1. [Tao] Ex 5.4.3,
2. [Tao] Ex 5.4.5 (you may assume result in 5.4.4)
3. [Tao] Ex 5.4.7,
4. [Tao] Ex 5.4.8
5. Let $A$ be the subset of $\Q$ consisting of rational numbers with denominators of the form $2^m$. Prove that for any $x \in \R$, there is a Cauchy sequence $(a_n)$ in $A$, such that $x = LIM a_n$.

Problem: Show that for any real number $x$, there is exactly one integer $N$, such that $N \leq x \leq N+1$.

Sol: If $x$ is an integer, then let $N=x$. Assume $x$ is not an integer now. Then $|x| > 0$, hence by Prop 5.4.12, there exists a positive integer $N_1$ such that $|x| < N_1$. Consider the set $S = \{-N_1, -N_1 + 1, \cdots, N_1 \}$ and the subsets $S_- = \{ n \in S \mid n < x \}$, $S_+ = \{n \in S \mid n > x\}$. Since $x$ is not an integer, then $S = S_- \sqcup S_+$ (disjoint union, meaning $S_- \cap S_+ = \emptyset$). Let $N = \max (S_-)$, since $S_-$ is a finite set, the max element exists in $S_-$.

Remark: we will call such $N$ the 'floor' of $x$, and write $N = \lfloor x \rfloor$.

Given any real numbers $x < y$, show that there is a rational number $q$ such that $x < q < y$.

Sol: Since $0 < y-x$, we can find a large enough positive integer $N$, such that $N (y-x) > 1$ (by Archimedian property). Let $m$ be the unique integer, such that $m \leq N x < m+1$ (proven in the problem 1), then $m + 1 \leq N x + 1 < N y$, hence we have $N x < m+1 < N y$. Let $q = (m+1)/N$.

Let $x,y$ be real nubmers. Show that $x \leq y+\epsilon$ for all real $\epsilon>0$ if and only if $x \leq y$. Show that $|x-y| < \epsilon$ for all real $\epsilon > 0$ if and only if $x=y$.

Solution: If $x \leq y$, then for any $\epsilon > 0$, we have $x \leq y < y + \epsilon$, hence $x < y + \epsilon$. In the other direction, if $x \leq y +\epsilon$ for all $\epsilon > 0$, we claim that $x - y \leq 0$. If not, say $c = x-y > 0$, then, we can let $\epsilon = c / 2$, then we get $c = (x-y) \leq \epsilon = c / 2 \Rightarrow c - c/2 \leq 0 \Rightarrow c \leq 0$ which contradict with assumption $c > 0$, hence the claim is true.

For the second statement, if $|x-y| < \epsilon$ for all real $\epsilon > 0$, then $|x-y| \leq 0$ by the same argument. Since we also have $|x-y| \geq 0$ by definition of absolute value, hence $|x-y|=0$. Thus $x = y$.

Let $(a_n)$ be a Cauchy sequence of rationals, and let $x$ be a real number. Show that, if $a_n \leq x$ for all $n$, then $LIM a_n \leq x$. Similarly, if $a_n \geq x$ for all $n$, then $LIM a_n \geq x$.

Solution: We only prove the first half of the statement, since the other half is similar. We prove by contradiction, if $LIM a_n > x$, then there is a rational number $q$, such that $LIM a_n > q > x$. Since we have $a_n < x < q$, we have $LIM a_n \leq LIM q = q$ (Corollary 5.4.10), contradict with the condition of $LIM a_n > q.

Let $A$ be the subset of $\Q$ consisting of rational numbers with denominators of the form $2^m$. Prove that for any $x \in \R$, there is a Cauchy sequence $(a_n)$ in $A$, such that $x = LIM a_n$.

Solution: Let $x \in \R$, and let $(b_n)$ be a Cauchy sequence in $\Q$, such that $x = LIM b_n$, we are going to construct $(a_n)$ in $A$, such that $(a_n)$ is equivalent to $(b_n)$, hence $x = LIM a_n$. For $n \geq 1$, we let $a_n = \frac{\lfloor 2^n b_n \rfloor}{2^n}$, where $\lfloor .. \rfloor$ is the floor function (see Problem 1). Then $|a_n - b_n| = | \frac{\lfloor 2^n b_n \rfloor}{2^n} - \frac{2^n b_n }{2^n} | = | \frac{\lfloor 2^n b_n \rfloor - b_n}{2^n}| \leq 1 / 2^n$ hence $(a_n)$ is equivalent to $(b_n)$.