# Math Help - Bijection in infinite sets

1. ## Bijection in infinite sets

Okay this is just a musing I'm having and it might be already talked about or proven or whatever.
If G is an infinite set and H is an infinite subset of G, can a bijection occur between G and H?
In this case G does not equal H.

For example the set R, of real numbers and Z the set of integers. Both sets are infinite, however R has a higher level of infinity than Z, obviously. So can a bijection exist between one such infinite set and its infinite subset?

2. Originally Posted by Zener
Okay this is just a musing I'm having and it might be already talked about or proven or whatever.
If G is an infinite set and H is an infinite subset of G, can a bijection occur between G and H?
In this case G does not equal H.

For example the set R, of real numbers and Z the set of integers. Both sets are infinite, however R has a higher level of infinity than Z, obviously. So can a bijection exist between one such infinite set and its infinite subset?
Yes. If $X$ is infinite and $A$ is any finite subset define $X - A = \{ x\in A | x\not \in A \}$.
Then there is bijection from $X-A$ to $X$.

3. Ah but for example:
0<x<1 where x is an element of R, is an infinite set. Then consider N, the subset of R of natural numbers. N has a cardinal number d, but the infinite set 0<x<1 has a cardinal number c that is greater than d as the set is nondenumerable. Therefore there couldn't be a bijection between these two sets.
I get what you are trying to say about X-A also being infinite like X, and an infinite set was actually defined once as a set that is equivalent to a proper subset of itself.
My point is simply that the argument doesn't always work. A scale of arithmetic infinites exists.

4. Originally Posted by Zener
Ah but for example:
0<x<1 where x is an element of R, is an infinite set. Then consider N, the subset of R of natural numbers. N has a cardinal number d, but the infinite set 0<x<1 has a cardinal number c that is greater than d as the set is nondenumerable. Therefore there couldn't be a bijection between these two sets.
I understand up to here correctly.However.....

I get what you are trying to say about X-A also being infinite like X, and an infinite set was actually defined once as a set that is equivalent to a proper subset of itself.
My point is simply that the argument doesn't always work. A scale of arithmetic infinites exists.
What do you mean the argument doesnt work? If there is a bijection between two sets, doesnt it mean that they have the identical cardinal numbers?

I dont understand what doesnt work... Can clearly spell it out for us?

Thanks

5. I never said there was a bijection between the two sets. I'm just having musings upon bijcetions between infinite sets.
Here's an example: The set of all algebraic numbers is denumerable. The set of complex numbers is not denumerable. Thus there must exist complex numbers that are not algebraic. These are transcendental numbers. One example is e.
I'm just voicing some inner thoughts on the crazy infinite set theory.

6. My point was that a bijection can occur across some infinite sets, but not all.

7. Originally Posted by Zener
My point was that a bijection can occur across some infinite sets, but not all.
Thats precisely why we have grouped the sets that have bijective correspondence into different classes. And we designate different cardinal numbers to it.

8. Exactly. As I said at the start this really was just a musing. So thanks for the help in allowing me to 'see' it clearly.

9. Originally Posted by Zener
Ah but for example:
0<x<1 where x is an element of R, is an infinite set. Then consider N, the subset of R of natural numbers. N has a cardinal number d, but the infinite set 0<x<1 has a cardinal number c that is greater than d as the set is nondenumerable. Therefore there couldn't be a bijection between these two sets.
I get what you are trying to say about X-A also being infinite like X, and an infinite set was actually defined once as a set that is equivalent to a proper subset of itself.
My point is simply that the argument doesn't always work. A scale of arithmetic infinites exists.
You are not following. I am saying if you remove finitely many elements from an infinite set then you can always form a bijection.

10. Actually I get exactly what you mean. You just misunderstand me, I'm not looking how to always form a bijection. Merely thinking aloud about the whole thing. I didn't really make this clear at the start but thanks for the help.