Two metrics p and T on the same set X are equivalent if and only if there exists a constant c > 0 such that for all u, v in X, the inequality (1/c)T(u,v) ≤ p(u,v) ≤ cT(u,v) holds. This condition is known as strong equivalence. The usual definition of metric equivalence, which states that convergence in one metric implies convergence in another, does not suffice to establish this strong equivalence. Understanding these definitions is crucial for grasping the nuances of metric spaces in Real Analysis.
PREREQUISITESStudents and educators in mathematics, particularly those focusing on Real Analysis and metric spaces, as well as researchers exploring the properties of different metrics.
What definition of equivalence for metrics are you using? The usual definition is that two metrics are equivalent if convergence of a sequence in one metric implies convergence in the other metric. That does not imply the condition $(1/c)T(u,v) \leqslant p(u,v) \leqslant cT(u,v)$ (for all $u,v\in X$), which is usually called strong equivalence of the two metrics (see the discussion in the Wikipedia page that I linked to above).wonguyen1995 said:Show that two metrics p and T on the same set X are equivalent if and only if there is a c > 0
such that for all u,v belong to X,
(1/c)T(u,v)=<p(u,v)=<cT(u,v)
Please help me , I'm so confused about Real Analysis.