Prove Hausdorff is a Topological Property

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SUMMARY

The discussion centers on proving that the Hausdorff property is a topological property, specifically under homeomorphisms. It establishes that if (X,T) is a Hausdorff space, then for any topological space (Y,U) with a homeomorphism f, (Y,U) must also be Hausdorff. The key argument involves demonstrating the existence of disjoint neighborhoods around points f(x1) and f(x2) in (Y,U) by leveraging the disjoint neighborhoods O1 and O2 in (X,T). A contradiction approach is suggested to solidify the proof.

PREREQUISITES
  • Understanding of topological spaces and their properties
  • Familiarity with homeomorphisms and continuous functions
  • Knowledge of neighborhood systems in topology
  • Basic proof techniques, including contradiction
NEXT STEPS
  • Study the properties of Hausdorff spaces in detail
  • Learn about homeomorphisms and their implications in topology
  • Explore neighborhood systems and their role in topological proofs
  • Practice proof techniques, particularly proof by contradiction
USEFUL FOR

Mathematics students, particularly those studying topology, educators teaching topological concepts, and anyone interested in understanding the implications of the Hausdorff property in various topological contexts.

tylerc1991
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Homework Statement



Prove that Hausdorff is a topological property.

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The Attempt at a Solution



For showing that a quality transfers to another space given a homeomorphism, we must show that given a Hausdorff space (X,T) and a topological space (Y,U), that (Y,U) is Hausdorff. So given two points in (Y,U), say f(x1) and f(x2), does there exist a pair of disjoint neighborhoods around both f(x1) and f(x2), call them D1 and D2. Well since (X,T) is Hausdorff, there exists a pair of disjoint neighborhoods around x1 and x2, call them O1 and O2. I am really just stuck with this problem and could use a slight push in the right direction. Any help would be greatly appreciated.
 
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As you have a homeomrophism, you have continuous bijection between spaces, with continuous inverse

maybe try and assume the 2 disjoint sets containing x1 & x2 are mapped to a non-disjoint set and look for a contradiction
 

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