Jump to content
BrainDen.com - Brain Teasers
  • 0
Anza Power

[0,1] to (0,1)

Question

9 answers to this question

Recommended Posts

  • 0

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.

Share this post


Link to post
Share on other sites
  • 0

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...

Yup, in fact that's the assumption I made.

Share this post


Link to post
Share on other sites
  • 0

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.

Share this post


Link to post
Share on other sites
  • 0

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

Share this post


Link to post
Share on other sites

Join the conversation

You can post now and register later. If you have an account, sign in now to post with your account.

Guest
Answer this question...

×   Pasted as rich text.   Paste as plain text instead

  Only 75 emoji are allowed.

×   Your link has been automatically embedded.   Display as a link instead

×   Your previous content has been restored.   Clear editor

×   You cannot paste images directly. Upload or insert images from URL.

Loading...

  • Recently Browsing   0 members

    No registered users viewing this page.

×
×
  • Create New...