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It is known that each $x,y \in (0,1)$ has a unique decimal expansion $ x = 0.x_1x_2x_3\ldots $ and $y = 0.y_1y_2y_3\ldots$ with the requirement that there are infinitely many indices different from $9$. Show that the following function $f : (0,1)×(0,1) → (0,1)$ is injective: $f(x,y) = 0.x_1y_1x_2y_2\ldots$ Is $f$ also surjective? Using the Schröder-Bernstein theorem show that $(0,1)×(0,1)$ and $(0,1)$ have the same cardinality.

For the injectivity part of the question, this is what I have so far. I have defined 2 sets $S=\{x_n : n \in \mathbb{N}\}$ and $T=\{y_n : n \in \mathbb{N}\}$. Then I have said $f(x,y)=\sum_{n\geq0}x_n10^{-2n+1}+y_n10^{-2n}$. Then I have concluded that since $x$ has a unique decimal expansion and so does $y$ then it follows that the sum of 2 unique expansions is itself unique, but I feel that this is missing a great deal of rigour and would like to know if it's even right or how to improve it. Also, I am unsure as to how to deal with the surjectivity part of the question. Any help would be greatly appreciated.

Roger
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  • Since $f:(0,1)^2\to (0,1)$ is injective, and $g:(0,1)\to (0,1)^2,$ where $g(x)=(x,1/2)$, is injective, the S-B Theorem (a.k.a. Cantor-Bernstein or Cantor-Schroeder-Bernstein) implies there is a bijection $h:(0,1)^2\to (0,1).$ – DanielWainfleet Apr 17 '17 at 17:52

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I guess that you mean $x = 0.x_1x_2x_3 . . .$ and $y = 0.y_1y_2y_3 . . .$

Injectivity is easy. If you have two different points $(u,v)$ and $(x,y)$ then either $u \neq x$ or $v \neq y$. If it is $u \neq x$ then they have at least one different digit $u_i \neq x_i$. So, you can point to a differing digit in the images of these points under your map. Similarly if $v \neq y$.

However, the map is not surjective. Nothing maps to $0.09090909 . . .$.

So, you need to think of of another map which is surjective. That should be very easy.

badjohn
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  • I don't see how one number can be a surjective map. But the surjection should be easy; you just want a map from a 2D space to a 1D space. However, to use Schröder–Bernstein directly, you actually want an injection both ways. You've got the hard one know. An injection in the other direction should be easy. I think that I have you that in an answer to another related question. (This was a reply to a comment that seems to have been deleted.) – badjohn Apr 17 '17 at 16:18
  • It would need some adjustment for closed intervals but that should not be hard. – badjohn Apr 17 '17 at 16:20
  • Sorry yes that reply was deleted because I realised what I said was nonsense – Roger Apr 17 '17 at 16:25
  • @Roger No problem but I was a bit puzzled. – badjohn Apr 17 '17 at 16:28
  • I would get the map from 2 dimensional space into one dimensional using the function given in the question? Then how would I deal with 0.9090909... and 0.0909090..? – Roger Apr 17 '17 at 16:32
  • Above, you have an injection from the 2D space to the 1D space. An injection in the other direction, 1D to 2D, is very simple: just map $x$ to $(x,0.5)$. Now you can invoke Schröder–Bernstein. That's the beauty of this theorem: you don't need to figure out the bijection, injections in both directions are sufficient. In many cases, such as this, it is much easier to get an injection each way then to get an explicit bijection. – badjohn Apr 17 '17 at 16:36
  • This got me interested. I suspect that the purpose of the question was to illustrate the SB theorem so just an injection in each direction was the expectation. However, it got me interested in the subject and I found this earlier post on an explicit bijection: https://math.stackexchange.com/questions/183361/examples-of-bijective-map-from-mathbbr3-rightarrow-mathbbr?lq=1 – badjohn Apr 18 '17 at 15:41