Previously in this question, we have found that $\mathbb{R^Z}$ is uncountable and its multiset of components, denoted by

$$K = \{ (..., 0, 0, w, 0, 0, ... ) : w \in \mathbb{R} \}$$

where for each real number $x$ there's a sequence $(k_m)$ in $K$ with $k_0 = w$ and $k_j = 0$ for all $j \ne 0$.


$$A = \{ Y_j : j \in \mathbb{Z} \}$$

where $$Y_j = \{ (..., 0, x_j, 0, 0, ... ) : x_j \in \mathbb{R} \}$$

is also uncountable and a sub(multi)set of of $\mathbb{R^Z}$


Now I am wondering about the cardinality of $\mathbb{Z^R}$ and $\mathbb{R^Z}$

Using my half baked understanding of cardinal exponentiation and multiplication, (assuming the Axiom of Choice), learnt from here, and also some properties of cardinality here, I use the following formulae

  1. If 2 ≤ κ and 1 ≤ μ and at least one of them is infinite, then: $$\max (κ, 2^μ) ≤ κ^μ ≤ \max (2^κ, 2^μ)$$
  2. If either κ or μ is infinite and both are non-zero, then $$\kappa\cdot\mu=\max\left\{\kappa,\mu\right\}$$
  3. $$|\mathbb{R}|=2^{\aleph_0}>\aleph_0=|\mathbb{N}|=|\mathbb{Z}|$$

and get

$$\max (|\mathbb{Z}|, |2^\mathbb{R}|) ≤ |\mathbb{Z^R}| ≤ \max (|2^\mathbb{Z}|, |2^\mathbb{R}|)$$ $$\max (|\mathbb{R}|, |2^\mathbb{Z}|) ≤ |\mathbb{R^Z}| ≤ \max (|2^\mathbb{R}|, |2^\mathbb{Z}|)$$ Simplifying using the formulae $$|2^\mathbb{R}| ≤ |\mathbb{Z^R}| ≤ |2^\mathbb{R}|$$ $$\max (|\mathbb{R}|, |2^\mathbb{Z}|) ≤ |\mathbb{R^Z}| ≤ |2^\mathbb{R}|$$ $\hspace{1mm}$ $$|\mathbb{Z^R}| = |2^\mathbb{R}|=2^{2^{\aleph_0}}$$ $$\max (2^{\aleph_0}, 2^{\aleph_0}) ≤ |\mathbb{R^Z}| ≤ 2^{2^{\aleph_0}}$$ $\hspace{1mm}$ $$|\mathbb{Z^R}| = |2^\mathbb{R}|=2^{2^{\aleph_0}}$$ $$2^{\aleph_0} ≤ |\mathbb{R^Z}| ≤ 2^{2^{\aleph_0}}$$ And finally $$|\mathbb{R}| < |\mathbb{R^Z}| ≤ |\mathbb{Z^R}|$$

But then what are the approaches I should use in order to seek for a bijective function (if it exists) that maps elements of $\mathbb{Z^R}$ to $\mathbb{Z^R}$ and vise versa, so that I can work out whether $|\mathbb{R^Z}| < |\mathbb{Z^R}|$ or $|\mathbb{R^Z}| = |\mathbb{Z^R}|$?

  • $\begingroup$ $|\mathbb{Z}^{\mathbb{R}}|= 2^{|\mathbb{R}|}$, and (I think) as the cofinality of the continuum is uncountable $|\mathbb{R}^{\mathbb{Z}}|= |\mathbb{R}|$. You can probably even see the second by doing something with decimal expansions (e.g. map $f:\mathbb{Z}\rightarrow [0,1]$ to $x$ whose $2^n\cdot 3^m$th decimal place is the $m$th decimal place of $f(n)$ and $0$ in all other places). Consequently one is strictly bigger than the other. $\endgroup$
    – James
    Jun 27, 2015 at 14:45

2 Answers 2


No. We have $$|\mathbb Z^{\mathbb R}|\ge |2^{\mathbb R}| > |\mathbb R| $$ but $$ |\mathbb R^{\mathbb Z}| = |(2^{\mathbb Z})^{\mathbb Z}| = |2^{(\mathbb Z\times\mathbb Z)}| = |2^{\mathbb Z}| = |\mathbb R|$$


As Henning explains, they're different. Do you know about the beth numbers? We have:

Proposition 0. $$|\mathbb{Z}| = \beth_0, \qquad |\mathbb{R}| = \beth_1$$

From this, we can answer your question:


$$|\mathbb{R}^\mathbb{Z}| = \beth_1, \qquad |\mathbb{Z}^\mathbb{R}| = \beth_2$$

Since the beth numbers are a strictly increasing family of cardinal numbers (this follows immediately from Cantor's theorem), the above claim implies that

$$|\mathbb{R}^\mathbb{Z}| < |\mathbb{Z}^\mathbb{R}|,$$

so they're different.

Lets begin by proving $|\mathbb{R}^\mathbb{Z}| = \beth_1$. Essentially we just copy Henning's proof, in different notation:

$$|\mathbb{R}^\mathbb{Z}| = \beth_1^{\beth_0} = (2^{\beth_0})^{\beth_0} = 2^{\beth_0 \cdot \beth_0} = 2^{\beth_0} = \beth_1$$

So much for that! Proving the latter is just a little more involved. We're trying to simplify the expression:


The intuition is that since $\beth_0$ is quite small compared to $\beth_1$, hence raising $\beth_0$ to the power of $\beth_1$ gives the same result as raising $2$ to the power of $\beth_1$. Much more precisely, we use the following theorem about cardinal numbers.

Proposition 1. If $2 \leq \kappa \leq 2^\nu,$ then $\kappa^\nu = 2^\nu$. (Assuming that $\nu$ is infinite.)

Hence since $2 \leq \beth_0 \leq 2^{\beth_1}$ (and $\beth_1$ infinite) we deduce that $\beth_0^{\beth_1} = 2^{\beth_1}$ which implies that: $$\beth_0^{\beth_1} = \beth_2.$$

Exercise. Prove Proposition 1 using only the basic principles of cardinal arithmetic. (It isn't hard!)

  • $\begingroup$ $$2\geq\kappa\geq 2^{\nu}$$ Assuming the Axiom of Choice Exponentiating both sides by $\nu$ $$2^{\nu}\geq\kappa^{\nu}\geq (2^{\nu})^{\nu}$$ Using the multiplication rule of Cardinal aeithmetic $$2^{\nu}\geq\kappa^{\nu}\geq 2^{{\nu}\cdot{\nu}}$$ $$2^{\nu}\geq\kappa^{\nu}\geq 2^{max\left\{\nu,\nu\right\}}$$ $$2^{\nu}\geq\kappa^{\nu}\geq 2^{\nu}$$ Thefore by some form of sandwich theorem $$\kappa^{\nu}= 2^{\nu}$$ as required? $\endgroup$
    – Secret
    Jun 27, 2015 at 16:11
  • $\begingroup$ @Secret, pretty much, but you've got the inequalities backward :) $\endgroup$ Jun 27, 2015 at 16:19
  • $\begingroup$ oops, that's because I typed "\geq" while I meant $\leq$ $\endgroup$
    – Secret
    Jun 27, 2015 at 16:21
  • $\begingroup$ @Secret, by the way, the "sandwich theorem" you're looking for is called the Schroder-Bernstein theorem. $\endgroup$ Jun 27, 2015 at 16:23
  • 1
    $\begingroup$ @Secret, also, the claim that $|\mathbb{R}| < |\mathbb{R}^\mathbb{Z}|$ is incorrect, since from my answer, we have: $|\mathbb{R}| = \beth_1$ and $|\mathbb{R}^\mathbb{Z}| = \beth_1$. $\endgroup$ Jun 27, 2015 at 17:10

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