Bijective mapping between $\mathbb{R}^2$ and $\mathbb{R}$?

Question

Please don't sign this as a duplicate of this, they aren't. I am interested about the actual algorithm, not a proof of the existence.

Does a such mapping exists? I think, it must, because both of them have the same cardinality ($\aleph_1$).

My actual suggestion for a such mapping were simply merge the digits of the real numbers. For example, from $\pi$ and $e$ we could got

$3.1415...$ and $2.7182...$ would lead to $32.17411852...$ .

Although I am not sure it were ok. Maybe a better, clearer solution also exists?

possible duplicate of existence of a map between $\mathbb R^2$ and $\mathbb R$ — Brad, Jun 18 '14 at 16:19
The assertion that $\mathbb{R}$ has cardinality $\aleph_1$ is the continuum hypothesis and is independent of ZFC, we say it has cardinality $2^{\aleph_0}$ — James, Jun 18 '14 at 16:20
Both have the same cardinality, but it may not be $\aleph_1$. — Michael Greinecker, Jun 18 '14 at 16:20
@Brad No, it is not. I am interested about a clear, simple algorithm, and not the proof of the existence. — peterh, Jun 18 '14 at 16:20
@MichaelGreinecker Yes, but I like to have a simple, clean mapping function, and not a proof of that exists (I know that already). — peterh, Jun 18 '14 at 16:21
@Omnomnomnom Thank you - although it doesn't give me an answer, it gives very useful information to find for me a better one. — peterh, Jun 18 '14 at 16:25
@PeterHorvath in fact, you can use the answer to build a bijection from $3$ others. Namely, we can create bijective maps that take us through $$ \mathbb{R}\times \mathbb{R} \to [0,1] \times [0,1] \to [0,1] \to \mathbb{R} $$ — Ben Grossmann, Jun 18 '14 at 16:27
How about using Cantor-Schroeder-Bernstein to construct an actual bijection, given injections in both directions? — user99680, Jun 18 '14 at 16:34
Take two real numbers in rational representation, and mix them by intertwining. Of course you need to treat cases with 2 different representations by fixing one. Another method is to mix blocks of digits. — mez, Jun 18 '14 at 16:36
possible duplicate of Bijection from $\mathbb R$ to $\mathbb {R^N}$ — mez, Jun 18 '14 at 16:37
@mez No, they aren't. Please, read the first line of the question. Maybe it is enough for a vote change? — peterh, Jun 18 '14 at 16:58

score 3 · Accepted Answer · edited Apr 13 '17 at 12:20

The method you describe (interleaving the digits) doesn't work, because it maps $(0.4999999\ldots, 0.99999\ldots)$ to $0.4999\ldots$, but it also maps $(0.50000\ldots, 0.00000)$ to $0.50000\ldots$, which is the same number as $0.4999\ldots$.

Maybe you want to say that we will forbid the use of funny-looking numbers like $0.4999999\ldots$. No, that does not work, because then there is no pair of numbers that is mapped to $ \frac{81}{198} = 0.409090909\ldots $.

I described in Examples of bijective map from $\mathbb{R}^3\rightarrow \mathbb{R}$ how to fix this, and at least one different method for constructing an explicit mapping.

Bijective mapping between $\mathbb{R}^2$ and $\mathbb{R}$?

1 Answers1

Linked

Related