in a subspace and not in the whole space;
• say which properties a subspace inherits and which it does not.