Show cardinality between two sets $x=(0,x_1x_2x_3...)_{10}=\sum _{k=1}^{\infty }x_k 10^{-k}$

Question

For every real number $x \in [0,1]$ can be written in decimal form:

$$x=(0,x_1x_2x_3...)_{10}=\sum _{k=1}^{\infty }x_k 10^{-k}$$ where $x_i \in \{0,1,2,3...,9\}$ for every $i$. Because of uniqueness we disallow expansions that end with an infinite number of $9$s. Let $A$ be the set of all $x \in [0,1]$ whose decimal expansion only has even digits. Show that $A$ and $R$ has the same cardinality, i.e. $|A| = |R| $.

I am lost.

• What does $(0,x_1x_2x_3...)_{10}$ mean?
• What is $x_i$ which is suddenly defined? Or what is $i$?
• What is the implication of disallowing expansions that end in $9$s?
• What is $x$ in the sigma notation as I do not see it defined anywhere?
• How do I calculate the first number in this sequence? $k$ is one. What is $x$?
• More importantly, where do I learn enough to understand this question myself?

Answer 1

To begin with, $x=\sum_k^\infty x_k$ is just the digital representation of a real number in $[0,1]$.

According to definition, the cardinality of two sets is $|X|\leq|Y|$ iff there is an injective mapping from $X\to Y$.

We have sets $A$ and $R$ as defined in the question. The elements of $A$ and $R$ are modeled by infinite sequences of digits $x_k\in\{0,\ldots,9\}$ with some additional restrictions.

As all elements of $A$ are obviously elements of $R$ we have $|A|\leq |R|$.

The startling thing and the point of the question is, that albeit $R$ has elements not fount in $A$ (such as $\frac19=0,1111111\ldots$), and all elements of $A$ are also in $R$, both have the same cardinality.

To show that wen need to show $|R|\leq |A|$ by finding an injection from $R\to A$.

As there is a countable infinity of indices, there is an easy trick:

Let $x=\sum_k^\infty x_k\in R$. Just map it to $y=\sum_k^\infty y_k$ with $y_{2k+1}=2*x_k \mod 10$ and $y_{2k}=2*\lfloor{2*x_k/10}\rfloor$.

In simpler word, take any digit $x_k$, double it resulting in two digits, the first one $0,1$, the last one even, Correct the the first one frm $1$ (which is not allowed in $A$ to $2$, if needed, and allign those digits at the end of the previously computed digits.

Answer 2

All your bulleted questions refer to the notations connected with the following fact: The set of real numbers $x\in[0,1[\>$ is in bijective correspondence with the set of all infinite decimal fractions $0.x_1x_2x_3\ldots$ with $x_i\in\{0,1,2,\ldots,9\}$, whereby $$0.x_1x_2x_3\ldots\quad \leftrightarrow \quad x=\sum_{k=1}^\infty x_k\,10^{-k}\ .$$ Some exception handling has to be done concerning the fact that, e.g., $0.39999\ldots=0.40000\ldots\ $. Therefore decimal fractions ending with all nines have been excluded in your source. I shall not deal with this.

Now the actual problem is the following: You have the set $R$ of all sequences $$x:\quad{\mathbb N}\to\{0,1,2,\ldots,9\},\qquad k\mapsto x_k\ ,$$ (omit the sequences ending with all nines, if you like) and the subset $A\subset R$ of all sequences $$y:\quad{\mathbb N}\to\{0,2,4,\ldots,8\},\qquad k\mapsto y_k\ .$$ It is claimed that $|R|=|A|$, even though it seems that $A$ has much fewer elements than $R$. For the proof we need the Schroeder-Bernstein Theorem:

• Given two sets $R$ and $A$, and we can invent injective maps $f:A\to R$, $\ g:R\to A$, then $|R|=|A|$.

Of course the injection map $f:A\to R$ is injective. To construct the $g:R\to A$ we have to injectively encode each sequence $x\in R$ as a new sequence $g(x)=:y\in A$. Let $x=(x_1,x_2,\ldots)\in R$. Define $$y_{2i-1}:=2\left\lfloor{x_i\over2}\right\rfloor, \quad y_{2i}:=2(x_i-y_{2i-1})\in\{0,2\}\qquad(i\geq1)\ .\tag{1}$$ It is easy to see that $y=(y_1,y_2,y_3,\ldots)\in A$, and that the sequence $x$ can be reconstructed uniquely from the $y$. Therefore the $g$ constructed in this way is injective.

Example: $$x=(3,4,1,6,6,5,7,9,\ldots), \quad y=g(x)=(2,2,4,0,0,2,6,0,6,0,4,2,6,2,8,2,\ldots)\ .$$