Is $|\mathbb{R}^{\infty}| =|\mathbb{R}|$

Question

Is $|\mathbb{R}^{\infty}| =|\mathbb{R}|$

I started reading Rudin's principle of mathematical analysis and I read this theorem

$2.13$ Theorem Let $A$ be a countable set, and let $B_n$ be the set of all $n$-tuples $(a_1 , ..., a_n)$, where $a_k \in A (k = 1, ... , t_i)$, and the elements $a_1 , .. , an$ need not be distinct. Then $B_n$ is countable.

but interestingly if we let $B$ be the set of countably infinite tuples of countable sets (loosely speaking) then $B$ is not countable in fact in this case $|B|=|\mathbb{R}| $ and this is the same as Theorem $2.14$

$2.14$ Theorem Let $A$ be the set of all sequences whose elements are the digits $0$ and $1$. This set $A$ is uncountable. The elements of A are sequences like $1, 0, 0, 1, 0, 1, 1, 1, ....$

and this got me thinking since any set of n-tuples of real numbers has the same cardinality as the real numbers (example $|\mathbb{C}| =|\mathbb{R}|$)then what about the countably infinite tuples of $\mathbb{R}$ does this set also have the same cardinality or bigger? and if it has bigger cardinality do this mean we can with this method by using countably infinite tuples of the previous infinite set can generate much bigger infinite sets ? to explain what I mean I will denote any countably infinite set like $\mathbb{N}$ by $S_1$ and by using countably infinite tuples of elements of $S_1$ (loosely speaking) we get $S_2 $ and $|S_2|=|\mathbb{R}|>|S_1|$ and by using countably infinite tuples of $S_2$ we get $S_3$ and $|S_3|>|S_2|$ and so on......

Answer 1

If by $\mathbb{R^\infty}$ you mean $\mathbb{R}^\mathbb{N}$, then $\vert \mathbb{R}^\infty\vert= \vert \mathbb{R}\vert$. This is simply by abstract cardinality arguments and is a pretty common exercise in set theory.

$\vert \mathbb{R}^{\mathbb{N}} \vert= \vert \mathbb{R}\vert^{\vert \mathbb{N} \vert} =(2^{\aleph_0})^{\aleph_0}= 2^{\aleph_0\cdot \aleph_0}\overset{\aleph_0\cdot \aleph_0=\aleph_0}{=}2^{\aleph_0}=\vert \mathbb{R}\vert$.