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If we take a function that maps all $\mathbb{N}$ to $\mathbb{R}$, I suspect that that can be encoded using just one number in $\mathbb{R}$, the question is, can we constructively prove that, or is it the case of we know such a mapping exists, but must appeal to e.g. the Axiom Of Choice?

One may try to use the answer from https://math.stackexchange.com/a/243689/227162, however that appears to appeal to AOC: "we take a sequence of irrationals...".

micsthepick
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    The basic argument is: $|\mathbb R|=|2^{\mathbb N}|$ so $$|\mathbb R^{\mathbb N}|=|2^{\mathbb N\times\mathbb N}|$$ and $|\mathbb N\times\mathbb N|=|\mathbb N|.$ – Thomas Andrews Apr 09 '25 at 22:52
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    Your title is backwards. – Ted Shifrin Apr 09 '25 at 22:52
  • I think a mapping would go like this: Represent each $r_1, r_2, \dots$ as a continued fraction, that is as a sequence of positive integers. We now have a countable sequence of sequences of integers to combine into a single sequence. That is, we have an array of $\Bbb N^2$ integers to turn into a single sequence. But we can do this using the same diagonal-path method that we use to show $|\Bbb Q| = |\Bbb N|$. – MJD Apr 09 '25 at 22:54
  • Yeah, "as big as" implies "greater than of equal to." The title should be "the same cardinality as..." – Thomas Andrews Apr 09 '25 at 22:54
  • I see that my continued-fraction suggestion is exactly the technique used by Asaf Kargila at https://math.stackexchange.com/a/243689/25554 – MJD Apr 09 '25 at 23:41
  • This question is confusing because the body of the question is quite obviously different from the title: the title asks about bijection between $\mathbb{R}$ and $\mathbb{R}^N$, but the body of the question is asking about mappings from $\mathbb{N}$ to $\mathbb{R}$. Can you clarify what exactly it is that you want to ask? – NikS Apr 10 '25 at 00:21
  • @NikS The question asks about taking a function "that maps all $\Bbb N $ to $\Bbb R$", and encoding it as "just one number in $\Bbb R$". That is, OP wants a bijection $\Bbb R^{\Bbb N} \leftrightarrow \Bbb R$. – MJD Apr 10 '25 at 01:02
  • @micsthepick You are concerned about "take a sequence of irrationals" in Asaf's answer. But Asaf only needs to supply a single specific sequence of irrationals. (He suggests $(1+n^2)^{1/2}$.) This does not require the axiom of choice or anything like it. If this is now your main concern, I suggest you ask another question about that issue specifically. – MJD Apr 10 '25 at 01:06
  • @MJD : I think non-standard notation is leading to (my) confusion here. “$\mathbb{N}$” (in “mathbb” font) normally means “the set of all natural numbers”, but the OP seems to also be using “$\mathbb{N}$” as a variable indicating a particular natural number (and maybe also to indicate a finite subset of natural numbers?) I’m still not entirely sure I know what the OP’s intended meaning was here, but I guess some people feel they understand it, so maybe that’s good enough. – NikS Apr 10 '25 at 21:48
  • @NikS N represents the set of natural numbers as usual, it's a mapping of countably infinite Reals to a single Real. – micsthepick Apr 24 '25 at 08:52

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A way to prove this is by cardinal arithmetic.

You want to see that $\#\mathbb{R}^{\mathbb{N}}=\#\mathbb{R}=c$. Let’s see it: $$\#\mathbb{R}^{\mathbb{N}}=c^{\aleph_0}=(2^{\aleph_0})^{\aleph_0}=2^{\aleph_0 \cdot \aleph_0}=2^{(\aleph_0)^2}=2^{\aleph_0}=c$$

If you don’t know cardinal arithmetic you could prove that if $\#A=a$ and $\#B=b$, then $\#A^B=\#\{f:B\to A \}=a^b$.

Despite this, there probably is a way to prove this by mapping both spaces, so maybe you could give it a try too.

Aaron
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