MHB Real Analysis Help: Metric Spaces

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Two metrics p and T on the same set X are equivalent if and only if there exists a constant c > 0 such that (1/c)T(u,v) ≤ p(u,v) ≤ cT(u,v) for all u, v in X. The discussion highlights confusion regarding the definition of metric equivalence, clarifying that the usual definition involves the convergence of sequences rather than the strong equivalence condition. It emphasizes that the strong equivalence condition is more stringent than the basic convergence definition. The conversation references a Wikipedia page for further clarification on the topic. Understanding these distinctions is crucial for grasping the concepts in Real Analysis.
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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.
 
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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.
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).
 
We all know the definition of n-dimensional topological manifold uses open sets and homeomorphisms onto the image as open set in ##\mathbb R^n##. It should be possible to reformulate the definition of n-dimensional topological manifold using closed sets on the manifold's topology and on ##\mathbb R^n## ? I'm positive for this. Perhaps the definition of smooth manifold would be problematic, though.

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