1-1 mapping question

1. Feb 3, 2012

cragar

1. The problem statement, all variables and given/known data
Consider the open interval (0,1), and lets S be the set of points in the open unit square
that is, S={(x,y):0<x,y<1}.
Find a function that maps (0,1) into S. but not necessarily onto.
3. The attempt at a solution
so I can describe any point in my square with x and y coordinates like (x,y)
so for all of my points ill just take so I will map the coordinate (x,y) to $2^x3^y$ .
so now each coordinate goes to one real. and (0,1) can be mapped to the whole real line so I shouldn't have problems with some of the numbers being bigger than one.
But I guess I could just map the coordinates to $2^{-x}3^{-y}$
and I think this function might be onto.

2. Feb 3, 2012

HallsofIvy

You appear to be completely misunderstanding the problem. There is NO "(x, y)". The problem asks you to map (0, 1), a subset of R, onto the square in R2.

You need to find a function that maps a single number, x, to a pair, (a, b).

3. Feb 3, 2012

micromass

Not even that. You just need to map into the square. It doesn't even need to be surjective.

4. Feb 3, 2012

Yes, thanks.

5. Feb 3, 2012

cragar

ok , so I just need to map the points from (0,1) into the square, but not nessicarily all the points in the sqaure to the line segmenet. but does my original statement work.

6. Feb 3, 2012

HallsofIvy

No, because there is NO "(x, y)" to begin with. If x is a number in (0, 1), what does x map to?

7. Feb 3, 2012

cragar

I dont understand what you mean. why cant you map x to x on the line.
what do you mean there is no (x,y) to begin with.
thanks for the help by the way.

8. Feb 3, 2012

Deveno

the domain of your map (function) is 1-dimensional (a line segment, minus its endpoints).

the co-domain (set that contains the range) is 2-dimensional (interior of a square).

your map should look something like this:

f(t) = (x(t),y(t))

i can think of a possible map that has a fairly simple form for both x(t) and y(t).

9. Feb 4, 2012

cragar

so could I just map my my line into the square using f(t)=(x,1/2)
where x varies and y is always at 1/2

10. Feb 4, 2012

Deveno

that's closer, but how is "x" determined....?

11. Feb 4, 2012