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$\mathbb R^\mathbb N$ is the set of all functions from the naturals to the reals.

I have to prove that $\mathbb R^\mathbb N$ has the same cardinality as $\mathbb R$. I found an injective function from $\mathbb R$ to $\mathbb R^\mathbb N$, but can't seem to find one for $\mathbb R^\mathbb N$ to $\mathbb R$.

Ittay Weiss
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Ryan
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4 Answers4

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$\mathbb{R} \cong 2^{\mathbb{N}}$ and $\mathbb{N}\times \mathbb{N} \cong \mathbb{N}$.

So $\mathbb{R}^{\mathbb{N}} \cong ( 2 ^ { \mathbb{N}})^{\mathbb{N}} \cong 2^{\mathbb{N} \times \mathbb{N}} \cong 2^{\mathbb{N}} \cong \mathbb{R}$

francis-jamet
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    This would be much better if you could explain and justify your steps. – Fly by Night Apr 05 '13 at 18:40
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    @user57 The comments box is for suggesting ways to improve an answer. I wasn't asking for an explanation. I was satying that the answer "would be much better if [he] could explain and justify [his] steps". – Fly by Night Apr 05 '13 at 18:51
  • Great answer, I didn't think about it that way. – Ryan Apr 05 '13 at 18:52
  • @FlybyNight Sorry, I think, it sounds disrespectful. I deleted my comment. –  Apr 05 '13 at 19:00
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It is sufficient to show that there is a bijection between $[0,1]$ and $[0,1]^\mathbb N$. Let $r \in [0,1]$ and take the decimal form $r = 0.a_1 a_2 ...$ where $(a_n)$ is a sequence in $\{0,1,...,9 \}^\mathbb N$. Then take an infinity of disjoints subsequences of disjoints integers $(n^1_k)_{k \in \mathbb N}$, $(n^2_k)_{k \in \mathbb N}$,... You can form an infinity of reals written in decimal form $r_j = 0.a_{n^j_1} a_{n^j_2}...$, one for each sequence $(n^j)_{k \in \mathbb N}$, $j \in \mathbb N$. Here's your bijection :).

roger
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  • Can you please explain what the bijection precisely is? – Ittay Weiss Apr 05 '13 at 18:54
  • Could you please explain more? How do you know it is a bijection? – Ryan Apr 05 '13 at 18:59
  • When you unwind all the arguments, you will see that the proofs by francis-jamet and roger coincide. In fact, roger's bijection (or rather injection, we have to be careful with periodic decimal forms!) is what comes out concretely from francis-jamet's abstract proof. – Martin Brandenburg Apr 05 '13 at 19:03
  • where is the number one (with its two decimal expansions) mapped to? – Ittay Weiss Apr 05 '13 at 19:09
  • see http://math.stackexchange.com/questions/243590/bijection-from-mathbb-r-to-mathbb-rn for a correct construction of a bijective function. – Ittay Weiss Apr 05 '13 at 19:19
  • The map $r \mapsto (r_j)_{j \in \mathbb N}$ is the injection (which I mistakenly refered to as a bijection, as pointed out). The image of $1 = 0.99...$ is actually the constant sequence $(1,1,....)$. – roger Apr 05 '13 at 19:21
  • @roger you can edit your post to make the correction then. To make the image of other numbers with two expansions, you need to fix some convention of which expansion to use. – Ittay Weiss Apr 05 '13 at 19:26
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Just an augmentation to @francis-jamet's answer with a solution that avoids cardinal arithmetic and instead uses CSB: the equality of cardinalities to be shown is $|[0,1]^\mathbb N|=|[0,1]|$. An injection $[0,1]\to [0,1]^\mathbb N$ is clear. To form an injection in the other direction, consider a sequence $(r_i)_{i\in \mathbb N}$ with $r_i\in [0,1]$. Write the $r_i$ in an infinite column and expand each in decimal form without repeating '9's. Traverse the digits as in the proof of the countability of $\mathbb N \times \mathbb N$ to form a single real number in $[0,1]$. This is an injection $[0,1]^{\mathbb N}\to [0,1]$.

Ittay Weiss
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More generally, if $\kappa,\lambda$ are cardinals and $\kappa$ is infinite, then $\mu:=\lambda^{\kappa}$ satisfies $\mu^{\aleph_0}=\mu$. The complete answer is given in Jech's Set theory: The exponential function $x \mapsto x^{\aleph_0}$ on cardinals is constant on each interval $\kappa \leq x \leq \kappa^{\aleph_0}$. This means that if $\lambda<\kappa$ with $\lambda^{\aleph_0} \geq \kappa$, then $\kappa^{\aleph_0}=\lambda^{\aleph_0}$. Now assume that $\lambda^{\aleph_0}<\kappa$ for all $\lambda<\kappa$. If the cofinality of $\kappa$ equals $\aleph_0$, then $\kappa^{\aleph_0} > \kappa$. Otherwise, $\kappa^{\aleph_0}=\kappa$.

Just a remark: Actually from this it follows that $\mathbb{R}^{\mathbb{N}}$ and $\mathbb{R}$ are isomorphic as abelian groups, since the $\mathbb{Q}$-dimensions agree with the cardinalities here. Of course no explicit isomorphism can be written down.

Martin Brandenburg
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  • How is it obvious that no explicit isomorphism can be given? – Ittay Weiss Apr 05 '13 at 19:27
  • @Ittay: Because it is false in some models of $\sf ZF$. It should be remarked that the bijection between the two sets exists regardless to the axiom of choice. – Asaf Karagila Apr 05 '13 at 19:36
  • @Ittay: To elaborate on my iPhone comment, in models of $\sf ZF+DC+BP$ (the latter is "all sets of real numbers have Baire property") every homomorphism of Polish groups is continuous, so there is certainly no isomorphism between $\Bbb{R^N}$ and $\Bbb R$, because no such bijection would be continuous. – Asaf Karagila Apr 05 '13 at 21:23