You can use a bijection to prove the interval A=[1,∞) is the same size as B=[2,∞)...

The bijection from set A to set B requires each set has the same number of elements. Show me how to prove that the non integer reals can be split in to two countable sets and I'll show you a field medal.

You can't map it to itself. You have to have two different sets.

You sure showed him senpai.

He thinks bijections must be continuous lol

>Show me how to prove that the non integer reals can be split in to two countable sets
No one made this claim.

What the fuck are you talking about? OP asked for a bijection from [1, infinity) to (1, infinity), the latter is just the former without the element 1, so we map any non-integer real number x in [1, infinity) to x in (1, infinity) and the map integers n to n+1

What are you talking about? He's splitting them into a countable set and an uncountable one, mapping the uncountable one to itself and translating the countable one by 1.

It's fine and dandy for you to say it. But why don't you write a formal proof and I'll tell you why you're wrong.

Again, it's impossible to map the in betweens to themself because there's no way to prove what they are.

>But why don't you write a formal proof and I'll tell you why you're wrong.
If you don't already understand the bijection then you're probably not cut out for math.

Is this some Wildburger crap?

No, a bijection is injective and surjective by definition.

The non integer reals are uncountably infinite. This means that no injection and or surjection can exist.

Therfore, there is never a bijection for an uncountably infinite set.

Go back to wiki and read the definitions for each and then read about cantors diagonal argument.

Are you out of your fucking mind? You're saying that uncountable sets can't have surjections on them? The identity function is a bijection on any set...

>not paying attention in analysis

He's a troll. If you were the
, then you were right. Fuck people like that guy, though, for making others not want to answer innocent questions.

Let f:[1, infinity) -> (1, infinity) be defined as:
x -> x, if x is not an integer
x -> x+1, if x is an integer
We want to show that f is a bijection, so we need to show that f is injective and surjective.
Note that [1, infinity) = (1, infinity) [math]\cup[/math] {1}.
Let x,y [math]\in[/math] [1, infinity), then if x=y, f(x)=f(y) and if x [math]\neq[/math] y then we have 4 cases:
1) x,y are both integers, then f(x)=x+1, f(y)=y+1, since x [math]\neq[/math] y, x+1 [math]\neq[/math] y+1, so f(x) [math]\neq[/math] f(y).
2) x is an integer, and y is not an integer, then f(x)=x+1 is an integer and f(y)=y is not an integer, thus f(x) [math]\neq[/math] f(y).
3) x is not an integer and y is an integer, then f(x)=x is not an integer and f(y)=y+1 is an integer, thus f(x) [math]\neq[/math] f(y).
4) x,y are not integers, so f(x)=x and f(y)=y, since x [math]\neq[/math] y, f(x) [math]\neq[/math] f(y).
Thus f is injective.
Now, let z [math]\in[/math] (1,infinity), then either:
1) z is an integer, then z>=2, so z-1 [math]\in[/math] [1,infinity) and z-1 is an integer, so f(z-1)=z-1+1=z, or:
2) z is not an integer, then since z [math]\in[/math] (1,infinity) and (1,infinity) [math]\subset[/math] [1,infinity), then z [math]\in[/math] [1,infinity), so f(z)=z.
Thus f is surjective.
Therefore f is a bijection.

What's wrong with this function?

[math]f : [1,\infty) \to (1,\infty) , f(x) = \begin{cases}x ~ &\text{if} ~x \in \mathbb{R}\setminus\mathbb{Z}\\ x + 1 ~ &\text{if} ~ x \in \mathbb{Z} \end{cases} [/math]

It is surjective since for any [math] y \in (0,\infty)\setminus\mathbb{Z} [/math], f(y) = y and any [math] y \in \mathbb{Z}\cap (0,\infty) [/math], f(y-1) = y. Now for injectivity, [math] f(x_1) = f(x_2) \implies x_1 = x_2 ~\text{or} ~ x_1 + 1 = x_2 + 1 \implies x_1 = x_2.[/math]

Woops, replace the zeros with 1's.

ignore this guy. either he's trolling or he's genuinely retarded.

Their equal as sets, but they are not equal topologically

I know this is the correct answer, but would it also be right to use the fact that the Reals can be well ordered?

Well order both [1,inf] and (1,inf) and put least->least, second least->second least and so on.

Does this definition have some issue with the uncountability of the reals? Since it seems to only define a countably finite subsection of reals.

The problem is not countability, as it does not work even for rationals.

Claim: There is no least [math]x\in\mathbb{R}:x>1[/math].
Proof: The average [math]y=\frac{1+x}{2}[/math] is real and has [math]1

the usual order on reals is not in question. by the axiom of choice, reals can be well ordered by some other relation

Are you retarded? In (R, ≤), that is, reals with usual order you're right, there are subsets with no least element, but every set, including set of reals, admits a well order, and if we well order reals then it's always possible to find the least element in any nonempty subset of R

By Schröder-Bernstein theorem, if there are injections between the two sets, there is a bijection. The injection from (1,∞) to [1,∞) is obvious. The function [math]1+e^x[/math] is injective in the reals and has image (1,∞), so its restriction to [1,∞) is an injection from [1,∞) to (1,∞).

>can you prove [1,∞) = or ≠ (1,∞)?
yes? rationals to rationals map, and an irrational identity map.

What do you mean by =?

So does my original question have any issues or is it good?

saying "second least to second least etc" is kinda misleading since the reals cannot be enumerated in the first place. what is your general formula ?

I guess that comes down to the heart of my issue. I don't understand how well ordering doesn't imply that enumeration is possible.

Let's say that we well order the reals. Then any set contained within the reals will have a least element.

Let X1 be the least element of the reals
Let X2 be the least element of the reals-{X1}
Let X3 be the least element of the reals-{X1,X2}
.....
Let Xn bet the least element of the reals-{X1,X2,....Xn-1}

Now it seems like the reals have been enumerated.

I know it is impossible to do this, so where is the issue in my logic?

nowhere does it imply that this assignment N -> R is surjective

Well-ordering does not preclude the possibility of elements with no unique predecessor. An element may have infinitely many elements less than it. For example, [math]\mathbb{N}\cup \left \{ \infty \right \}[/math] with [math]\forall n . n < \infty[/math] is well-ordered.

You can just use the Cantor Bernstein theorem to show that they are of the same caridnality. No bijection required.

Guys am I wrong or can you do this and preserve the standard ordering as well? I think I did this proof during undergrad but I forget it

Not if you want a bijection. Since (1,∞) is open, for whatever value of f(1) there will be an element of (1,∞) less than that, but it cannot be mapped to any other element of [1,∞) which would be greater than 1.

Thankyou! That was the piece that I was missing

[math]\left[1,+\infty\right)=\left(1,+\infty\right)\cup \{1\}[/math], and the fact that [math]\#\left(A\right)=\#\left(A\cup\{x\}\right)[/math] for any [math]x\not\in A[/math] is pretty elementary fact

Help a brainlet out here. I don't understand set notation and I'm scared to write down my thoughts using them.
Is it correct to write {1}U{2} = {1,2} ?

Or is it {{1},{2}}

And what does spamming brackets do anyway

1 is an element of {1, 2}
{1} is an element of {{1}, 2}
1 is not an element of {{1}, 2}

So the first guess then?