Denote by $S$ the set of all convergent sequences $(a_n)_{n=0}^\infty$, where each $a_n\in\mathbb{R}$. What is $|S|$?
It must be the case that $|S|\geq|\mathbb{R}|$. For each $x\in\mathbb{R}$, given the decimal expansion $x_0.x_1x_2x_3\ldots$ of $x$, the mapping $x\mapsto(\sum_{k=0}^nx_k10^{-k})_{n=0}^\infty$ is an injection $\mathbb{R}\to S$.
I would guess that $|S|=|\mathcal{P}(\mathbb{R})|$. Is this true? If so, what might be an approach to prove it?
Given any countable subset of $\mathbb{R}$, I believe the following construction generates a unique convergent sequence and hence an injection into $S$. Order the elements of the subset as $a_1,a_2,\ldots$, and consider the following map $f:\mathbb{R}\to\mathbb{R}$:
$$f(a_k)=\begin{cases}
|a_k|,&|a_k|\leq1 \\
|1/a_k|,&|a_k|>1 \\
\end{cases}$$
Then the sequence $\big(\sum_{k=0}^n(-1/2)^kf(a_k)\big)_{n=1}^\infty$ converges by the alternating series test. I'm not sure of any construction for uncountable subsets of $\mathbb{R}$, as I don't think that this approach would help. Additionally, I would still need to show either an injection from $S$ into $\mathcal{P}(\mathbb{R})$ or that some map $g:\mathcal{P}(\mathbb{R})\to S$ is a bijection, and I don't currently have any ideas for either of these.
Best Answer
Sequences are elements of $\Bbb R^{\Bbb N}$, therefore the cardinality of $S$ is bounded above by $$\left\lvert \Bbb R^{\Bbb N}\right\rvert=\left\lvert (2^{\Bbb N})^{\Bbb N}\right\rvert=\left\lvert 2^{\Bbb N\times\Bbb N}\right\rvert=2^{\lvert \Bbb N\rvert}=\lvert\Bbb R\rvert$$
Moreover, there are several easy injections of $\Bbb R$ in $S$.