The cardinality of $P(\mathbb Z_+)$ is the same as $\mathbb R$ using decimal expansion.

Question

The problem is as follows:

Show that $|P(\mathbb Z_+)| = |\mathbb R|$ using the fact that every real number has a decimal expansion that is unique if expansions that end in an infinite sequence of $9$'s are forbidden. The question is from Munkres Topology.

I've seen a multitude of answers to this question on MATH.SE and other resources, but almost all of them utilized unique binary expansions of $x \in [0,1)$ (the existence of which I'd prefer not to assume, hence the question). I understand how the solution goes if unique binary expansions are allowed. However, I'd like to know how this is done using just decimal expansions.

If this question is a duplicate, I'll gladly delete it. This question is for self study purposes.

Answer 1

Injection $[0,1)\to P(\Bbb Z_+)$:

Given a real number $0.a_1a_2a_3\ldots\in [0,1)$ (where $a_1, a_2, a_3, \ldots$ are the digits in the decimal expansion), consider the set $$ \{10\cdot i + a_i\mid i\in \Bbb Z_+\} $$ This is a subset of $\Bbb Z_+$, and no two different numbers make the same set.

Example: $\pi - 3 = 0.14159\ldots$ is taken to $$ \{11, 24, 31, 45, 59, \ldots\} $$

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Injection $P(\Bbb Z_+)\to \Bbb [0,1)$:

Given a set $S\subseteq \Bbb Z_+$, write the list of positive integers in ascending order (in base ten). Swap out every $n$-digit number not in $S$ with $n$ zeroes, then concatenate the list.

Example: The set of even numbers is sent to $$ 0.0\,2\,0\,4\,0\,6\,0\,8\,0\,10\,00\,12\,00\,14\ldots $$ while the set of odd numbers is sent to $$ 0.1\,0\,3\,0\,5\,0\,7\,0\,9\,00\,11\,00\,13\,00\ldots $$

Answer 2

Well, you could use the bijection $(0,1)\rightarrow{\Bbb R}: x\mapsto\tan(\pi x-\pi/2)$ and the representation of the numbers in $(0,1)$ by decimal expansion.