Real Number

main def: 记运算$L_{n\to\infty}:=\{(a_{n},b)\in(a_{n})^{cauchy}\times R:a_{n}L_{n\to\infty}b\}$.

def 8: 两实数相等, 当且仅当其domain是等价的柯西序列.

• 该相等是定义明确的, 即:$x= L(a_{n})$, $y=L(b_{n})$, $z=L(c_{n})$. 首先等号是reflexive的(定义得), 其次, 该等号也是transitive和symmetric的.

  • 证: sym: $\forall a_{n},b_{n}(d(a_{n},b_{n})=d(b_{n},a_{n}))\to\bigg( (a_{n})\sim(b_{n})\to(b_{n})\sim(a_{n})\bigg) Q.E.D$.
  • transitive: $L(a_{n})=L(b_{n})\wedge L(b_{n})=L(c_{n})\to \left\{\begin{matrix}\forall \epsilon \exists N\forall n(|a_{n}-b_{n}|<\epsilon/2 ) \\ \forall \epsilon \exists N\forall n(|b_{n}-c_{n}|<\epsilon/2 ) \end{matrix}\right.$. $\to (|a_{n}-c_{n}|<|a_{n}-b_{n}|+|b_{n}-c_{n}|<\epsilon)$. Q.E.D

• 该等式满足替换公理.

  • 证: 略. 用有理数序列的接近性性质来证会很快. def 14 in From Natural to Rational.

Add

def 9 (加法): 令$x=L(a_{n}),y=L(b_{n})$, 则$x+y:=L(a_{n}+b_{n})$.

lemma 3: 如果x,y是实数, 那么x+y也是实数.

• 证: $x=L(a_{n}), y=L(b_{n})$. if x and y are all real numbers, then $\forall \epsilon \exists N_{1}\forall n_{1}\forall k_{1}(|a_{n_{1}}-a_{n_{1}+k_{1}}|<\epsilon)\wedge \forall \epsilon \exists N_{2}\forall n_{2}\forall k_{2}(|b_{n_{2}}-b_{n_{2}+k_{2}}|<\epsilon)$. We have that: $\forall \epsilon_{1}+\epsilon_{2}(|a_{n_{i}}+b_{n_{i}}-a_{n_{i}+k_{i}}-b_{n_{i}+k_{i}}|<...<\epsilon_{1}+\epsilon_{2})$. For $n_{i}=max(n_{1},n_{2}), \epsilon_{1},\epsilon_{2}$ are threshold for a and b under $n_{i}$ Q.E.D

Multiple

def 10 (乘法): 令$x=L(a_{n}),y=L(b_{n})$, 则$xy:=L(a_{n}b_{n})$.

• 易证, 乘法也是定义明确的且满足替换公理. 用def 14 in From Natural to Rational 来证会很快.

Minus

def 11 (负运算): 令x为实数, 则$-x:=(-1)\cdot x$. (注意到-1是有理数, 所有有理数的柯西序列是0接近的)

def 11$\to$ def 12 (减法): $x-y:=x+(-y)$.

Inverse

为了避免除以0 (包括任何和$(0)_{n=1}^{\infty}$等价的柯西序列以及包含0的柯西序列), 我们先定义一些别的:

def 13 (远离0): 称$(a_{n})$是远离0的, iff $\exists c>0\bigg(\forall n\ge1(|a_{n}|\ge c)\bigg)$.

lemma 4: 设x是不为0的实数, 则存在某个远离0的柯西序列$(a_{n})$使得$x=L(a_{n})$.

• 证: 什么b证明. $\forall x\not= 0\wedge x\in R(\exist(b_{n})(x=L(b_{n})))$. This means that each real number has infinite correspond cauchy series. Thus if x is not 0, then $b_{n}$ is not $\epsilon$ close to $(0)_{n=1}^{\infty}$, which means $b_{n}\not \sim (0)$. This means $\forall \epsilon>0(\exists n(\forall i,j>n(|b_{i}-b_{j}|\le\frac{\epsilon}{2})\wedge \exists k>n(|b_{k}-0|\ge\epsilon)))$.

  $\to |b_{i}-b_{k}|<\frac{\epsilon}{2}\wedge|b_{k}|>\epsilon\to |b_{i}|>\frac{\epsilon}{2}$. (三角不等式把$b_{i}$移到等式另一边).

  Besides, we could find another cauchy series $(a_{n})\sim (b_{n})(\forall s<n(|a_{s}|>0))$. Q.E.D

  思路就是先假设x对应一个柯西序列, 然后该柯西序列不接近0即意味着对任意小正数,序列中总存在某项与0的距离大于该小正数, 最后证明存在其他项的绝对值全大于0的序列接近该序列.

lemma 5: 如果序列$(b_{n})$是远离0的柯西序列, 那么$(b_{n}^{-1})$也是远离0的柯西序列.

• 证: for the cauchy part: assume $(a_{n})$ is cauchy away from 0, then we have: $\forall \epsilon\exists N\forall n\forall j,k(|a_{j}-a_{k}|<\epsilon)$.

  for $\forall \epsilon\exists N\forall n\forall j,k(|a_{j}^{-1}-a_{k}^{-1}|=|\frac{a_{j}-a_{k}}{a_{j}a_{k}}|)$. To prove the result, we just need to find a number smaller than $\epsilon$ that makes $|a_{j}a_{k}|\cdot n\ge 1\to |\frac{a_{j}-a_{k}}{a_{j}a_{k}n}|\le \epsilon$. Or we could use another method: as $(a_{n})$ is cauchy, then $|a_{j}|,|a_{k}|$ are larger than some numbers c. So we have: $|\frac{a_{j}-a_{k}}{a_{j}a_{k}}|\le |\frac{a_{j}-a_{k}}{c^{2}}|$. This means if we want the result holds, we just need to make sure $|a_{j}-a_{k}|\le c^{2}\epsilon$, for $c^{2}>0$ as it is far from 0. The final result is: $\forall \epsilon\exists N\forall j,k(|a_{j}-a_{k}|<c^{2}\epsilon\to|\frac{a_{j}-a_{k}}{a_{j}a_{k}}|<\epsilon)$.

   

  for the far from 0 part: suppose $\forall c>0(\exists n\ge 1 (|a_{n}^{-1}-0|<c))\to\forall c>0(\exists n\ge 1 (|a_{n}-0|>\frac{1}{c})\to (a_{n})$ is not cauchy.

  Q.E.D

def 14: suppose $x=L(a_{n})$, for $a_{n}$ is cauchy series far from 0. 那么我们定义$x^{-1}:=L(a_{n}^{-1})$.

Lemma 6: 倒数的定义是明确的.

• 证明略.

Division

def 15 (除法): $x/y=x\cdot y^{-1}$. 通过除法易证乘法有消去律.

Problem Set for Real Number

5.3.3: a, b为有理数, 证: a=b当且仅当L(a)=L(b).

• 证: if: $L(a)=L(b)\to \forall \epsilon \forall N\ge 1\forall i(|a_{i}-b_{i}|<\epsilon)\to$

  suppose $|a_{i}-b_{i}|>0\to\exists \epsilon'(\epsilon'>|a_{i}-b_{i}|)\to L(a)\not=L(b)\to |a_{i}-b_{i}|=0\to a_{i}=b_{i}$.

  Q.E.D

5.3.4 - 5.3.5: easy.