# Example of h: R->R^2 such that h is onto

1. Jun 5, 2007

### kkitkat

example of h: R-->R^2 such that h is onto

Hello everyone,

can anyone suggest an example of a function h: R --> R^2 such that h is onto. All I could come up with is the following:

h: R --> R^2 ==>

f(x) = y1
g(x) = y2

==> for h to be onto I need to find a function that gives me the following:

f^(-1)(y1) = g^(-1)(y2),

where f^(-1) is the inverse of f and g^(-1) is the inverse of g,

but I can't think of any. Can anyone please give me a hint in the right direction.

Thanks a lot

2. Jun 5, 2007

### matt grime

What relation do f and/or g have to do with h?

It is trivial to find examples like this, as long as you 'do it by force'. The two sets clearly have the same cardinality, so they are in bijection.

Or do you actually want a continuous surjection? Try the Peano space filling curve.

3. Jun 5, 2007

### kkitkat

Well, as far as I understand R-->R^2 functions it goes like this: h(t)=(3t, t) so in this example f(t) is 3t and g(t) is t. But obviously h(t) is not onto. And I'm sure that the solution is trivial, but for some reason I just can't see it. I'm not a mathematician so most of the things you've mentioned are away over my head :( If anyone could dumb it down for me a bit, I would really appreciate it

4. Jun 5, 2007

### matt grime

You're thinking of far too nice functions. Functions typically don't look like t-->3t.

How about this one: take a real number, write it in decimal notation, then do something with the digits?

You are never going to get a function by thinking about t-->(f(t),g(t)). For a start, fix an x and consider {t: f(t)=x}. That will have to be an uncountable set: h has to map to every pair (x,y), and there are uncountably many y to hit for each x.