Exercise 9.1

Question

Define an injective map $f: {{\mathbb{Z}}_+} 	o X^{\omega}$, where $X$ is the two-element set $\left\{0,1ight\}$, without using the choice axiom.

Dan Whitman · Accepted Answer

ewcommand\seqinf[1]{\left(#1_1, #1_2, \ldots ight)}
\def{ho}

For any $n \in {{\mathbb{Z}}_+}$, define
\begin{gather*}
x_i = \begin{cases}
0 & i 
eq n \
1 & i = n
\end{cases}
\end{gather*}
for $i \in {{\mathbb{Z}}_+}$.
Then set $f(n) = \mathbf{x}  = \seqinf{x}$ so that clearly $f$ is a function from ${{\mathbb{Z}}_+}$ to $X^{\omega}$.
It is easy to show that $f$ is injective.
\begin{proof}
Consider $n,m \in {{\mathbb{Z}}_+}$ where $n 
eq m$.
Then let $\mathbf{x}  = f(n)$ and $\mathbf{y}  = f(m)$.
Then we have that $x_n = 1$ while $y_n = 0$ by the definition of $f$ since $n 
eq m$.
It thus follows that $f(n) = \mathbf{x}  
eq \mathbf{y}  = f(m)$, which shows that $f$ is injective since $n$ and $m$ were arbitrary.
\end{proof}

Exercise 9.1

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Exercise 9.1

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