Answer
See explanation
Work Step by Step
Let
\[
A \;=\;\{0,1,\dots,9\}^{\mathbb{Z}^+}
\quad\text{and}\quad
B \;=\;\{0,1\}^{\mathbb{Z}^+}.
\]
In other words, \(A\) is the set of all functions from \(\mathbb{Z}^+\) (the positive integers) into the 10‐element set \(\{0,\dots,9\}\), and \(B\) is the set of all functions from \(\mathbb{Z}^+\) into the 2‐element set \(\{0,1\}\). We want to show \(\lvert A\rvert = \lvert B\rvert\). By the **Schroeder–Bernstein Theorem**, it suffices to find:
1. An **injection** \(f\colon B \to A\).
2. An **injection** \(g\colon A \to B\).
Once we have injections both ways, the theorem guarantees a bijection exists, so \(A\) and \(B\) have the same cardinality.
---
## 1. Injection \(f\colon B \to A\)
This one is **straightforward**. Each function in \(B\) (i.e., each infinite sequence of 0’s and 1’s) is already a function into \(\{0,\dots,9\}\) if we simply regard 0’s and 1’s as valid decimal digits. Formally:
- If \(\phi \in B\), then \(\phi(n) \in \{0,1\} \subset \{0,\dots,9\}\) for each \(n\).
- Define \(f(\phi) = \phi\) itself (but now viewed as a function into the larger set \(\{0,\dots,9\}\)).
Clearly \(f\) is **one‐to‐one**: two distinct binary sequences remain distinct as decimal‐digit sequences.
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## 2. Injection \(g\colon A \to B\)
Now let \(\psi\in A\). So \(\psi\) is an infinite sequence \(\psi(1)\,\psi(2)\,\psi(3)\dots\) where each \(\psi(n)\) is a digit in \(\{0,\dots,9\}\). We must encode each decimal digit as a (fixed‐length) binary block to get a sequence of 0’s and 1’s.
A simple scheme:
1. **Block size**: 4 bits can represent \(2^4 = 16\) possible patterns, which is more than enough to encode 10 decimal digits (0 through 9).
2. **Assign** each digit \(d\in \{0,\dots,9\}\) to a distinct 4‐bit code, e.g.
\[
0 \mapsto 0000,\;
1 \mapsto 0001,\;
2 \mapsto 0010,\;
\dots,\;
9 \mapsto 1001.
\]
3. **Translate** \(\psi(n)\) into its 4‐bit code and concatenate these codes in order.
Concretely, define \(g(\psi)\) to be the infinite binary sequence obtained by writing down the 4‐bit code of \(\psi(1)\), then the 4‐bit code of \(\psi(2)\), and so on, back to back. Because different decimal sequences yield different blocks of bits, this process is **injective**.
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### Why \(g\) is Injective
If \(\psi_1 \neq \psi_2\), then at the first index \(n\) where they differ, \(\psi_1(n)\) and \(\psi_2(n)\) are distinct digits in \(\{0,\dots,9\}\). Their 4‐bit encodings are also distinct, so the resulting binary sequences differ in at least one bit. Hence \(g(\psi_1)\neq g(\psi_2)\).
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## 3. Conclusion by Schroeder–Bernstein
We have:
- An injection \(f\colon B \to A\).
- An injection \(g\colon A \to B\).
By the **Schroeder–Bernstein Theorem**, there must be a **bijection** between \(A\) and \(B\). Therefore,
\[
\bigl|\{0,1,\dots,9\}^{\mathbb{Z}^+}\bigr|
\;=\;
\bigl|\{0,1\}^{\mathbb{Z}^+}\bigr|.
\]
Equivalently, “there are as many functions from \(\mathbb{Z}^+\) to \(\{0,\dots,9\}\) as there are functions from \(\mathbb{Z}^+\) to \(\{0,1\}\).”
