ParadigmShifter
Random Nonsense Generator
The same way... find a bijective mapping from one set to the other.
Let's discuss Mathematics
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The word "bigger" is ambiguous. You can use it to say that at least in the sense of measure or number of elements (which is a measure too, btw), or if you have order. Thus even though set [0,1]U{2} isn't bigger in the sense of (Lebesgue) measure than [0,1], you can say it's bigger because inclusion of sets is a partial order. At least so I think.
1-to-1 correspondence between [0,1] and [0,2] is pretty easy,
f:[0,1] -> [0,2]: f(x)=2x
will do
In general, there is no one to one correspondence between uncountable sets, since they aren't necessarily of the same cardinality. There is no bijection from R to it's power set, although both of them are uncountable.
1-to-1 correspondence between [0,1] and [0,2] is pretty easy,
f:[0,1] -> [0,2]: f(x)=2x
will do
In general, there is no one to one correspondence between uncountable sets, since they aren't necessarily of the same cardinality. There is no bijection from R to it's power set, although both of them are uncountable.
ParadigmShifter
Random Nonsense Generator
What mapping is it that maps [0,1] onto R?
ParadigmShifter
Random Nonsense Generator
Never mind. You can use the tangent function for that. EDIT: Obviously you need to scale and translate the domain.
f:[0,1]-> R:
f(x) = 1/x when 0<x<½,
f(x) = 4 + 1/(x-1) when ½=<x<1,
f(0)=f(1)=0.
Bijection is of course little harder
f(x) = 1/x when 0<x<½,
f(x) = 4 + 1/(x-1) when ½=<x<1,
f(0)=f(1)=0.
Bijection is of course little harder
ParadigmShifter
Random Nonsense Generator
Hmm yeah tangent only works for an open interval as well.
But the bijection is easier
If f(0) = f(1) doesn't that rule out a bijection anyway?
But the bijection is easier
If f(0) = f(1) doesn't that rule out a bijection anyway?
Yes it does. (It's enough to show that there are surjections (or ontos as you call them) [0,1]->R and R->[0,1] to prove that they are of the same cardinality).
I can't quickly come up with any bijections.
EDIT: Did you happen to confuse bijection and injection (in to)? It can't be easier to come up with bijection, since every time you come up with one, you'll have a surjection too!
EDIT2: corrected the mis-wording...
I can't quickly come up with any bijections.
EDIT: Did you happen to confuse bijection and injection (in to)? It can't be easier to come up with bijection, since every time you come up with one, you'll have a surjection too!
EDIT2: corrected the mis-wording...
ParadigmShifter
Random Nonsense Generator
I call them surjections too!
Sometimes we say a function is onto as well.
Sometimes we say a function is onto as well.
ParadigmShifter
Random Nonsense Generator
Yeah I think you have a point.
We don't need a 1-1 correspondence anyway (since when we do the diagonal argument showing there are the same number of rationals as natural numbers we cover some rationals more than once as well).
We don't need a 1-1 correspondence anyway (since when we do the diagonal argument showing there are the same number of rationals as natural numbers we cover some rationals more than once as well).
Neither do we need a continuous mapping.
The two conditions below are equivalent:
1. There are injections X->Y and Y->X
2. There is a bijection X->Y.
It's called Schröder-Bernstein theorem, and I'm about to look up proof from Introduction to Topology and Modern Analysis by Simmons, pp 29-30.
The two conditions below are equivalent:
1. There are injections X->Y and Y->X
2. There is a bijection X->Y.
It's called Schröder-Bernstein theorem, and I'm about to look up proof from Introduction to Topology and Modern Analysis by Simmons, pp 29-30.
I know the bijection doesn't have to be continuous, I just wanted to point out that we won't be able to find a "nice" function that works, since [0,1] is closed (well, since it's not open). For the open interval (-1,1) a scaled tangent function is an easy example, as PS showed.
Here's a bijection F between [0,1] and (0, 1):
Define F(1) = 1/2
F(1/2) = 1/4
F(1/4) = 1/8
F(1/8) = 1/16, and so on.
Define F(0) = 1/3
F(1/3) = 1/9
F(1/9) = 1/27
F(1/27) = 1/81, and so on.
For all other x in [0,1], define F(x) = x.
Also, if I've done the math right, G(x) = tan(pi(x - (1/2)) defines a bijection between (0,1) and R. Compose F and G to get a bijection between [0,1] and R. Not continuous, of course.
Define F(1) = 1/2
F(1/2) = 1/4
F(1/4) = 1/8
F(1/8) = 1/16, and so on.
Define F(0) = 1/3
F(1/3) = 1/9
F(1/9) = 1/27
F(1/27) = 1/81, and so on.
For all other x in [0,1], define F(x) = x.
Also, if I've done the math right, G(x) = tan(pi(x - (1/2)) defines a bijection between (0,1) and R. Compose F and G to get a bijection between [0,1] and R. Not continuous, of course.
That's pretty clever! 
I'm sooooo jealous for not coming up with that. Especially since the previous page was all about how a subset can be equal in cardinality to the set itself.
I'm sooooo jealous for not coming up with that. Especially since the previous page was all about how a subset can be equal in cardinality to the set itself.
IdiotsOpposite
Boom, headshot.
What is the difference in cardinality between [0,1] and [0,1)?
ParadigmShifter
Random Nonsense Generator
They have the same cardinality.
Just rip off Petek's argument, define
f(1) = 1/2
f(1/2) = 1/4
f(1/4) = 1/8
etc.
and all the other f(x) = x
Then you have a mapping from [0,1] to [0,1)
Just rip off Petek's argument, define
f(1) = 1/2
f(1/2) = 1/4
f(1/4) = 1/8
etc.
and all the other f(x) = x
Then you have a mapping from [0,1] to [0,1)
IO, everything in R^n you can come up with has either the cardinality of R or is countable.
There is a little twist to it called the continuum hypothesis, which says that there are no cardinalities between those two. It is known that this hypothesis can't be proven to be true or false (Gödel& Cohen, 30s and 60s).
Anyway, as a result, you won't come up with a set in R^n that has other cardinality than the mentioned two, since otherwise you would have proven the hypothesis to be false.
There is a little twist to it called the continuum hypothesis, which says that there are no cardinalities between those two. It is known that this hypothesis can't be proven to be true or false (Gödel& Cohen, 30s and 60s).
Anyway, as a result, you won't come up with a set in R^n that has other cardinality than the mentioned two, since otherwise you would have proven the hypothesis to be false.
Here's another cardinality problem:
Let S be an infinite set. Let S' be the set defined by removing a single element from S. Prove that S' is infinite.
Let S be an infinite set. Let S' be the set defined by removing a single element from S. Prove that S' is infinite.
ParadigmShifter
Random Nonsense Generator
Well that's easy if S is countable. EDIT: I mean countably infinite, of course
Does the rest of the proof use transfinite induction?
Does the rest of the proof use transfinite induction?
IdiotsOpposite
Boom, headshot.
Holy hell, CFC has moderators?
I never understood Hilbert's Hotel. Could someone explain it to me in proper math notation?
I never understood Hilbert's Hotel. Could someone explain it to me in proper math notation?
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