Build a bijection $\mathbb{R}^{\mathbb{N}} \to \mathbb{R}$ by using the two following known biections $\varphi:{\mathbb{N}} \to {\mathbb{N}} \times {\mathbb{N}}$ and $\psi:{\mathbb{R}} \to \{0,1\}^{{\mathbb{N}}}$.


My solution.

Use the classical bijection $\varphi: \mathbb{R}^2 \to \mathbb{R}.$ Now construct a bijection $\Phi: \mathbb{R}^{\mathbb{N}} \to \mathbb{R}$ by $$ \Phi(x_1,x_2,\ldots,x_n, \ldots)=\varphi(...\varphi(\varphi(x_1,x_2),x_3)...) $$

I know that it doesnt use those two proposed bijections but is it correct?

  • 1
    $\begingroup$ I don't understand the downvote. I like this question. $\endgroup$ – Kyle Dec 27 '14 at 21:35
  • 5
    $\begingroup$ @KyleGannon: "The question does not show any research effort". I can completely understand the downvote. $\endgroup$ – Michael Albanese Dec 27 '14 at 21:37
  • $\begingroup$ @Michael: Tons of questions on this site don't show any effort :) $\endgroup$ – Kyle Dec 27 '14 at 21:43
  • 1
    $\begingroup$ I think that I posted at least three or four answers to that question. Either in form of a hint more relevant for actual bijections; or in the form of cardinal arithmetic (which are just neat ways for talking about bijections). $\endgroup$ – Asaf Karagila Dec 27 '14 at 21:59
  • 3
    $\begingroup$ @KyleGannon Therefore, tons of questions on this site get downvoted, closed, and deleted. $\endgroup$ – user147263 Dec 27 '14 at 22:14

Hint: think about the problem from a different angle; you're looking for a bijection

$$(2^{\Bbb N})^{\Bbb N}\xrightarrow{\sim} 2^{\Bbb N}\dots $$


Hint: Note that $\mathbb R^\omega=(2^\omega)^\omega=2^{\omega^2}=2^\omega=\mathbb R$. If you can find a bijection for each equality, composing them should give you your desired bijection.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.