Function continuity in metric spaces

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SUMMARY

The discussion centers on proving the equivalence of continuity in metric spaces, specifically between the statements that a function f: X → Y is continuous and that the closure of the inverse image of any subset B in Y is contained within the inverse image of the closure of B. The participants confirm that proving statement (1) leads to statement (2) is straightforward, while demonstrating the reverse requires applying the theorem that the inverse image of every closed set in Y is closed in X. The use of closed sets and their closures is emphasized as a key approach in the proof.

PREREQUISITES
  • Understanding of metric spaces and their properties
  • Familiarity with the concept of continuity in topological spaces
  • Knowledge of closed sets and closures in metric spaces
  • Ability to work with inverse images of functions
NEXT STEPS
  • Study the properties of closed sets in metric spaces
  • Learn about the concept of closure in topological spaces
  • Explore the implications of continuity in different types of functions
  • Investigate the relationship between topological spaces and metric spaces
USEFUL FOR

Students of mathematics, particularly those studying topology and metric spaces, as well as educators seeking to deepen their understanding of function continuity and its proofs.

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Homework Statement



Let [tex](X,d_X)[/tex] and [tex](Y,d_Y)[/tex] be metric spaces and let [tex]f: X \to Y[/tex].

Homework Equations



Prove that the following statements are equivalent:

1. [tex]f[/tex] is continuous on [tex]X[/tex],
2. [tex]\overline{f^{-1}(B)} \subseteq f^{-1}(\overline{B})[/tex] for all subsets [tex]B \subseteq Y[/tex]

The Attempt at a Solution



I an prove that (1) leads to (2) but don't know how to show (2) leads to (1). Can you give me some hint?
 
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A function f: X -> Y between topological spaces is continuous if and only if the inverse image of every closed set in Y is a closed set in X.

Let F be an arbitrary closed set in Y. Then F = cl(F), where cl(F) is the closure of F. Now apply (2) with F in place of B.
 
Thanks, I didn't realize I should use this theorem and was trying to prove it using any point x and its neighborhood approach.
 

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