TOPOLOGY 541 Let M be an arbitrary set and define D M M as

TOPOLOGY 541

Let M be an arbitrary set and define D : M ×M as follows: D(x,x) = 0 for all x M; for x = y,D(x,y) = D(y,x) = t where t [1, 2]. prove that (M, D) is a metric space.

Solution

To prove that (M,D) is a metric space, we must check the four axioms of metric space for M.

1) d(x,y) = t >0, since 0<t<1. Hence first axiom holds good.

2) d(x,x) =0 given. Hence if d(x,y) =0 then x =y thus second axiom holds good.

3) Given that d(x,y) = d(y,x) . Hence third axiom also holds good.

4) Let d(x,z) = t3. and d(x,y) = t1, d(y,z) = t2

By triangle inequality we get d(x,z) <= d(x,y) +d(y,z)

Hence M,D is a metric space