Here is the bottom line: if there is a one to one correspondence between all the odd numbers and all the whole natural numbers, then there is a ONE to ONE correspondence, and nothing missing.
Correct. Every natural number n corresponds with a single odd number 2n-1.
Therefore if I add one single number to the odd numbers
Above you said “all the odd numbers” so you have no more odd numbers to add. If you add a single non-odd number, say 4, then you no longer have the set of odd numbers, but a different set.
That new set is still in a one-to-one correspondence with the natural numbers. What you can do is to shift every odd number along one, leaving a space at the start for your extra number.
Before we have:
1 → 1
2 → 3
3 → 5
4 → 7
…
n → 2n-1
…
With a shift of 1 and inserting 4 at the start, we have:
1 → 4
2 → 1
3 → 3
4 → 5
5 → 7
…
n → 2n-3
…
Both sets are still in a one-to-one correspondence. Note that the formula changes, because of the shift of one position, and the formula no longer works for the first entry, since that is the inserted extra even number.
that set will be greater than the whole natural numbers, because there is NO NUMBER for it to correspond with, since there is already a complete one to one correspondence.
That correspondence can be shifted by 1, so leaving a space at the beginning to insert the additional number, 4 in my example.
I am pretty sure that the “Caesar shift encryption” is just blurring the regular 1 2 3 4 5 and to infinity set, and not one of its subsets.
It was an attempt to get you thinking in terms of sets, rather than in terms of numbers. This concept needs to be seen in terms of sets and their elements.
rossum