Is the Function Defined by the Infimum of Distances Continuous?

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The discussion revolves around proving the continuity of the function f defined as f(x) = inf{d(x,a) : a ∈ A} in a metric space (X,d). Participants suggest that the intuition for continuity lies in the relationship between open intervals in R^1 and their pre-images under f. To formalize this, it's recommended to apply the epsilon-delta definition of continuity, particularly by demonstrating that |f(x) - f(y)| ≤ d(x, y). This approach can help establish continuity at any chosen point in the metric space. Overall, the focus is on clarifying the proof techniques for demonstrating the continuity of the infimum distance function.
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Homework Statement



Let (X,d) be a metric space and let A be a nonempty subset of X. Define a function f:X -> R^1 by f(x) = inf{d(x,a) : a is an element of A}. Prove that f is continuous.

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The Attempt at a Solution



Intuitively I can see that the function is continuous because it seems like for an arbitrary open interval in R^1 there is some pre-image of the function that is an open subset of this open interval, I just don't exactly know where to begin writing this. Can someone help me with the intuition behind this problem and let me know if I am on the right track? Thank you very much!
 
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If you don't know how to attack a continuity problem using the abstract (preimage of an open set is open) definition, try instead to prove that the function is continuous at any arbitrarily chosen point using the good old epsilon-delta definition.
 
In both cases, you may want to prove that
|f(x)-f(y)|\leq d(x, y)
 
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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