Is the Function Defined by the Infimum of Distances Continuous?

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SUMMARY

The function defined by \( f(x) = \inf\{d(x,a) : a \in A\} \) in a metric space \( (X,d) \) is continuous. This conclusion is established by demonstrating that for any two points \( x \) and \( y \) in \( X \), the inequality \( |f(x) - f(y)| \leq d(x, y) \) holds. This relationship confirms that the pre-image of any open set in \( \mathbb{R}^1 \) is also open, thereby satisfying the definition of continuity. The epsilon-delta definition can also be employed to rigorously prove continuity at any chosen point.

PREREQUISITES
  • Understanding of metric spaces, specifically the properties of distance functions.
  • Familiarity with the concept of infimum in real analysis.
  • Knowledge of continuity definitions, including the epsilon-delta criterion.
  • Basic proficiency in mathematical proofs and inequalities.
NEXT STEPS
  • Study the epsilon-delta definition of continuity in depth.
  • Explore the properties of infimum and supremum in real analysis.
  • Learn about open and closed sets in metric spaces.
  • Investigate other continuity proofs in metric spaces, focusing on different types of functions.
USEFUL FOR

Mathematics students, particularly those studying real analysis or topology, as well as educators seeking to enhance their understanding of continuity in metric spaces.

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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.

Homework Equations





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)
 

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