Exercise 5.5

Proof. Consider any $x\in X$ so that $x_i\in \mathbb{Z}$ for any $i\in {\mathbb{Z}}_{\text{+}}$. It is then immediately obvious that $x\in {\mathbb{Z}}^{\omega }=Y$. Hence $X\subset Y$ since $x$ was arbitrary.

Now consider any $x\in Y={\mathbb{Z}}^{\omega }$ so that $x_i\in \mathbb{Z}$ for every $i\in {\mathbb{Z}}_{\text{+}}$. Again it is obvious by the definition of $X$ that $x\in X$. Hence $Y\subset X$ since $x$ was arbitrary. This shows that $X=Y$, as desired. □

Proof. First consider $x\in X$ so that $x_i\ge i$ for any $i\in {\mathbb{Z}}_{\text{+}}$. Then, for any $i\in {\mathbb{Z}}_{\text{+}}$ clearly $x_i\in Y_i$ by definition since $x_i\ge i$ (and also $x_i\in ℝ$). Hence it follows that $x=\left(x_1,x_2,\dots \right)\in Y_1\times Y_2\times \cdots =Y$. Since $x$ was arbitrary, this shows that $X\subset Y$.

Now suppose that $x\in Y$ so that $x_i\in Y_i$ for every $i\in {\mathbb{Z}}_{\text{+}}$. Consider any such $i\in {\mathbb{Z}}_{\text{+}}$ so that $x_i\in Y_i$. Then, by definition $x_i\ge i$. Since $i$ was arbitrary, this shows that $x\in X$ by definition. Hence $Y\subset X$ since $x$ was arbitrary so that $X=Y$. □

Proof. Consider any $x\in X$ so that $x_i\in \mathbb{Z}$ for all $i\ge 100$. Suppose $i\in {\mathbb{Z}}_{\text{+}}$. If $i<100$ then clearly $x_i\in ℝ=Y_i$ since $x\in {ℝ}^{\omega }$. If $i\ge 100$ then we have that $x_i\in \mathbb{Z}=Y_i$. Hence in either case $x_i\in Y_i$ so that $x\in Y_1\times Y_2\times \cdots =Y$ since $i$ was arbitrary. Since $x$ was arbitrary, this shows that $X\subset Y$.

Now consider any $x\in Y$ and any $i\in {\mathbb{Z}}_{\text{+}}$ where $i\ge 100$. Then $x_i\in Y_i=\mathbb{Z}$ so that $x_i$ is an integer. From this it follows that $x\in X$ by definition since obviously $x\in {ℝ}^{\omega }$ (since $x_i\in Y_i=ℝ$ when $i<100$). Hence $Y\subset X$ since $x$ was arbitrary. This completes the proof that $X=Y$. □

Proof. Suppose to the contrary that there are $X_i\subset ℝ$ for $i\in {\mathbb{Z}}_{\text{+}}$ where $X=X_1\times X_2\times \cdots$. Let $(a,a,\dots )$ denote the sequence $\left(x_1,x_2,\dots \right)$ where $x_i=a$ for all $i\in {\mathbb{Z}}_{\text{+}}$. We then have that $(1,1,\dots )$ and $(2,2,\dots )$ are both in $X$ since clearly $x_2=x_3$ in both. Hence we have that $1$ and $2$ are both in $X_i$ for every $i\in {\mathbb{Z}}_{\text{+}}$ since $X=X_1\times X_2\times \cdots$. Now define

for $i\in {\mathbb{Z}}_{\text{+}}$. Clearly $y=\left(y_1,y_2,\dots \right)\in X_1\times X_2\times \cdots$ since both $1$ and $2$ are in each $X_i$. However, it is also clear that $y\notin X$ by definition since $y_2=2\ne 1=y_3$. This contradicts the fact that $X=X_1\times X_2\times \cdots$, which shows the desired result. □