let (X,d) be a metric space which is not totally bounded. prove that these exists a positive A and a sequence {xn} in X such that d(xm,xn)>= A whenever m not equal to n
Question
let (X,d) be a metric space which is not totally bounded. prove that these exists a positive A and a sequence {xn} in X such that d(xm,xn)>= A whenever m not equal to n
Solution
The proof is by contradiction.
Step 1: Assume the opposite, i.e., assume that for every positive A, there exists a sequence {xn} in X such that d(xm, xn) < A whenever m ≠ n.
Step 2: This means that for any positive A, we can cover X with a finite number of balls of radius A. This is the definition of a totally bounded space.
Step 3: But we are given that X is not totally bounded. This contradicts our assumption in Step 1.
Step 4: Therefore, there must exist a positive A and a sequence {xn} in X such that d(xm, xn) ≥ A whenever m ≠ n.
This completes the proof.
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