Let's discuss Mathematics

[Souron]

I just mean the quantity by which we judge something to be equal by. If that quantity is cardinality, then {1} is non zero. If that quantity is length, then a point is zero.

Yes, it's arbitrary that we compare numbers by cardinality, and curves by length, but we do most commonly.

 

[ParadigmShifter]

Well that's because the number of points on a continuous curve over an interval in R is infinite, so we need something else to use as a measure of its size.

 

[IdiotsOpposite]

Well that's because the number of points on a continuous curve over an interval in R is infinite, so we need something else to use as a measure of its size.

What, I can't claim that [0,2] is bigger than [0,1]?

 

[ParadigmShifter]

Yes, using measure instead of dimension. They both have the same number of points though Just like the number of points in N, Z and Q are the same.

 

[Souron]

What, I can't claim that [0,2] is bigger than [0,1]?

Well there is a 1 to 1 correspondence between these two sets...

Speaking of which, how does one prove that for uncountable sets?

 

[ParadigmShifter]

The same way... find a bijective mapping from one set to the other.

 

[Atticus]

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.

 

[ParadigmShifter]

What mapping is it that maps [0,1] onto R?

 

[ParadigmShifter]

Never mind. You can use the tangent function for that. EDIT: Obviously you need to scale and translate the domain.

 

[Atticus]

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

 

[ParadigmShifter]

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?

 

[Atticus]

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...

 

[ParadigmShifter]

I call them surjections too!

Sometimes we say a function is onto as well.

 

[dutchfire]

It should generally be impossible to find a continuous (assuming the normal topology) bijection [0,1]-> R since R is open while [0,1] is not and the inverse image of a open set is open. Right?

 

[ParadigmShifter]

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).

 

[Atticus]

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.

 

[dutchfire]

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.

 

[Petek]

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.

 

[Atticus]

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.

 

[IdiotsOpposite]

What is the difference in cardinality between [0,1] and [0,1)?