You can use a bijection to prove the interval A=[1,∞) is the same size as B=[2,∞), as any number n of A has an equivalent n+1 in B.
Question is: can you prove [1,∞) = or ≠ (1,∞)? If you try to make a bijection you would have to add an infinitesimal, which is more of a concept than a number you would add to a number set, so i'm really confused on this one
You can use a bijection to prove the interval A=[1,∞) is the same size as B=[2,∞)...
Daniel Johnson
Lincoln Walker
>can you prove [1,∞) = or ≠ (1,∞)?
They're equal since both have cardinality of the continuum.
Brody Cook
The difference is in the first case you look at whole numbers only, while in the second case you look at all the reals. Otherwise (1, inf) would just be [2, inf) lol.
Andrew Campbell
map each non-integer to itself and each integer n to n+1, there, a bijection
Cooper Mitchell
Doesn't work because there's countable infinite integers and uncountable infinite R\Z by cantors proof. So no bijection exists.
Brayden Rogers
read my post again, all non-integers (real numbers between the integers) MAP TO THEMSELVES. That leaves us with only having to map the integers 1,2,3,... to 2,3,4,..., use the bijection n->n+1 and we get the bijection from [1, infinity] to (1, infinity) defined as:
x -> x, if x is not an integer
x -> x+1, if is an integer.
Nathaniel Wilson
What? ]1,inf[ is just [1,inf[\{1}
Jaxon James
In order to do this, you have to prove there's the same amount of numbers between 0 and 1 as there are between 1 and 2. But I can prove that it's impossible for you to prove this is the case. Since you can't count all the numbers, you can't make the bijection. It's that simple.
Jose Williams
>Since you can't count all the numbers, you can't make the bijection.
Wrong.
Dylan Hill
>you have to prove there's the same amount of numbers between 0 and 1 as there are between 1 and 2
Why do I have to do that? all numbers between n and n+1 get mapped to themselves, please read my post again you braindead fuck