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[0,1] to (0,1)

Anza Power


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If x=0, f(x)=1/2;

if x≠0 and 1/x is an integer, f(x)=x/(1+2x);

otherwise, f(x)=x.

Very impressive! Did you just devise this yourself?

I first thought to make room in the range of the function for the extra two (images of 0 and 1)

by moving a subset of the rationals up two spots (as Zeno would have done) thusly:

Going 1/2 way the distance each time, the subset of the rationals is

{1/2,3/4,7/8,15/16,31/32,...}. So, mapping 0 to 1/2 and 1 to 3/4, I just had to

map everything in the subset up two positions forward in the set. Thus, I had

to map 1/2→ 7/8, 3/4→ 15/16, 7/8→ 31/32, 15/16→ 63/64,... I could leave everything

else not in the subset to map to itself. This seemed a bit cumbersome, so I set out

to simplify. At some point I realized than the numerators could all be the same and

I could have used the set {1/2,1/4,1/8,1/16,1/32,...} instead. Then, it hit me that

I didn't need powers of two at all, just increasing denominators, which led me to

use the set {1/2,1/3,1/4,1/5,1/6,...}. From the original Zeno idea to that final

subset of the rationals only took 5 minutes or so, but there was no particular rush

as I noticed that there was no activity on this OP.

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Are there conditions other than being one-to-one and onto?

No other conditions, although if you want something really interesting try having a continuous function...

continuous image of compact space has to be compact.

Oh, that's even simpler than what I had in mind:

Since the function is continuous and reversible, it has to be monotonic, assume it's monotonically rising, by definition of "continuous" lim

x->1f(x) = f(1), but f(1) is somewhere in the open segment (0,1), so let's say f(1)=1-epsilon, but since f is monotonically rising that means there is nothing mapped to the segment (1-epsilon,1)

Edited by Anza Power
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