- State the approximation property of the supremum.

For $S⊂ℝ$ nonempty and bounded above, let $s=\sup S$; then for all $ϵ>0$, there exists $x∈S$ such that $x>s-ϵ$. - Prove that if $f: S→T$ and $g: T→U$ are injective, then so is $g∘f: S→U$.

For any $s$ and $s'∈S$, suppose that $g∘f(s)=g∘f\left(s'\right)$. Then $g(f(s))=g\left(f\left(s'\right)\right)$ by definition of composition. Since $g$ is injective, $f(s)=f\left(s'\right)$. Then since $f$ is injective, $s=s'$. Hence $g∘f$ is injective. - Prove that for all $n∈ℕ, 2^{n+1} ≥ 2 n+2$.

Proof by induction on $n$. For $n=0$, clearly $2^{0+1}=2≥2=2⋅0+2$. Now assume that for a given $n, 2^{n+1}≥2 n+2$. Then $2^{n+2}=2⋅2^{n+1}=2^{n+1}+2^{n+1}≥2 n+2+2 n+2=2(n+1)+2+2n≥2(n+1)+2$ where the last inequality is because $n≥0$ for all natural $n$. - For any $S⊂ℝ$ containing 0 , let $f:[0,∞)→ℝ$ be $f(x)=\sup (S∩[0, x])$.
(a) State a formula for $f$ if $S=ℤ$, the integers (no proof necessary).(a) $f(x)=[x]$; (b) $f(x)=x$; (c) If $x≤y$, then $S∩[0, x]⊂S∩[0, y]$, so any upper bound (such as sup $S∩[0, y]$ ) for the latter is an upper bound for the former, so the least upper bound must satisfy $\sup S∩[0, x]≤\sup S∩[0, y]$, so $f(x)≤f(y)$. Hence $f$ is monotonic, hence integrable on $[0,10]$.

(b) State a formula for $f$ if $S=ℝ$, all real numbers (no proof necessary).

(c) Prove that any such $f$ is integrable on $[0,10]$. - For $a< b ∈ ℝ$, if $g:[a, b]→ℝ$ is integrable and $|g|≤1$, prove that $\left|\int_a^b g(x) d x\right|≤b-a$.

Proof 1: For all $x∈ℝ,|g(x)|≤1$, so $-1≤g(x)≤1$, so by the comparison theorem for integrals, $-(b-a)=\int_a^b-1 d x≤\int_a^b g(x) d x≤\int_a^b 1 d x=b-a$, so $\left|\int_a^b g(x) d x\right|≤b-a$.

Proof 2: By an assigned problem (A5#6 in fact) $g$ integrable implies $|g|$ integrable, and by another assigned problem (A4#5) and the comparison theorem for integrals, $\left|\int_a^b g(x) d x\right|≤\int_a^b|g(x)| d x≤\int_a^b 1 d x=b-a$.

(Proof 1 is perhaps better than Proof 2 since it doesn't use $g$ integrable implies $|g|$ integrable, which is relatively hard to prove.) - Prove that $\lim _{x→0} \frac{1}{x}$ does not exist.

Proof 1: If there were a limit, say $K∈ℝ$, then by the definition of limit, for all $ϵ>0$ there exists $\delta>0$ such that $0<|x|<\delta$ implies $|1 / x-K|< ϵ$, that is, $-ϵ< 1 / x-K< ϵ$. For any such $ϵ$ and $\delta$, by the Archimedean property there exists an integer $n>\max (1 / \delta, ϵ+K)$, that is, $n>1 / \delta$ and $n>ϵ+K$. Then $x=1 / n$ satisfies $0<|x|<\delta$ but $1 / x-K>ϵ$, contradiction.

Proof 2: If there were a limit, say $K∈ℝ$, then basic limit theorem 1c implies $1=\lim _{x→0} x \frac{1}{x}=\left(\lim _{x→0} x\right)\left(\lim _{x→0} \frac{1}{x}\right)=0 K=0$, contradiction. - Suppose $f, g: ℝ→ℝ$ are any two functions such that for all $x∈ℝ,|f(x)|≤|x|$ and $|g(x)|≤1$. Prove that the product function $f g$ is continuous at 0 .

Proof 1: since all terms in the inequalities are nonnegative, $|f g(x)|=|f(x)||g(x)|≤|x|⋅1=|x|$, so $-|x|≤f g(x)≤|x|$. Since $\lim _{x→0}{-|x|}=\lim _{x→0}|x|=0$, by the squeezing theorem $\lim _{x→0} f g(x)=0$. But $|f(0)|≤|0|$, so $f(0)=0$ and $f g(0)=0 g(0)=0$. Hence $\lim _{x→0} f g(x)=f g(0)$

Proof 2: Given $ϵ>0$, take $\delta=ϵ$. Then $0<|x-0|<\delta$ implies $|f g(x)|=|f(x)||g(x)|≤|x|⋅1<\delta=ϵ$, so $\lim _{x→0} f g(x)=0$. But $|f(0)|≤|0|$, so $f(0)=0$ and $f g(0)=0 g(0)=$ 0. Hence $\lim _{x→0} f g(x)=f g(0)$