# Show continuously uniform function

• aeronautical
In summary, the conversation discusses the problem of showing that the function f(x) defined as the infimum of the distances between x and a non-empty subset F of a metric room (X,d) is continuously uniform in the entire X. The conversation also presents a general solution for this problem, which involves proving a stronger inequality. The key insight is to consider the special case where F consists of only two points, and then generalize the solution to the case of arbitrary F.

#### aeronautical

Let F be a non-empty subset of a metric room (X,d) and define the function f: X→R through f(x)= inf_{y∈F} d(x,y)= inf {d(x,y):y ∈ F}. Please show that f is continuously uniform in the entire X.

Please note that f(x)= inf_{y∈F} d(x,y) means that y∈F is written below the inf...it is hard to show it on one line, so I chose to show as a subscript...Can anyone tell me how I should begin this problem?

Strange coincidence, I've been trying to solve the same problem! Well, not exactly the same... I have been trying to show that the map $x\mapsto d(x, F)$ is Lipschitz continuous, meaning that there exists a constant C such that |d(x,F)-d(y,F)|≤Cd(x,y) for all x,y in X. Obviously Lipschitz continuity is stronger than uniform continuity since given ε>0, it suffices to take δ=ε/C. As it turns out, there is an exercise in the book Topology and Geometry of G. Bredon that asks the reader to show that the map $x\mapsto d(x, F)$ is (merely!) continuous. And a hint is provided in suggesting to show that this map is actually Lipschitz of constant C=1. Meaning that, somewhat surprisingly, the easiest way to solve your problem and mine is probably to demonstrate the stronger inequality |d(x,F)-d(y,F)|≤d(x,y) for all x,y in X. So let's try to show this.

The key for me was to consider first the very special case in which F consists of only two points: F={p,q}. In this scenario, the problem is not so overwhelming psychologically and after solving it, I realized that its solutions admits an immediate generalization to the case of arbitrary F.

The general solution I found is as follows. Tell me what you think.

First, note that we can assume without loss of generality that d(y,F)≤d(x,F) [if not, interchange the labels of x and y...]. In this case, |d(x,F)-d(y,F)|=d(x,F)-d(y,F) and so we must prove that d(x,F)≤d(x,y)+d(y,F) [i.e., we must prove a kind of triangle inequality in which sets are allowed to occupy one of the variable position].

Next, consider {e_n}, {f_n} two sequences of points in F such that d(x,F)=lim d(x,e_n) and d(y,F)=lim d(y,f_n) and further assume that d(x,e_n)≤d(x,f_n) for all n in N.

Then, we have d(x,y)+d(y,f_n) ≥ d(x,f_n) ≥ d(x,e_n) for all n in N and passing to the limit yields the desired conclusion.