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Is there an injective function $f :\mathbb{R}^2\rightarrow\mathbb{R}$?

I approached this problem from the perspective of $f^{-1}$, from which I showed there exists a surjective function $f^{-1}:\mathbb{R}\rightarrow\mathbb{R}^2$. Would this imply that there exists an injective function for the inverse mapping?

Rod Carvalho
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James
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    This is the kind of question that allows one to find out who is a constructivist, and who is not. I love it! – Rod Carvalho Sep 16 '12 at 21:46
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    There is such an injective function, but not a continuous one. – GEdgar Sep 16 '12 at 22:34
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    I answered this in considerable detail here. – MJD Sep 17 '12 at 06:04
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    There is another approach to this problem: If $|\mathbb{R^2}| = |\mathbb{R}|$ then, by definition of cardinality, there is a bijection between $\mathbb{R^2}$ and $\mathbb{R}$. It is known, that $|\mathcal{P}(\mathbb{N})| = |\mathbb{R}|$. Now you can take two functions $g, h : \mathbb{N} \to \mathbb{N}$, defined by $g(n) = 2n, \ h(n) = 2n - 1$. And then take $f : \mathcal{P}(\mathbb{N})^2 \to \mathcal{P}(\mathbb{N})$ defined by $f(A,B) = g(A) \cup h(B)$. This is a bijection. Therefore $|\mathbb{R^2}| = | \mathcal{P}(\mathbb{N})^2| = |\mathcal{P}(\mathbb{N})| = |\mathbb{R}|$. –  Aug 02 '17 at 08:50

3 Answers3

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Let's construct an injective function $f : (0,1) \times(0,1) \to (0,1)$. Since there exist bijections between $\mathbb{R}$ and $(0, 1)$, the proposed function $f$ is sufficient to prove the existence of an injective function from $\mathbb{R}^2$ to $\mathbb{R}$.

Let the decimal representation of $x$ be $0.x_1x_2x_3\cdots$, and that of $y$ be $0.y_1y_2y_3\cdots$. Let $f(x, y)$ be $0.x_1y_1x_2y_2x_3y_3\cdots$.

To make this function well-defined, we should avoid decimal representations that end with infinite successive $9$s. Once this is taken care of, it's easy to show that this function is injective.

Ayman Hourieh
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Hint: Take the interval $[0, 1]$ and think how we might try to map pairs of numbers from that interval 1-1 into the interval by 'interleaving' binary representations. In other words if $a = \cdot a_1a_2a_3a_4\ldots$ and $b = \cdot b_1b_2b_3b_4\ldots$, then send $\langle a, b \rangle$ to $\cdot a_1b_1a_2b_2a_3b_3\dots$ ....

Peter Smith
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    Fumbling on an iPad, I've just deleted someone's comment by accident. Sorry! The comment was that "this doesn't work" because of the non-uniqueness of binary representations. And my comment on the comment was going to be that this is a hint, an idea to start with! (And I thought that hints are the name of the game when it comes to homework-style questions.) – Peter Smith Sep 17 '12 at 06:25
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Whether or not this is sufficient will depend on what framework you are working in. The axiom of choice is equivalent to the statement that every surjective function $g$ has a right inverse. If you can find a surjection $g: \mathbb{R} \rightarrow \mathbb{R}^2$, you can just take $f$ to be a right inverse of $g$, which will necessarily be injective since $g \circ f$ is injective. If you don't have the axiom of choice, then you can't do this.

However, the assertion can be proved without appealing to the axiom of choice. As a hint, think about interweaving the digits of the number. You might have to do a bit of work to deal with non-uniqueness of decimal representations.

Logan M
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