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#1 2007-02-09 08:22:39

rockysheedy
Member
Registered: 2006-06-11
Posts: 12

Metric Space d*(x,y) <= d(x,y) then how {xeX: d(p,x)<r} is the subset

In Metric Space if d and d* are two Metrics.

If d*(x,y) <= d(x,y) then how {x belongs to X: d(p,x)<r} is the proper subset of  {x belongs to X: d*(p,x)<r}. Is is correct? If yes could you please explain?

Shouldn't it be ---------------
If d*(x,y) <= d(x,y) then {x belongs to X: d*(p,x)<r} is the proper subset of  {x belongs to X: d(p,x)<r}.

The exact question and solution is as follows:

Question:- Let (X,d) be a Metric Space and let d*(x,y) = min {1,d(x,y)}. Even d* is a Metric. Show that d and d* are equivalent?

Solution:- Inorder to prove d and d* are equivalent it suffices to show that d-open sphere about a point p belongs to X contains a d*-open sphere about p and conversely.

Since d*(x,y) <= d(x,y) we have {x belongs to X: d(p,x)<r} is the proper subset of  {x belongs to X: d*(p,x)<r}. -------------------->This is where I am stucked how is it possible. How can {x belongs to X: d*(p,x)<r} be bigger than {x belongs to X: d(p,x)<r} when d*(x,y) is smaller or equal to d(x,y).

for every p belongs to X and every r > 0. On the other hand, if r is any positive number, let
q=min(r,1). Then {x belongs to X: d*(p,x)<q} is the proper subset of  {x belongs to X: d(p,x)<q}

For every p belongs to X.

It follows that d-open set is a d*-open and every d*-open set is d-open. Hence both are equivalent.

Last edited by rockysheedy (2007-02-09 08:23:34)

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