Existence of bijection between $\mathbb{R}$ and $\mathbb{R^2}$

Question

I would like to show the existence of a bijection between $(0,1)$ and $(0,1) \times (0,1)$. Would $f(x)=(x,x)$ work ($x \in (0,1)$)? It seems to be injective, but am I okay in using this even if its domain and range can go outside of $(0,1)$?

Notes:

Define a second function $g: (0,1) \times (0,1) ↦ (0,1)$ which takes two decimal expansions $0.(x_1)(x_2)(x_3)...$ and $0.(y_1)(y_2)(y_3)...$, the first expansion coming from the first component of $(0,1) \times (0,1)$ and the second coming from the second component and maps them to $0.(x_1)(y_1)(x_2)(y_2)(x_3)(y_3)...$. For example, take $0.04265$ and 0.83169. The function g maps the two decimal expansions to 0.0843216659. This seems like it could be an injective function because it takes two numbers and maps them to a unique number. However, I'm not sure how I would formally prove it...

This implies that, by Cantor-Schroder-Bernstein theorem, there exists a bijection between $(0,1)$ and $(0,1) \times (0,1)$.

Fixed, thanks! The more important question is, can I use the function even though its domain/range technically can extend out beyond $(0,1)$? — Timor12, Nov 27 '16 at 18:48
Yes, it is ok. It is the restriction of the map $\Bbb{R} \to \Bbb{R}^2$ defined by $x \mapsto (x,x)$, but it is a (well-defined) function anyway. — Crostul, Nov 27 '16 at 18:51
Note that $\text{dom }g = {(x,x) : x \in (0,1)} \neq (0,1)^2$, for example, $(\frac13,\frac12) \not\in \text{dom }g$. — Vitor Borges, Nov 27 '16 at 19:00
If you want such a function, you can find it here: http://math.stackexchange.com/questions/183361/examples-of-bijective-map-from-mathbbr3-rightarrow-mathbbr/183383#183383 — Vitor Borges, Nov 27 '16 at 19:04
@user8485, hold on, you define x to be in the interval $(0,1)$, and $12 ∉ (0,1)$, so I'm not really sure what you are getting at? — Timor12, Nov 27 '16 at 19:04
@user8485, the domain of g is the unit square (without the borders), so ($\frac{1}{3},\frac{1}{2}$) seems to be in the domain. We can pick any point in the unit square and the output should be in the $(0,1)$ interval... Which changes my answer to $g(x,x)=x$, for $(x,x) \in (0,1)^2$. — Timor12, Nov 27 '16 at 19:12
@Timor12 And how do you think we should define $g({1\over 3}, {1\over 2})$? Your function needs to be defined on all of $(0, 1)^2$, not just some of it. — Noah Schweber, Nov 27 '16 at 19:16
@Timor12 I meant that you only defined $g$ for the points of the form $(x,x)$. We cannot calculate $g(\frac13,\frac12)$ using your definition. — Vitor Borges, Nov 27 '16 at 19:17
To your suggestion of the different methods of finding a bijection, I think CBS is the most straightforward method, so I'll stick to that. Thanks though! — Timor12, Nov 27 '16 at 19:17
Two things: Your function $g$ is not defined on $(0,1)^2$. It's defined only on ${ (x,y) \in (0,1) : x=y}$. Also, do you actually need to construct a bijection, or just show one exists? CBS doesn't help you construct. — , Nov 27 '16 at 19:26
@NoahSchweber thanks, I got it. I'm now hard pressed to find a trivial injective function $(0,1)^2 → (0,1)$ function, though... — Timor12, Nov 27 '16 at 19:26
@tilper, sorry, I should've been more clear. I need to show the existence of a bijection. — Timor12, Nov 27 '16 at 19:29
@Timor12, what about $|\ln (xy)|$? Haven't checked it myself but it's at least defined on $(0,1)^2$. — , Nov 27 '16 at 19:38
@tilper wouldn't work; counterexample: $g(\frac{1}{2}, \frac{1}{2})=\frac{1}{4}$ and $g(\frac{1}{3.8}, \frac{3.8}{4})=\frac{1}{4}$ — Timor12, Nov 27 '16 at 19:51
@Timor12 There isn't a trivial injection - in particular, there is no continuous injection from $(0, 1)^2$ to $(0, 1)$. You're going to have to do some work. — Noah Schweber, Nov 27 '16 at 20:03
@NoahSchweber, I think taking two decimal expansions and then mapping them to a different decimal expansion could work (see edited above), but I'm not sure how I would go about proving it formally. Intuition says it's an injective function, but I'd like to prove it. Any ideas? — Timor12, Nov 27 '16 at 22:47
@Timor12 You are starting to get the right idea, but it's going to be more complicated than what you've written. How do you deal with a number with an infinite decimal expansion - say, ${1\over 3}=0.333333...$ or ${1\over 7}=0.142857142857...$? What should $g({1\over 3}, {1\over 7})$ be? (There's a secondary difficulty, of how to deal with numbers with two decimal expansions - e.g. $0.2999999...=0.30000000....$ - but this can be handled fairly easily.) — Noah Schweber, Nov 27 '16 at 23:03
@NoahSchweber, I see. If we have an irrational number as a first input, then the second input's decimal expansion would never start (theoretically) since it has an infinite expansion. I suppose I could make the output be of the form $0.(x_1)(y_1)(x_2)(y_2)(x_3)(y_3)...$ to avoid the problem. — Timor12, Nov 27 '16 at 23:24
@Timor12 That is a much better approach! (And note that it's not just irrationals that pose a problem - rationals whose denominator isn't a power of $10$, like ${1\over 3}$, also pose problems.) It needs a bit of work - what do you do if the input is a real with two decimal expansions? - but it's essentially the right map. — Noah Schweber, Nov 27 '16 at 23:35
@NoahSchweber not sure what you are getting at... the $0.29999....$ expansion can be represented one way, namely by inputs $0.299999...$ and $0.9999...$. $0.3000000...$ does not have any two inputs (since $0.000000000...$ is not part of our domain), but that's okay because we are looking for an injective function, not a surjective function. The question is how do I argue formally, or is logic an argument enough? — Timor12, Nov 28 '16 at 00:18
@Timor12 I was thinking of $0.29999...$ as one of the two inputs we feed to $g$. That is, what should we do with $g(0.3, 0.4)=g(0.29999..., 0.399999)$? And no, this isn't formal enough - you need to show that $g$ is actually well-defined. So you do need to say something about what you want $g$ to do when one or both of its inputs have multiple decimal expansions. — Noah Schweber, Nov 28 '16 at 00:33
So let's get this straight: you mean $0.299...$ with 9s continuing on and on for infinity is equal to $0.3$? If we chose to represent $g(0.3, 0.4)=g(0.299..., 0.399...)$ as you suggested, then we wouldn't have a function because in that case, g(0.3,0.4) would be either 0.34 or 0.239999, and that violates the definition of a function. — Timor12, Nov 28 '16 at 00:51
@NoahSchweber To avoid this, I would define 0.2999 repeating and 0.3 as two distinct numbers, because if they were equal, $g$ wouldn't be a function. — Timor12, Nov 28 '16 at 00:57
@Timor12 Well, too bad - they are equal. You can't just change what a real number is. You need to fix your function, $g$, so that it is well-defined. — Noah Schweber, Nov 28 '16 at 01:02
@NoahSchweber can I then define $0.2999...$ to always be represented as $0.3$, $0.3999...$ to be represented as $0.4$, and so on, or is that breaking the rules? Otherwise I don't really see how I can redefine my function to fit the parameters. — Timor12, Nov 28 '16 at 01:28
@Timor12 Well, $0.29999...$ is equal to $0.300000...$ - there's no need for you to "define" them to be equal since they already are. The issue, again, is how are you defining $g$: what should $g(0.29999..., 0.39999...)=g(0.3, 0.4)$ be? (HINT: you can pick a decimal representation . . . ) — Noah Schweber, Nov 28 '16 at 01:32
@NoahSchweber right, so whenever I see $0.299...$ I can say that it will be always $0.3$ in my function, so $g(0.2999..., 0.39999...)=g(0.3, 0.4)=0.34$. That's what I meant, though probably worded it poorly. — Timor12, Nov 28 '16 at 01:53
@Timor12 Yup, that's it! Always pick the finite decimal expansion, if there's a choice. (Or make the other choice - the point is, specify how you're going to pick the decimal expansion to use.) Now, can you show that this is in fact an injection? — Noah Schweber, Nov 28 '16 at 02:02

Noah Schweber · Accepted Answer · 2016-11-28T02:58:40.340

1

Are you sure it's surjective? What gets sent to $({1\over 2}, {1\over 3})$?

EDIT: In the comments below the OP, we've hashed out the following injection from $(0, 1)^2$ to $(0, 1)$ (which, in conjunction with the "diagonal" injection $x\mapsto(x, x)$ given by the OP originally, will establish that $(0, 1)$ and $(0, 1)^2$ have the same cardinality via Cantor-Shroeder-Bernstein): let the standard decimal expansion of a real $r$ be the decimal expansion of $r$ if $r$ has only one decimal expansion, and otherwise the decimal expansion with only finitely many $9$s. So e.g. the standard decimal expansion of $\pi$ is $3.14159...$, and the standard decimal expansion of ${1\over 2}$ is $0.5$ rather than $0.499999...$. Then, if $r, s$ are real numbers in $(0, 1)$ with standard decimal expansions $$r=0.a_1a_2a_3...\quad s=0.b_1b_2b_3,$$ we let $$g(r, s)=0.a_1b_1a_2b_2a_3b_3...$$

We're not done: we still need to show that this is injective. It's usually harder to prove something if you think it's trivial, so: how could $g$ not be injective? Well, for instance, if we had two numbers $r$ and $s$ such that $g(r, s)=0.221999999...$ and two other numbers $r'$ and $s'$ such that $g(r', s')=0.22200000...$, then this would be a problem. So: can you see why this sort of issue can't happen?

edited Nov 28 '16 at 02:58

answered Nov 27 '16 at 18:42

Noah Schweber

245,398

Thanks for the heads up! How about I invoke the Cantor-Schroder-Bernstein theorem and define two functions f and g? Then can I say that there exists a bijective function? – Timor12 Nov 27 '16 at 18:45
However, the main question is, can I use that function even though its domain/range can go beyond $(0,1)$? – Timor12 Nov 27 '16 at 18:46
@Timor12 You can define a function on a restricted domain: let $f: (0, 1)\rightarrow (0, 1)^2: x\mapsto (x, x)$. There's no obligation to always have a function be defined on the largest possible domain! – Noah Schweber Nov 27 '16 at 18:47
@Timor12 Re: your first comment, yes, and you have an $f$ going one way; the difficulty is finding a $g$ to go the other way. – Noah Schweber Nov 27 '16 at 18:48
@BjörnFriedrich The author's original question was whether a given function provided a bijection between $(0, 1)$ and $(0, 1)^2$; this indeed answers it. – Noah Schweber Nov 27 '16 at 20:28
Wait, this issue cannot happen since one of the inputs is not in our domain, i.e. $0.19999999...=g(0.1999..., 0.999....)$ and we've established that by our function $0.999...=1$, which is not in our domain $(0,1)$. Thus, there are no two inputs for that one output $(0.2)$; in fact, in this case, there are 0 inputs for 1 output, implying that the function is not surjective, but injection still holds? – Timor12 Nov 28 '16 at 02:32
@Timor12 I've tweaked the example so it's relevant. The point is that reals in the domain could (in principle) yield failures of injectivity, and you need to argue that they don't. – Noah Schweber Nov 28 '16 at 02:59

Existence of bijection between $\mathbb{R}$ and $\mathbb{R^2}$

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