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.