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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 $\psi:{\mathbb{R}} \to \{0,1\}^{{\mathbb{N}}}$.

Edit.

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?

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    $\begingroup$ I don't understand the downvote. I like this question. $\endgroup$ – Kyle Dec 27 '14 at 21:35
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    $\begingroup$ @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
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    $\begingroup$ 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
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    $\begingroup$ @KyleGannon Therefore, tons of questions on this site get downvoted, closed, and deleted. $\endgroup$ – user147263 Dec 27 '14 at 22:14
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Hint: think about the problem from a different angle; you're looking for a bijection

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

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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.

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