Binary intersection property, and not fixing what isn’t broken

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 {x_\alpha\in X} and numbers {r_\alpha>0} such that {d(x_\alpha,x_\beta)\le r_\alpha+r_\beta} for all {\alpha,\beta} there exists {x\in X} such that {d(x_\alpha,x)\le r_\alpha} for all {\alpha}.

For example, {\mathbb R} has this property: {x=\inf_\alpha (x_\alpha+r_\alpha)} works. But {\mathbb R^2} does not:

Failure of the binary intersection property
Failure of the binary intersection property

The space of bounded sequences {\ell^\infty} has the binary intersection property, and so does the space {B[0,1]} of all bounded functions {f:[0,1]\rightarrow\mathbb R} with the supremum norm. Indeed, the construction for {\mathbb R} generalizes: given a family of bounded functions {f_\alpha} and numbers {r_\alpha>0} as in the definition, let {f(x)=\inf_\alpha (f_\alpha(x)+r_\alpha)}.

The better known space of continuous functions {C[0,1]} has the finite version of binary intersection property, because for a finite family, the construction {\inf_\alpha (f_\alpha(x)+r_\alpha)} produces a continuous function. However, the property fails without finiteness, as the following example shows.

Example. Let {f_n\in C[0,1]} be a function such that {f_n(x)=-1} for {x\le \frac12-\frac1n}, {f_n(x)=1} for {x\ge \frac12+\frac1n}, and {f_n} is linear in between.

Since {\|f_n-f_m\| \le 1} for all {n,m}, we can choose {r_n=1/2} for all {n}. But if a function {f} is such that {\|f-f_n\|\le \frac12} for all {n}, then {f(x) \le -\frac12} for {x<\frac12} and {f(x) \ge \frac12} for {x>\frac12}. There is no continuous function that does that.

More precisely, for every {f\in C[0,1]} we have {\liminf_{n\to\infty} \|f-f_n\|\ge 1 } because {f(x)\approx f(1/2)} in a small neighborhood of {1/2}, while {f_n} change from {1} to {-1} in the same neighborhood when {n} 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 {B[0,1]} onto {C[0,1]}, that is a map {\Phi:B[0,1]\rightarrow C[0,1]} such that {\Phi(f)=f} for all {f\in C[0,1]}.

The failure of binary intersection property, as demonstrated by the sequence {(f_n)} above, implies that {\Phi} cannot be a contraction. Indeed, let {f(x)= \frac12 \,\mathrm{sign}\,(x-1/2)}. This is a discontinuous function such that {\|f-f_n\|\le 1/2} for all {n}. Since {\liminf_{n\to\infty} \|\Phi(f)-f_n\|\ge 1}, it follows that {\Phi} cannot be {L}-Lipschitz with a constant {L<2}.

It was known for a while that there is a retraction from {B[0,1]} onto {C[0,1]} with the Lipschitz constant at most {20}: see Geometric Nonlinear Functional Analysis by Benyamini and Lindenstrauss. In a paper published in 2007, Nigel Kalton proved the existence of a {2}-Lipschitz retraction.

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