A metric space has the *binary intersection property* if every collection of closed balls has nonempty intersection unless there is a trivial obstruction: the distance between centers of two balls exceeds the sum of their radii. In other words, for every family of points and numbers such that for all there exists such that for all .

For example, has this property: works. But does not:

The space of bounded sequences has the binary intersection property, and so does the space of all bounded functions with the supremum norm. Indeed, the construction for generalizes: given a family of bounded functions and numbers as in the definition, let .

The better known space of *continuous* functions has the *finite* version of binary intersection property, because for a finite family, the construction produces a continuous function. However, the property fails without finiteness, as the following example shows.

**Example.** Let be a function such that for , for , and is linear in between.

Since for all , we can choose for all . But if a function is such that for all , then for and for . There is no continuous function that does that.

More precisely, for every we have because in a small neighborhood of , while change from to in the same neighborhood when is large.

Given a discontinuous function, one can approximate it with a continuous function in some way: typically, using a mollifier. But such approximations tend to change the function even if it was continuous to begin with. Let’s try to not fix what isn’t broken: look for a **retraction** of onto , that is a map such that for all .

The failure of binary intersection property, as demonstrated by the sequence above, implies that cannot be a contraction. Indeed, let . This is a discontinuous function such that for all . Since , it follows that cannot be -Lipschitz with a constant .

It is known that there is a retraction from onto with the Lipschitz constant at most : see Geometric Nonlinear Functional Analysis by Benyamini and Lindenstrauss. The gap appears to remain at present; at least I don’t know the smallest Lipschitz constant required to retract bounded functions onto continuous ones.