# 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:

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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