[DawkinsGawd]

Can someone explain this sentence to me:

"Consider now the subsequences starting at the smallest natural number: inclusion of the upper bound would then force the latter to be unnatural by the time the sequence has shrunk to the empty one."

The previous statement asserts that inclusion of the lower bound is preferred because the sequence then starts with a natural number (so for a set 1<=x<12 the sequence starts at 1 instead of 1.000000 .... 1 in 1<x<12). How is inclusion of the upper bound forcing the "latter" (what is he referring to by latter) to be unnatural by the time the sequence has shrunk to the empty one.

[to3m]

"The latter" here refers to the upper bound. By "unnatural" he means negative, as he is discussing the natural numbers (positive integers and zero). By convention these are the numbers used to index arrays. (So, for your examples: for the set 1<=x<12, the sequence is 1,2,3,4,5,6,7,8,9,10,11; for the set 1<x<12, the sequence is 2,3,4,5,6,7,8,9,10,11.)

As for what he's saying: suppose you decide your array indexing starts at the lowest natural number, i.e., 0. And imagine an array with 1 element. And assume you've adopted the notation a<=i<=b. This array's indexes are then 0<=i<=0.

But what if you have an array of 0 elements? What then? You can't say 0<=i<=-1, because -1 is not natural and therefore not a valid array index.

But if you adopt the convention that a<=i<b, then your 1-element array's indexes would be 0<=i<1, and your empty array would be 0<=i<0. (Or, indeed, 1<=i<1, or n<=i<n for any n you choose - the first section deals with how to denote the range, so the choice of n isn't the key thing.)

[DawkinsGawd]

Thank you for the explanation.