You can use a bijection to prove the interval A=[1,∞) is the

>>9112859
>can you prove [1,∞) = or ≠ (1,∞)?
They're equal since both have cardinality of the continuum.

>>9112859
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.

map each non-integer to itself and each integer n to n+1, there, a bijection

>>9112948
Doesn't work because there's countable infinite integers and uncountable infinite R\Z by cantors proof. So no bijection exists.

>>9112955
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.

>>9112955
What? ]1,inf[ is just [1,inf[\{1}

>>9112960
>>9112962
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.

>>9112969
>Since you can't count all the numbers, you can't make the bijection.
Wrong.

>>9112969
>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

>>9112975
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.

>>9112985
You can't map it to itself. You have to have two different sets.

>>9112948
You sure showed him senpai.

He thinks bijections must be continuous lol

>>9112986
>Show me how to prove that the non integer reals can be split in to two countable sets
No one made this claim.

>>9112993
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

>>9112986
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.

>>9113003
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.

>>9113010
>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.

>>9113010
Is this some Wildburger crap?

>>9113012
>>9113014
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.

>>9113019
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...

>>9113023
>not paying attention in analysis

>>9113023
He's a troll. If you were the
>>9112948, then you were right. Fuck people like that guy, though, for making others not want to answer innocent questions.

>>9113032
>>9113010
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.

>>9113019
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]

>>9113049
Woops, replace the zeros with 1's.

>>9112955
>>9112969
>>9112986
>>9113010
>>9113019
ignore this guy. either he's trolling or he's genuinely retarded.

>>9112859
Their equal as sets, but they are not equal topologically

>>9112948
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.

>>9114098
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<y<x[/math].

As such, you can't map least to least. Similar proof holds for rationals.

>>9114132
the usual order on reals is not in question. by the axiom of choice, reals can be well ordered by some other relation

>>9114132
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

>>9112859
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,∞).

>>9112859
>can you prove [1,∞) = or ≠ (1,∞)?
yes? rationals to rationals map, and an irrational identity map.

>>9112859
What do you mean by =?

>>9114148
>>9114152
So does my original question >>9114098 have any issues or is it good?

>>9114975
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 ?

>>9115039
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?

>>9115118
nowhere does it imply that this assignment N -> R is surjective

>>9115118
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.

>>9112859
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

>>9116864
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.

>>9116827
Thankyou! That was the piece that I was missing

>>9112859
[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

>>9112859
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} ?

>>9118232
Or is it {{1},{2}}

>>9118239
And what does spamming brackets do anyway

>>9118250
1 is an element of {1, 2}
{1} is an element of {{1}, 2}
1 is not an element of {{1}, 2}

>>9118258
So the first guess then?