Mapping Function for (0,1) into Open Unit Square

[cragar]

Homework Statement

Consider the open interval (0,1), and let's 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.

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 [itex] 2^x3^y [/itex] .
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 [itex] 2^{-x}3^{-y} [/itex]
and I think this function might be onto.

 

[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).

 

[micromass]

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).

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

 

[HallsofIvy]

Yes, thanks.

 

[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.

 

[HallsofIvy]

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

 

[cragar]

I don't understand what you mean. why can't 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.

 

[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).

 

[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

 

[Deveno]

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

 

[cragar]

how about y=x