# Lightness, hyperspace, and lower oscillation bounds

When does a map ${f\colon X\to Y}$ admit a lower “anti-continuity” bound like ${d_Y(f(a),f(b))\ge \lambda(d_X(a,b))}$ for some function ${\lambda\colon (0,\infty)\to (0, \infty)}$ and for all ${a\ne b}$? The answer is easy: ${f}$ must be injective and its inverse must be uniformly continuous. End of story.

But recalling what happened with diameters of connected sets last time, let’s focus on the inequality ${\textrm{diam}\, f(E)\ge \lambda (\textrm{diam}\, E)}$ for connected subsets ${E\subset X}$. If such ${\lambda}$ exists, the map f has the LOB property, for “lower oscillation bound” (oscillation being the diameter of image). The LOB property does not require ${f}$ to be injective. On the real line, ${f(x)=|x|}$ satisfies it with ${\lambda(\delta)=\delta/2}$: since it simply folds the line, the worst that can happen to the diameter of an interval is to be halved. Similarly, ${f(x)=x^2}$ admits a lower oscillation bound ${\lambda(\delta) = (\delta/2)^2}$. This one decays faster than linear at 0, indicating some amount of squeezing going on. One may check that every polynomial has the LOB property as well.

On the other hand, the exponential function ${f(x)=e^x}$ does not have the LOB property, since ${\textrm{diam}\, f([x,x+1])}$ tends to ${0}$ as ${x\to-\infty}$. No surprise there; we know from the relation of continuity and uniform continuity that things like that happen on a non-compact domain.

Also, a function that is constant on some nontrivial connected set will obviously fail LOB. In topology, a mapping is called light if the preimage of every point is totally disconnected, which is exactly the same as not being constant on any nontrivial connected set. So, lightness is necessary for LOB, but not sufficient as ${e^x}$ shows.

Theorem 1: Every continuous light map ${f\colon X\to Y}$ with compact domain ${X}$ admits a lower oscillation bound.

Proof. Suppose not. Then there exists ${\epsilon>0}$ and a sequence of connected subsets ${E_n\subset X}$ such that ${\textrm{diam}\, E_n\ge \epsilon}$ and ${\textrm{diam}\, f(E_n)\to 0}$. We can assume ${E_n}$ compact, otherwise replace it with its closure ${\overline{E_n}}$ which we can because ${f(\overline{E_n})\subset \overline{f(E_n)}}$.

The space of nonempty compact subsets of ${X}$ is called the hyperspace of ${X}$; when equipped with the Hausdorff metric, it becomes a compact metric space itself. Pass to a convergent subsequence, still denoted ${\{E_n\}}$. Its limit ${E}$ has diameter at least ${\epsilon}$, because diameter is a continuous function on the hyperspace. Finally, using the uniform continuity of ${f}$ we get ${\textrm{diam}\, f(E) = \lim \textrm{diam}\, f(E_n) = 0}$, contradicting the lightness of ${f}$. ${\quad \Box}$

Here is another example to demonstrate the importance of compactness (not just boundedness) and continuity: on the domain ${X = \{(x,y)\colon 0 < x < 1, 0 < y < 1\}}$ define ${f(x,y)=(x,xy)}$. This is a homeomorphism, the inverse being ${(u,v)\mapsto (u, v/u)}$. Yet it fails LOB because the image of line segment ${\{x\}\times (0,1)}$ has diameter ${x}$, which can be arbitrarily close to 0. So, the lack of compactness hurts. Extending ${f}$ to the closed square in a discontinuous way, say by letting it be the identity map on the boundary, we see that continuity is also needed, although it’s slightly non-intuitive that one needs continuity (essentially an upper oscillation bound) to estimate oscillation from below.
Theorem 2: If ${I\subset \mathbb R}$ is a bounded interval, then every light map ${f\colon I\to Y}$ admits a lower oscillation bound.
Proof. Following the proof of Theorem 1, consider a sequence of intervals ${(a_n, b_n)}$ such that ${b_n-a_n\ge \epsilon}$ and ${\textrm{diam}\, f((a_n,b_n))\to 0}$. There is no loss of generality in considering open intervals, since it can only make the diameter of the image smaller. Also WLOG, suppose ${a_n\to a}$ and ${b_n\to b}$; this uses the boundedness of ${I}$. Consider a nontrivial closed interval ${[c,d]\subset (a,b)}$. For all sufficiently large ${n}$ we have ${[c,d]\subset (a_n,b_n)}$, which implies ${\textrm{diam}\, f([c,d])\le \textrm{diam}\, f((a_n,b_n))\to 0}$. Thus ${f}$ is constant on ${[c,d]}$, a contradiction. ${\quad \Box}$