• 设 $a \in \mathbb{R}$ 。 $a$ 的邻域是 $\mathbb{R}$ 中一个包括一段含有 $a$ 的线段的子集。
• $\mathbb{R}$ 是可分空间。
• $\mathbb{R}$ 的开集是开区间的并集。
• $\mathbb{R}$ 的紧子集等价于有界闭集（Heine-Borel定理 $)$ 。特别是：所有含端点的有限线段都是紧子 集。
• $\mathbb{R}$ 是连通且单连通的。
• $\mathbb{Q}$ 在 $\mathbb{R}$ 中处处稠密。
• 每个 $\mathbb{R}$ 中的有界序列都有收敘子序列 (Bolzano-Weierstrass定理)
• $\mathbb{R}$ 中的连通子集是线段、射线与 $\mathbb{R}$ 本身。由此性质可迅速导出中值定理（MVT定理）。。。。
• 设 $\left(F_{n}\right)_{n \in \mathbb{N}}$ 是一个有界闭集的序列, 且 $\forall n \in \mathbb{N}\left(F_{n} \supseteq F_{n+1}\right),$ 则其交集非空。 (区间
套定理）

Problem 1. Let $F$ be an ordered field. Prove directly from the axioms of that if $x, y, z \in F, y<z$ and $x<0,$ then $x y>x z$

Proof . All numbers in this problem refer to the notes. We will first show that $0 \cdot x=0$ for any $x$ in an ordered field. To see this, by $(\mathrm{D})$ and $(\mathrm{A} 4)$ of Definition 1.1 .5 we have
$$0 \cdot x+0 \cdot x=(0+0) x=0 \cdot x$$
Using
(ii) of Definition 1.1 .1 we have
$$0 \cdot x+0 \cdot x+0 \cdot(-x)=0 \cdot x+0 \cdot(-x)$$
Using $(\mathrm{D}),(\mathrm{A} 4),(\mathrm{A} 5)$ we have
$$0 \cdot x+0 \cdot(x+(-x))=0 \cdot(x+(-x)) \Longleftrightarrow 0 \cdot x=0$$
Now back to the original problem. Since $x<0$ we have $(-x)+x<0+(-x)$ by (ii), so $0<-x$ by
(A4) and (A5). Since $y<z$ we have $y+(-y)<z+(-y)$ by (ii) and so $0<z+(-y)$ by (A5). By (ii) of Definition 1.1 .7 we then have $(-x)(z+(-y))>0 .$ Using $(\mathrm{D})$ and (ii) of Definition 1.1 .1 we have
$$(-x) z+(-x)(-y)>0 \Longrightarrow(-x) z+(-x)(-y)+x(-y)>(-x)(-y)+x(-y)$$
Using $(\mathrm{D}),(\mathrm{A} 5)$ and $0 \cdot x=0$ we have
$$(-x) z+((-x)+x)(-y)>x(-y) \Longrightarrow(-x) z+0 \cdot(-y)>x(-y) \Longrightarrow(-x) z>x(-y)$$
Two applications of (ii) of Definition 1.1 .1 gives
$$(-x) z+x z+x y>(-x) y+x z+x y$$
Using (A2) and (D) we have
$$((-x)+x) z+x y>((-x)+x) y+x z$$
Finally using $($ A5 $)$ and $0 \cdot x=0$ we get
$$0 \cdot z+x y>0 \cdot y+x y \Longrightarrow x y>x z$$
which is what we want to prove.

Problem 2. Let $S$ be an ordered set. Suppose that $A \subset S$ is such that $s=\sup (A) \in S$ exists, but $s \notin A$. Show that $A$ must contain a countably infinite subset.
Proof . By definition of the supremum, we can find $a_{1} \in A$ such that $s>a_{1}>s-1 .$ Since $a_{1} \neq s,$ there is $a_{2} \in A$ such that $s>a_{2}>a_{1}$. More generally, having defined $a_{1}, \ldots, a_{n},$ we let $a_{n+1} \in A$ be such that $s>a_{n+1}>a_{n} .$ This gives a surjective map from $\mathbb{N}$ to $\left\{a_{1}, a_{2}, \ldots\right\} \subset A,$ so $A$ contains a countably infinite subset.
Problem 3. Let $S=\mathbb{N} \cup\{\infty\}$. Make this an ordered set by making $n<\infty$ for every $n \in \mathbb{N} \subset S$ and otherwise using the usual order on $\mathbb{N}$. Show that every $E \subset S$ is bounded and $\sup (S)$ and inf $(S)$ both exist.
Proof . If $E$ is empty then everything is vacously true, so suppose $E \neq \emptyset$. Every set $E \subset S$ is bounded below by 1 and above by $\infty$. If $\infty \notin E$, then $E$ has an infimum is well-defined by well-ordering principle. $E$ has a supremum if $E$ is finite, and if $E$ is infinite the supremum is $\infty$. If $\infty \in E$, then the infimum is again guaranteed by well-ordering principle, and the supremum is just $\infty$.
Problem 4.

Let $x, y \in \mathbb{R}$ satisfy $x^{2}+y^{2}=0$ show that $x=y=0$.

Proof . Since $x^{2} \geq 0$ we have $x^{2} \leq x^{2}+y^{2}=0 \Longrightarrow x^{2}=0 \Longrightarrow x=0 .$ Similarly $y=0$.
Problem 5.

Use induction to show that for any $x \in \mathbb{R}$ if $1+x>0,$ then $(1+x)^{n} \geq 1+n x$.

Proof .

When $n=1$ we have $(1+x)^{n}=1+x$ so the inequality holds (and is actually a equality). Suppose the claim is true for $n,$ then
$(1+x)^{n+1}=(1+x)^{n}(1+x) \geq(1+n x)(1+x)=1+n x+x+n x^{2}=1+(n+1) x+n x^{2}>1+(n+1) x$
so the claim holds true for $n+1$ as well.

Problem 6.

For $x, y \in \mathbb{R}$ show that
$$\max \{x, y\}=\frac{1}{2}(x+y+|x-y|) \text { and } \min \{x, y\}=\frac{1}{2}(x+y-|x-y|)$$

Proof . If $x \geq y,$ then $\max \{x, y\}=x$ and
$$\frac{1}{2}(x+y+|x-y|)=\frac{1}{2}(x+y+x-y)=\frac{1}{2}(2 x)=x$$
Similarly if $y \geq x$ then $\max \{x, y\}=y$ and
$$\frac{1}{2}(x+y+|x-y|)=\frac{1}{2}(x+y+y-x)=\frac{1}{2}(2 y)=y$$
The second statement is proven similarly.
Problem 7.

Let $A$ be a non-empty set and $f, g: D \rightarrow \mathbb{R}$ be two bounded functions. Show that
$$\sup _{x \in D}(f(x)+g(x)) \leq \sup _{x \in D} f(x)+\sup _{x \in D} g(x)$$
Give an example to show the inequality can be strict.

Proof . Given $\varepsilon>0,$ by definition there is $x_{1} \in D$ such that $f\left(x_{0}\right)+g\left(x_{0}\right)>\sup _{x \in D}(f(x)+g(x))-\varepsilon .$ Then
$$\sup _{x \in D}(f(x)+g(x))<f\left(x_{0}\right)+g\left(x_{0}\right)+\varepsilon \leq \sup _{x \in D} f(x)+\sup _{x \in D} g\left(x_{0}\right)+\varepsilon$$
Since this is true for any $\varepsilon$ we see that
$$\sup _{x \in D}(f(x)+g(x)) \leq \sup _{x \in D} f(x)+\sup _{x \in D} g\left(x_{0}\right)$$
Let $D=\{1,-1\}, f(1)=1, f(-1)=-1$ and $g(1)=-1, g(-1)=1,$ then $f+g$ is identically $0,$ so
$0=\sup _{x \in D}(f(x)+g(x)) \leq \sup _{x \in D} f(x)+\sup _{x \in D} g\left(x_{0}\right)=1+1=2$

real analysis代写， math 115A代写，实分析代写， 数学分析代写请认准UpriviateTA. UpriviateTA为您的留学生涯保驾护航。