A reason for writing $\mathfrak{c}=2^{\aleph_{0}}$

Question

In the book Measures, Integrals and Martingales the author asks to show that $\# \{1,2\}^{\Bbb{N}} \le \# (0,1) \le \# \{0,1\}^{\Bbb{N}}$ and to conclude that $\# (0,1)= \# \{0,1\}^{\Bbb{N}}$ . He also remarks that this is a reason for writing $\mathfrak{c}=2^{\aleph_{0}}$.

I am stuck on two parts showing $\# \{1,2\}^{\Bbb{N}} \le \# (0,1) \le \# \{0,1\}^{\Bbb{N}}$ the hint given is to "interpret $\{0,1\}^{\Bbb{N}}$ as base-$2$ expansions of all numbers in $(0,1)$ while $ \{1,2\}^{\Bbb{N}}$are all infinite base-$3$ expansions lacking the digit $0$" but I've never worked with base expansions or dyadic fractions so I'm trying to find another way(I wouldn't mind if someone could guide me through this, all I know is that a dyadic fraction is rational of the form $\displaystyle\frac{a}{2^q}$.)

For the second part I'm trying to show that $f: \{0,1\}^{\Bbb{N}} \to \{1,2\}^{\Bbb{N}}$ defined by $f((a_i)_{i \in N})=(b_1, b_2,\ldots)$ where $b_i =1$ if $a_i=0$ and $2$ otherwise, is a bijection and the result follows by the Cantor-Bernestein theorem. Concerning the remark it seems that $\{0,1\}^\Bbb{N}=2^{\aleph_0}$ but I haven't been able to prove that.

Notation: $\{0,1\}^{\Bbb{N}}$ is the set of sequences $(x_j)_{j \in \Bbb{N}}$ such that $x_j \in\{ 0,1\}$

Edit: Is the reason for $\{0,1\}^\Bbb{N}=2^{\aleph_0}$ the following?

If we want to chose a sequence whose terms are $0$ or $1$ for each $n \in \Bbb{N}$ then each $(x_j)_{j \in \Bbb{N}}$ has a value for each $j$ that is it has $\aleph_0$ entries each of which have two choices then we have $2 \times 2 \times 2 \times \ldots$ choices $= 2^{\aleph_0}$ choices.

Answer 1

To explain what he means by the hint: for the right-half of that inequality, all you need to know is that every $r\in(0,1)$ can be expressed as $$ r = \sum_{n=1}^{\infty} \frac{a_n}{2^{n}} $$ Where each $a_n$ is equal to either $0$ or $1$. This is true for the same reason that each real number has some digital representation $$ r = \sum_{n=1}^{\infty} \frac{b_n}{10^{n}}=0.b_1b_2\dots; \quad b_n\in\{0,1,\dots,9\} $$ This gives us an injection from $(0,1)$ to $\{0,1\}^{\mathbb N}$, which gives us the right side of the desired inequality.

For the next bit, note that given a sequence $\{c_n\}_{n\geq1}$ where each $c_n$ is a $1$ or $2$, we note that $$ \sum_{n=1}^{\infty} \frac{c_n}{3^{n}} $$ Will always give us a unique number in $(0,1)$, giving us an injection from $\{1,2\}^{\mathbb N}$ to $(0,1)$. Here, we're using a base $3$ representation, but again as long as you accept decimal notation, this shouldn't be too much of a stretch.

Answer 2

This answer concerns injectivity of the first map. As Omnonnomnom says, this maps $\{c_n\}$ to $\sum \frac{c_n}{3^n}$. To really show injectivity we need to show there cannot be two ways to write a real number with no zeroes.

To take a more familiar case, in base 10 there are coincidences only for the "terminating decimals" e.g. $.237=.236999...$ where the first "terminating decimal" has all zeroes after the third place, while the second has all nines after the third place. The same thing happens in base 3, but here the highest digit is 2, so that e.g. $.122=.121222...$ where again the left "terminating trecimal" has all zeroes after the third position, while the right trecimal has all 2's after the third place.

By using only the digits $1,2$ as the $c_n$ we rule out any possible collisions of the above "terminating" type, and these being the only possible ones, the map is injective.

There is one more technical point: The case where all the $c_n=2$ winds up producing the real number $1$ which is not in $(0,1)$. This is no problem, since we may map this particular sequence of all 2's to some number like $1/9$ which has expansion $.01$ in base 3, and cannot be rewritten without any zeroes, so that none of the other sequences will hit $1/9$.