Product topology, closed subset, Hausdorff

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SUMMARY

The discussion centers on proving that the graph of a continuous function f : X → Y, where Y is a Hausdorff space, is a closed subset of the product space X × Y. Participants suggest using the properties of closed sets and the concept of limit points to establish the proof. The key approach involves demonstrating that for any point (x,y) not in the graph, there exist neighborhoods that separate (x,y) from points in the graph, leveraging the continuity of f and the Hausdorff property of Y.

PREREQUISITES
  • Understanding of topological spaces and their properties
  • Knowledge of continuous functions in topology
  • Familiarity with the Hausdorff condition in topological spaces
  • Concept of closed sets and limit points in topology
NEXT STEPS
  • Study the properties of closed sets in topological spaces
  • Learn about the implications of the Hausdorff property on continuity
  • Explore examples of continuous functions and their graphs in topology
  • Investigate the relationship between convergence and neighborhoods in topological spaces
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Students and researchers in mathematics, particularly those studying topology, as well as educators looking for insights on teaching concepts related to continuous functions and closed sets.

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Homework Statement



Let (X,\tau_X) and (Y,\tau_Y) be topological spaces, and let f : X \to Y be continuous. Let Y be Hausdorff, and prove that the graph of f i.e. \graph(f) := \{ (x,f(x)) | x \in X \} is a closed subset of X \times Y.

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The Attempt at a Solution



Which property of closed set should I use to prove this? Should I assume a sequence inside the graph set converging to some (x,y) \in X and then somehow show that this limit point belongs to the graph? Or should I prove that the complement of the graph set is not open? I don't know how to finish the proof with either approach. Please give me some hint.
 
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The idea here is to find (x,y) not on the graph such that every neighbourhood misses the graph. To do this, pick (x,y) not on the graph, so it is different than (x,f(x)) and separate them by neighbourhoods. You must also use continuity here.
 

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