A set can be enlarged to a closed one, and there is a smallest way to do it. Add exactly the points that every open set around them already touches the original.

Shrink-wrap a parcel. The wrapping takes the tightest shape that encloses the contents completely, and adds nothing it does not have to.

For $A\subseteq X$, the closure $\operatorname{cl}A$ is the intersection of all closed sets containing $A$. That intersection is closed by the previous lesson, and it contains $A$, so it is the smallest closed set that does. Throughout this lesson $U$ always denotes an open set.

• The collection being intersected is never empty: $X$ itself is closed and contains $A$, so there is always something to intersect.
• $A\subseteq\operatorname{cl}A$, and $A=\operatorname{cl}A$ exactly when is already closed.

The closure adds exactly the points that every open set around them already touches $A$ — the boundary, and nothing beyond it. It is the smallest closed set containing $A$, so $A$ is closed precisely when the two rings coincide.

Both directions are the same fact read two ways. If some open set around the point misses $A$ entirely, its complement is a closed set containing $A$, so the closure is inside that complement and the point is outside the closure. Turn it round: if the point is outside the closure, then the complement of the closure is itself an open set containing the point, and it misses $A$ because $A$ lies inside the closure. So failing to be in the closure and having an open set that misses $A$ are the same condition.

Suppose some open set around the point misses $A$ altogether.

Missing $U$ entirely means $A$ lies in its complement, which is closed because $U$ is open.

The closure is the smallest closed set containing $A$, so it is inside this one.

The point is in $U$ and the closure avoids $U$, so the point is outside the closure.

The other direction uses the same set: the complement of the closure is open, contains the point, and misses $A$ because $A$ is inside the closure.

The closure sits inside every closed set containing $A$.

Every interval around 0 contains positive numbers below 1.

Nothing is added, because nothing has to be.

Every interval contains a fraction, so no real number escapes.

The closure of a union is the union of the closures. For an intersection only one inclusion survives.