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

Question

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 $$(y_l)=(\dots 0,1,1,0\dots)\notin Y$$

Answer 1

It can't be countable since it contains an uncountable subset, e.g.

$$Y = \{ (..., 0, 0, x, 0, 0, ... ) : x \in \mathbb{R} \}$$

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.