Yeah, OP seems to be assuming a continuous mapping. It still works if you don’t, but the standard way to prove it is the more abstract “diagonal argument”.
Yeah, that was actually an awkward wording, sorry. What I meant is that given a non-continuous map from the natural numbers to the reals (or any other two sets with infinite but non-matching cardinality), there’s a way to prove it’s not bijective - often the diagonal argument.
For anyone reading and curious, you take advantage of the fact you can choose an independent modification to the output value of the mapping for each input value. In this case, a common choice is the nth decimal digit of the real number corresponding to the input natural number n. By choosing the unused value for each digit - that is, making a new number that’s different from all the used numbers in that one place, at least - you construct a value that must be unused in the set of possible outputs, which is a contradiction (bijective means it’s a one-to-one pairing between the two ends).
Actually, you can go even stronger, and do this for surjective functions. All bijective maps are surjective functions, but surjective functions are allowed to map two or more inputs to the same output as long as every input and output is still used. At that point, you literally just define “A is a smaller set than B” as meaning that you can’t surject A into B. It’s a definition that works for all finite quantities, so why not?
Give me an example of a mapping system for the numbers between 1 and 2 where if you take the average of any 2 sequentially mapped numbers, the number in-between is also mapped.
because I assumed continuous mapping the number c is between a and b it means if it has to be mapped to a natural number the natural number has to be between 22 and 23 but there is no natural number between 22 and 23 , it means c is not mapped to anything
Then you did not prove that there is no discontiguous mapping which maps [1, 2] to the natural numbers. You must show that no mapping exists, continugous or otherwise.
Your explanation is wrong. There is no reason to believe that “c” has no mapping.
Edit: for instance, it could map to 29, or -7.
Yeah, OP seems to be assuming a continuous mapping. It still works if you don’t, but the standard way to prove it is the more abstract “diagonal argument”.
But then a simple comeback would be, “well perhaps there is a non-continuous mapping.” (There isn’t one, of course.)
“It still works if you don’t” – how does red’s argument work if you don’t? Red is not using cantor’s diagonal proof.
Yeah, that was actually an awkward wording, sorry. What I meant is that given a non-continuous map from the natural numbers to the reals (or any other two sets with infinite but non-matching cardinality), there’s a way to prove it’s not bijective - often the diagonal argument.
For anyone reading and curious, you take advantage of the fact you can choose an independent modification to the output value of the mapping for each input value. In this case, a common choice is the nth decimal digit of the real number corresponding to the input natural number n. By choosing the unused value for each digit - that is, making a new number that’s different from all the used numbers in that one place, at least - you construct a value that must be unused in the set of possible outputs, which is a contradiction (bijective means it’s a one-to-one pairing between the two ends).
Actually, you can go even stronger, and do this for surjective functions. All bijective maps are surjective functions, but surjective functions are allowed to map two or more inputs to the same output as long as every input and output is still used. At that point, you literally just define “A is a smaller set than B” as meaning that you can’t surject A into B. It’s a definition that works for all finite quantities, so why not?
Give me an example of a mapping system for the numbers between 1 and 2 where if you take the average of any 2 sequentially mapped numbers, the number in-between is also mapped.
because I assumed continuous mapping the number c is between a and b it means if it has to be mapped to a natural number the natural number has to be between 22 and 23 but there is no natural number between 22 and 23 , it means c is not mapped to anything
Then you did not prove that there is no discontiguous mapping which maps [1, 2] to the natural numbers. You must show that no mapping exists, continugous or otherwise.