Real Analysis代写| Topology of the Real Number Line代写|R上的拓扑代写

证明 . 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.

证明 . 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.

证明 . 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 $\infty$.

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

证明 .

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.

证明 . 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.

证明 . 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
$0=\sup _{x \in D}(f(x)+g(x)) \leq \sup _{x \in D} f(x)+\sup _{x \in D} g\left(x_{0}\right)=1+1=2$