Exercise 6.3

Proof. Let $Y=\left\{x\in X^{}\mid x_1=0\right\}$, which is clearly a proper subset of $X^{}$ since, for example, $(1,1,\dots )$ is in $X^{}$ but not in $Y$. We construct a bijective function $f$ from $X^{}$ to $Y$. So consider any $x\in X^{}$ and define

for $i\in {\mathbb{Z}}_{\text{+}}$, noting that when $i\ne 1$ we have $i>1$ so that $i-1\ge 1$, and thus $y_i=x_{i-1}$ is defined. Now define $f(x)=y=y$ so that clearly $f$ is a function from $X^{}$ to $Y$, since $y_1=0$ for any input $x$.

To show that $f$ is injective, consider $x$ and $x^{\prime }$ in $X^{}$ where $x\ne x^{\prime }$, and let $y=f(x)$ and $y^{\prime }=f(x^{\prime })$. Now, since $x\ne x^{\prime }$, there is an $i\in {\mathbb{Z}}_{\text{+}}$ where $x_i\ne x_i^{\prime }$. Since $i>0$ (since $i\in {\mathbb{Z}}_{\text{+}}$) it follows that $i+1>1$ so that $i+1\ne 1$. We then have by the definition of $f$ that $y_{i+1}=x_{(i+1)-1}=x_i\ne x_i^{\prime }=x_{(i+1)-1}^{\prime }=y_{i+1}^{\prime }$ so that clearly $f(x)=y\ne y^{\prime }=f(x^{\prime })$. Since $x$ and $x^{\prime }$ were arbitrary, this shows that $f$ is indeed injective.

Now consider any $y\in Y$ so that $y_1=0$. Define $x_i=y_{i+1}$ for any $i\in {\mathbb{Z}}_{\text{+}}$ and let $x=x$. Then $x\in X^{}$ since clearly each $x_i=y_{i+1}\in X$. Now let $y^{\prime }=f(x)$ and consider any $i\in {\mathbb{Z}}_{\text{+}}$. If $i=1$ then clearly $y_i^{\prime }=y_1^{\prime }=0=y_1=y_i$ ($y_1^{\prime }=0$ since the range of $f$ is $Y$). If $i\ne 1$ then $y_i^{\prime }=x_{i-1}^{\prime }=y_{(i-1)+1}=y_i$. Hence $y_i^{\prime }=y_i$ in both cases so that $f(x)=y^{\prime }=y$ since $i$ was arbitrary. This shows that $f$ is surjective since $y$ was arbitrary.

Therefore $f$ is bijective as desired. □