this post was submitted on 23 Oct 2024
129 points (98.5% liked)
Asklemmy
43885 readers
1872 users here now
A loosely moderated place to ask open-ended questions
Search asklemmy ๐
If your post meets the following criteria, it's welcome here!
- Open-ended question
- Not offensive: at this point, we do not have the bandwidth to moderate overtly political discussions. Assume best intent and be excellent to each other.
- Not regarding using or support for Lemmy: context, see the list of support communities and tools for finding communities below
- Not ad nauseam inducing: please make sure it is a question that would be new to most members
- An actual topic of discussion
Looking for support?
Looking for a community?
- Lemmyverse: community search
- sub.rehab: maps old subreddits to fediverse options, marks official as such
- !lemmy411@lemmy.ca: a community for finding communities
~Icon~ ~by~ ~@Double_A@discuss.tchncs.de~
founded 5 years ago
MODERATORS
you are viewing a single comment's thread
view the rest of the comments
view the rest of the comments
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.
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.
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?
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.