Question
If S := {1/n ? 1/m : m, n ? N}, find sup S and inf S.
Explanation / Answer
To get an idea what inf and sup should be: Fixing m for a moment, we see that inf {1/n - 1/m} = 0 - 1/m (letting n get really large), and sup {1/n - 1/m} = 1 - 1/m (at n = 1). Now, letting n vary, we conjecture that inf S = -1 and sup S = 1 (in both cases, as m gets really large). As such, S has no max nor min. ----------------------- Now for the proofs. (i) inf S = -1. Since m and n are positive integers, 1/n - 1/m > 0 - 1/m = -1/m > -1/1 = -1; hence -1 is a lower bound for S. To show that this is the greatest lower bound, suppose that for some e > 0, -1+e is a lower bound for S. We want to show that this is not the case. To kill one variable, let m = 1 ==> 1/n - 1/m = 1/n - 1. Now, note that 1/n - 1 > -1+e is not true for all positive integers n, because 1/n - 1 > -1+e n < 1/e. So, if we choose n as a positive integer > 1/e, we have produced an element in S which is smaller than -1+e, yielding a desired contradiction. ---------------------- (ii) sup S = 1. This is done similarly. Since m and n are positive integers, 1/n - 1/m < 1/1 - 1/m = 1 - 0 = 1; hence 1 is an upper bound for S. To show that this is the least upper bound, suppose that for some e > 0, 1 - e is an upper bound for S. We want to show that this is not the case. To kill one variable, let n = 1 ==> 1/n - 1/m = 1 - 1/m. Now, note that 1 - 1/m < 1 - e is not true for all positive integers n, because 1 - 1/m < 1 - e m < 1/e. So, if we choose m as a positive integer > 1/e, we have produced an element in S which is smaller than 1 - e, yielding a desired contradiction.