Proving Metric Space Reflexivity with Three Conditions

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SUMMARY

The discussion focuses on proving that the three specified conditions of a metric space imply the symmetry condition, d(x, y) = d(y, x). The conditions are: (1) d(x, y) ≥ 0 for all x, y in R, (2) d(x, y) = 0 if and only if x = y, and (3) d(x, y) ≤ d(x, z) + d(z, y) for all x, y, z in R. The conclusion drawn is that these conditions collectively establish the reflexivity of the metric space without needing to state it explicitly.

PREREQUISITES
  • Understanding of metric spaces and their properties
  • Familiarity with the triangle inequality in mathematics
  • Knowledge of equivalence relations in set theory
  • Basic proficiency in mathematical proofs and logic
NEXT STEPS
  • Study the properties of metric spaces in detail
  • Learn about equivalence relations and their implications in metric spaces
  • Explore examples of metric spaces that satisfy the three conditions
  • Investigate the role of reflexivity in various mathematical contexts
USEFUL FOR

Mathematics students, educators, and anyone interested in the foundational concepts of metric spaces and their properties.

GridironCPJ
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Homework Statement



Show that the following three conditions of a metric space imply that d(x, y)=d(y, x):

(1) d(x, y)>=0 for all x, y in R
(2) d(x, y)=0 iff x=y
(3) d(x, y)=<d(x, z)+d(z, y) for all x, y, z in R

(Essentially, we can deduce a reduced-form definition of a metric space, one without explicitly stating the reflexivity condition because the other 3 conditions imply it)

Homework Equations



The three conditions above.

The Attempt at a Solution



I've gone in circles, getting nowhere.
 
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I would start looking for a counterexample of what you're trying to prove.
 

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