From: rusin@olympus.math.niu.edu (Dave Rusin)
Newsgroups: sci.math
Subject: Re: (0,1) <--> [0,1] bijection?
Date: 2 Dec 1994 20:52:54 GMT
In article <3bndmt$fdk@apple.com>, Robert Johnson wrote:
>
>>Can you figure out a bijection between the interval and the square? :-)
>
>n-dimensional Hilbert curves continuously map the unit interval onto the unit
>cube in R^n. Since these curves are 1-1, they are continuous bijections.
>However, they can not be bicontinuous as that would imply that the unit
>interval and the square are topologically the same.
Umm, that's not enough to save you from the topology. If f:X--> Y is
continuous, 1-1, and onto, then it will in fact be bicontinuous if X
is compact and Y is Hausdorff (which certainly applies here). For if
C is a closed subset of X, then it's also compact, hence f(C) is too;
but compact subsets of Hausdorff spaces are closed. Thus C closed
in X implies (f^-1)^-1(C) is closed in Y, so that
f^-1 : Y --> X is also continuous.
So it seems to me like you _can't_ have a map I-->I^2 which is
continuous, one-to-one, and onto (although you can have any two of the
three).
dave