# Is $\mathbb{R^Z}$ or its elements countable?

Continue on the self study on infinite vector spaces. According to this link, $$\mathbb{R^Z}$$ has elements of the following form:

$$(y_k)_{k\mathbb{\in Z}}=(\dots y_{-1},y_0,y_{1}\dots)$$

Or more concisely:

$$y: \mathbb{Z}\rightarrow \mathbb{R}$$

This basically look like the $$\mathbb{R}^\infty$$ defined in this link.

where its Hamel Basis is uncountable by Baire Theorem

Now recall that a sequence $$a$$ in a set $$S$$ is countable if there exists a bijective function $$g$$ such that:

$$g: (a_k)_{k\in S} \rightarrow \mathbb{Z}$$

At first glance, it seems I can uniquely assign each of the elements of $$\mathbb{R^Z}$$ with a number in $$\mathbb{Z}$$ but I am not too sure about each components $$y_k$$ of the elements in $$\mathbb{R^Z}$$ as it seemed all $$y_k$$ seem to grew faster than the natural numbers $$\mathbb{Z}$$ can enumerate (but that might be because I am still unfamiliar with the properties of infinite spaces)

1. Is $$\mathbb{R^Z}$$ countable, i.e. is the set of elements of $$\mathbb{R^Z}$$ countable?

2. Is the multiset of all components $$y_k$$ of $$\mathbb{R^Z}$$ countable? If it is uncountable, is it the same as $$\mathbb{R}$$ itself, or is it larger than that?

3. Is this multiset a sub(multi)set of $$\mathbb{Z^R}$$, or $$\mathbb{Z^R}$$ itself?

Elaboration on question 2:

Consider some elements $$(a_k)$$, $$(b_k)$$ $$\in \mathbb{R^Z}$$ where each component are not necessary distinct (i.e. there exists some k such that $$a_k=a_{k+w}$$ where $$w\in \mathbb{Z}$$, and similarly for $$(b_k),(c_k) \dots$$. Then whether it is possible to have a function $$h$$ such that it provides a bijection of each component of the elements of $$\mathbb{R^Z}$$ into the natural numbers

================================================================= Edit:

Expansion on question 3

Using mike4ty4's answer at 2015-05-25 07:15:10 and the geometric intuition of the cartesian product, it looks tempting to say the multiset of components of $$\mathbb{R^Z}$$ can be represented by

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

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

Then you can say there is a noninjective function $$m$$ where

$$m : \mathbb{R} \rightarrow \mathbb{Z}$$

Then it does naively look like a sub(multi)set $$\mathbb{Z^R}$$ since naively speaking there are '$$\mathbb{Z}$$ copies' of $$\mathbb{R}$$ where only one component is nonzero each. But I felt like I have misunderstood something (because infinite spaces are not straightforward). Surely $$A$$ is uncountable based on mike4ty4's answer but can it be built using this naive approach, and thus concluding it naively a sub(multi)set of $$\mathbb{Z^R}$$.

We also don't need to worry about convergence since we are not adding elements to each other

================================================================= Edit 2:

Sorry I was wrong. After reviewing mike4ty4's answer again, I now understood that $$Y$$ is a subset of $$\mathbb{R^Z}$$ (and possibly $$\mathbb{Z^R}$$, detailed discussion in another question) since elements such as

$$(y_l)=(\dots 0,1,1,0\dots)\in\mathbb{R^Z}$$ But $$(y_l)=(\dots 0,1,1,0\dots)\notin Y$$

• One comment/warning on notation: it is much more common to denote by $\mathbb{R}^\infty$ the space of all sequences in which only finitely many components are nonzero (at least in topology). This is of course not a Banach space, but has a countable basis. May 25, 2015 at 7:23
• @Peter Franek I think the $\mathbb{R^Z}$ mentioned in the link have not implied there are finite components nonzero, thus I guess it might be more complicated than the $\mathbb{R^\infty}$ as the space of all sequence with finite nonzero components May 25, 2015 at 7:28
• Yes, sure. My comments raferred to the mentioning of $\mathbb{R}^\infty$ (it is mentioned once in the question) May 25, 2015 at 7:34
• Note that $\mathbb{R}^\mathbb{Z}$ is isomorphic to $\mathbb{R}^\mathbb{N}$, as a real vector space, at least. May 25, 2015 at 8:42

$$Y = \{ (..., 0, 0, x, 0, 0, ... ) : x \in \mathbb{R} \}$$
where for each real number $x$ there's a sequence $(y_k)$ in $Y$ with $y_0 = x$ and $y_j = 0$ for all $j \ne 0$. There is an obvious bijection to $\mathbb{R}$, so this $Y$ is uncountable, thus $\mathbb{R}^{\mathbb{Z}}$ is uncountable.
The multiset of components can't be countable either, since it contains $\mathbb{R}$ as a sub(multi)set. In particular, $Y$ above contributes all of $\mathbb{R}$ to the multiset of components.