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The discussion focuses on demonstrating that the function f defined as f(x) = inf_{y∈F} d(x,y) is continuously uniform across the metric space (X,d). Participants suggest that proving Lipschitz continuity, specifically |d(x,F) - d(y,F)| ≤ d(x,y), is a strong approach to establish uniform continuity. A special case is considered where F consists of two points, simplifying the problem and allowing for generalization. The solution involves proving a triangle inequality for distances involving points in F. Ultimately, the discussion emphasizes the connection between Lipschitz continuity and uniform continuity in this context.
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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?
 
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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.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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