Continuous, Onto Function: (0,1)x(0,1)->R^2

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Main Question or Discussion Point

Continuous, Onto Function: (0,1)x(0,1)-->R^2

Hi, All:

Just curious about finding a continuous onto function from the open unit square (0,1)x(0,1)
into R^2. All I can think is that the function must go to infinity towards the edges, because
if it could be continued into the whole square, so that we would get a continuous bijection between compact [0,1]x[0,1] and Hausdorff R^2, which would then necessarily be a homeomorphism, which cannot happen for many reasons (e.g., square has boundary, R^2 does not; square has a cutset of two points and R^2 does not).

It seems at first that pairing up two bijections f,g:(0,1)-->R^2 into h=fxg would work, but it does not, I don't think; the technique does not work in the general case, e.g., f(x)=x and g(x)=2x, both from R-->R.

Any Ideas?

Thanks.
 

Answers and Replies

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[URL]http://latex.codecogs.com/gif.latex?f(z)=%5Cleft%20(%5Ctan%5Cleft%20%5Cpi%20(x-%5Cfrac{1}{2})%20%5Cright%20%5Cright%20)%20+%5Cmathbf{i}%5Cleft%20(%5Ctan%5Cleft%20%5Cpi%20(y-%5Cfrac{1}{2})%20%5Cright%20%5Cright%20)[/URL]
where x=Re{z} and y=Im{z}.
Does it work?
 
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