Number of points on the plane vs. number of points on the line

[Helicobacter]

Is the number of points/cardinality in R^2 greater than R, or are they equal?

if they are equal, how does one explain the following: R^2 is a superset of R, it has infinitely many R's...

 

[disregardthat]

Cardinality is a specific measure of the "size" of a set which only deals with the equivalence relation that is: two sets A and B have equal cardinality if there is a bijective function from A to B. In this context R and R^2 have equal size, because we can find such a bijection. In fact, A and A^2 will generally have the same cardinality for infinite sets A.

That R^2 may seem "bigger" in a different context doesn't matter when it comes to cardinality. In another sense this is true: if you consider R as the x-axis of R^2, R will have lebesgue-measure 0 while R^2 will have infinite lebesgue measure. Both are useful ways of interpreting size, but it's important to keep the two concepts separate.

 

[g_edgar]

Wait: there's a bijection between the integers and the even integers, even though the big one has TWO copies (evens and odds) of the small one. Cardinality does these strange things.

 

[zetafunction]

but this should imply that the set of REALS is isomorphic To the set of complex

since ther would always be a biyection between a real number 'x' and a complex number a+bi

 

[Mark44]

but this should imply that the set of REALS is isomorphic To the set of complex

since ther would always be a biyection between a real number 'x' and a complex number a+bi

Here's a bijection. We can, without loss of generality, assume that the real numbers are in the interval [0, 1].

Write the real number x in its decimal form, as x = 0.d₁d₂d₃d₄d₅d₆...

Form the complex number z = 0.d₁d₃d₅... + 0.d₂d₄d₆... i

This is basically the same way that R can be mapped one-to-one to R².

 

[micromass]

And, the reals are indeed isomorphic to the complex numbers as additive groups! They're not isomorphic as fields however...

 

[disregardthat]

And, the reals are indeed isomorphic to the complex numbers as additive groups! They're not isomorphic as fields however...

Under what 'addition'?

 

[micromass]

The usual addition. The point is that $$\mathbb{R}$$ and $$\mathbb{C}$$ are $$\mathbb{Q}$$-vector spaces of thesame dimension. Thus the two spaces are isomorphic as vector spaces, but this exactly means that the additive groups are isomorphic.

This of course uses the axiom of choice...

 

[dodo]

I'm also curious about Jarle's question. For example, one way to prove that the interval (-pi/2, pi/2) has the same cardinality as the whole R, is to use the bijection f(x) = tan(x). But, since tan(x+y) is not in general equal to tan(x) + tan(y), more work appears to be needed in order to prove that these sets are isomorphic.

 

[micromass]

I'm also curious about Jarle's question. For example, one way to prove that the interval (-pi/2, pi/2) has the same cardinality as the whole R, is to use the bijection f(x) = tan(x). But, since tan(x+y) is not in general equal to tan(x) + tan(y), more work appears to be needed in order to prove that these sets are isomorphic.

Well, obviously R and (-pi/2,pi/2) cannot be isomorphic as groups since (-pi/2,pi/2) isn't even a group... But R and (-pi/2,pi/2) do have the same number of elements though (= the same cardinality). To prove that two sets have the same cardinality, it suffices to find a bijection between them, you don't care whether the group property is satisfied.

It is easily seen that R and C have the same cardinality. Mark supplied a nice proof of this. All I said was that they are also isomorphic as groups, but this requires a different proof then what Mark supplied. In fact, the proof is a little harder, since it is nonconstructive (= it requires the axiom of choice!).

 

[dodo]

Hi, MM,
(-pi/2, pi/2) is not a group... under the usual addition. But it can be a group under the addition operator '*', defined as x*y = arctan(tan(x) + tan(y)). Hence Jarle's question (and mine).

Now, if you say the proof is non-constructive, then there's no point on asking what the addition operation is... but man... curiosity itches...

Of course you can always repeat the trick, and say that, if there exists a bijection between R and C, call it g:R -> C, and if you define the operation '@' defined as x@y = g^-1(g(x) + g(y)), then (R,@) is isomorphic to (C,+). But again, that was not the assertion of that non-constructive proof.

 

[micromass]

Hi, MM,
(-pi/2, pi/2) is not a group... under the usual addition. But it can be a group under the addition operator '*', defined as x*y = arctan(tan(x) + tan(y)). Hence Jarle's question (and mine).

Now, if you say the proof is non-constructive, then there's no point on asking what the addition operation is... but man... curiosity itches...

Of course you can always repeat the trick, and say that, if there exists a bijection between R and C, call it g:R -> C, and if you define the operation '@' defined as x@y = g^-1(g(x) + g(y)), then (R,@) is isomorphic to (C,+). But again, that was not the assertion of that non-constructive proof.

Oops, sorry if my answer was not clear. But the addition is the usual addition. Thus

in R: a+b is just the usual addition.
in C: (a+bi)+(c+di)=(a+c)+(b+d)i is again the usual addition.

That's the point, that we are dealing with the usual additions. Not some "exotic" addition like arctan(tan(x)+tan(y)). The nonconstructive part comes in actually defining the isomorphism between the two groups.

 

[dodo]

OK... I was just curious about that proof. I'll try to google harder. :)

 

[disregardthat]

The usual addition. The point is that $$\mathbb{R}$$ and $$\mathbb{C}$$ are $$\mathbb{Q}$$-vector spaces of thesame dimension. Thus the two spaces are isomorphic as vector spaces, but this exactly means that the additive groups are isomorphic.

This of course uses the axiom of choice...

I agree, nifty solution. In general we require that the dimensions are of equal cardinality, which clearly is satisfied here. The isomorphism function would be tricky to find, it is impossible to write it down.