两件不同的事

这门课有两个目标，本讲专门处理第二个：怎样把「为什么应该成立」写成一份正式的数学证明。

这两件事不一样。第 2 讲证明 $\sqrt 2$ 存在时，讲的是想法：假设上确界偏大或偏小，各挪一点制造矛盾。那是草稿纸上的推理。真正交上去的证明则是另一套写法——先把要用的量（比如 $h$）一次性定义好，再按固定结构走完。正式证明看起来「从天而降」，但读起来清楚得多。

课程用同一个 $\sqrt 2$ 例子演示两遍：一遍当草稿，一遍当正式证明。

时间轴

上界、下界与有界

定义

设 $S$ 是有序集，$A\subseteq S$。

• $A$ 有上界：存在 $M\in S$，使 $a\le M$ 对一切 $a\in A$ 成立。这样的 $M$ 叫上界。
• $A$ 有下界：存在 $m\in S$，使 $m\le a$ 对一切 $a\in A$ 成立。这样的 $m$ 叫下界。
• $A$ 有界：既有上界又有下界。

边界元不必属于子集

这是同学们在讨论区问得最多的一点：$M$ 必须在哪个集合里？

答案是：$M$ 只需在更大的集合 $S$ 里，不必属于 $A$。

例子：$A=(0,1)\subset\mathbb R$。$1$ 是上界，$0$ 是下界，但 $0,1$ 都不属于 $A$。而且 $A$ 中没有任何元素能当上界：任取 $x\in A$，总有 $x/2+1/2\in A$ 比它大。所以上界只能来自集合外部。

一般情形下，上确界与下确界也可能不在 $A$ 内。它们可以属于 $A$，但不是必须。

一个容易错的例子：A 是谁的子集

定义里有一个容易被忽略的动作：所有边界元都要取自 $S$，而不是取自整个 $\mathbb R$。

取 $S=(0,\infty)$，$A=(0,1)$，把 $A$ 看作 $S$ 的子集。

• $A$ 有上界：$1$ 是上界，因为 $x<1$ 对一切 $x\in A$ 成立，而 $1\in S$。
• $A$ 没有下界。任何下界 $m$ 都必须在 $S=(0,\infty)$ 中，因此 $m>0$。取 $x=\min\{m/2,\,1/2\}\in A$，则 $x<m$，与 $m$ 是下界矛盾。

同一个 $A$，若看作 $\mathbb R$ 的子集，$0$ 就是下界。差别不在 $A$，而在「谁允许提供边界元」。这说明谈上界、下界之前必须先说清楚环境集合 $S$。

课程还顺带澄清：$+\infty$ 与 $-\infty$ 不是实数，不能拿来当上界或下界。写「上界是 $\infty$」在实数系里没有意义。

完备性：上确界与下确界

设 $S$ 是有序集，$A\subseteq S$ 有上界。若存在 $M\in S$ 满足

1. $M$ 是 $A$ 的上界；
2. 对 $A$ 的任意上界 $M_1$，都有 $M\le M_1$，

则称 $M$ 为 $A$ 的最小上界，记 $\sup A$。

对称地，若 $A$ 有下界，存在最大的下界 $m$，即

1. $m$ 是下界；
2. 对任意下界 $m_1$，有 $m_1\le m$，

则称 $m$ 为最大下界，记 $\inf A$。

定义。 有序集 $S$ 称为完备的，如果它的每个有上界子集都有最小上界。

$\mathbb R$ 完备，因此每个有界子集既有上确界也有下确界。两者的关系是

$$\inf A=-\sup(-A), \qquad -A=\{-a:a\in A\},$$

所以只证上确界就够，下确界可以照搬。

正式证明的写法：以 $\sqrt 2$ 为例

定理与集合

定理。 存在 $\alpha\in\mathbb R$，$\alpha>0$，使 $\alpha^2=2$。

证明。 令

$$A=\{\,x\in\mathbb R : x>0,\ x^2<2\,\}.$$

注意这里把集合清楚写出，并注明 $x\in\mathbb R$——正式证明里不能省这一步。

第一步：$A$ 非空。 $1\in A$，因为 $1>0$ 且 $1^2=1<2$。

第二步：$A$ 有上界。 断言 $2$ 是上界。若 $x\in A$ 且 $x\ge2$，则 $x^2\ge4>2$，与 $x^2<2$ 矛盾。所以 $x<2\le2$，即 $2$ 是上界。

第三步：取定 $\alpha$。 由 $A$ 非空有上界与 $\mathbb R$ 完备，令

$$\alpha=\sup A .$$

第四步：$\alpha>0$。 因为 $1\in A$，而 $\alpha$ 是上界，所以 $1\le\alpha$，特别地 $\alpha>0$。同时 $\alpha\le2$，因为 $2$ 是上界而 $\alpha$ 是最小的。

剩下只需证 $\alpha^2=2$，分两部分。

第一部分：$\alpha^2\le2$

反设 $\alpha^2>2$。目标：构造一个比 $\alpha$ 小的上界，与最小性矛盾。

定义

$$h=\frac{\alpha^2-2}{4\alpha}, \qquad \alpha_1=\alpha-h .$$

$h>0$：分子 $\alpha^2-2>0$，分母 $4\alpha>0$。

$\alpha_1>0$：因为

$$h=\frac{\alpha^2-2}{4\alpha}<\frac{\alpha^2}{4\alpha}=\frac\alpha4,$$

所以 $\alpha_1=\alpha-h>\frac34\alpha>0$。

$\alpha_1^2>2$：展开并丢掉正项 $h^2$，

$$\alpha_1^2=(\alpha-h)^2=\alpha^2-2\alpha h+h^2\ge\alpha^2-2\alpha h .$$

代入 $h$：

$$2\alpha h=2\alpha\cdot\frac{\alpha^2-2}{4\alpha}=\frac{\alpha^2-2}{2},$$

于是

$$\alpha_1^2\ge\alpha^2-\frac{\alpha^2-2}{2}=\frac{\alpha^2+2}{2}>2,$$

最后一个不等号用了假设 $\alpha^2>2$。

现在任取 $x\in A$。由 $x>0$、$\alpha_1>0$、$x^2<2<\alpha_1^2$，开方得 $x<\alpha_1$。所以 $\alpha_1$ 是 $A$ 的上界，而 $\alpha_1<\alpha$，与 $\alpha$ 是最小上界矛盾。

故 $\alpha^2\le2$。

第二部分：$\alpha^2\ge2$

反设 $\alpha^2<2$。目标：找一个属于 $A$ 且大于 $\alpha$ 的数，与「$\alpha$ 是上界」矛盾。

定义

$$h=\frac{2-\alpha^2}{4\alpha}, \qquad \alpha_2=\alpha+h .$$

$0<h<\frac12$：分子为正，分母为正，故 $h>0$；又 $\alpha\ge1$，所以

$$h=\frac{2-\alpha^2}{4\alpha}<\frac{2}{4\alpha}\le\frac12<1 .$$

$\alpha_2^2<2$：展开，

$$\alpha_2^2=(\alpha+h)^2=\alpha^2+2\alpha h+h^2 .$$

因为 $0<h<1$，有 $h^2<h$，故

$$\alpha_2^2<\alpha^2+2\alpha h+h .$$

代入 $2\alpha h=\dfrac{2-\alpha^2}{2}$，并用 $h<\dfrac{2-\alpha^2}{2}$（由 $\alpha\ge1$ 得 $\frac1{2\alpha}\le1$，故 $h=\frac{2-\alpha^2}{4\alpha}\le\frac{2-\alpha^2}{2}$），得到

$$\alpha_2^2<\alpha^2+\frac{2-\alpha^2}{2}+\frac{2-\alpha^2}{2}=2 .$$

同时 $\alpha_2>\alpha>0$，所以 $\alpha_2\in A$。但 $\alpha_2>\alpha$，与 $\alpha$ 是 $A$ 的上界矛盾。

故 $\alpha^2\ge2$。

两部分合起来给出 $\alpha^2=2$，定理得证。$\square$

从草稿到正式证明

对比第 2 讲与上面的写法，差别在信息的顺序：

正式证明里，$h$ 看起来凭空出现。它的来源是草稿纸上的计算：要让 $(x-h)^2>2$，只需 $2xh<x^2-2$。正式写法把这次计算压缩成一句「定义 $h=\cdots$」，代价是读者看不到动机，好处是逻辑链条没有缺口。

课程给了一个实用建议：只写需要用到的性质。若为了「完整」把集合的其它性质也证一遍，反而让读者抓不住重点。

阿基米德性质的正式证明

定理。 自然数集 $\mathbb N$ 在 $\mathbb R$ 中无上界。

等价的说法是：对每个实数 $x$，存在 $n\in\mathbb N$ 使 $n>x$。

证明（反证）。 假设 $\mathbb N$ 有上界。由 $\mathbb R$ 完备，令

$$\alpha=\sup\mathbb N .$$

矛盾将来自「$\alpha-1$ 也是上界」。

因为 $\alpha$ 是上界，对每个 $n\in\mathbb N$ 有 $n\le\alpha$。又 $n+1\in\mathbb N$，所以

$$n+1\le\alpha \implies n\le\alpha-1 \quad\text{对一切 } n\in\mathbb N .$$

于是 $\alpha-1$ 也是 $\mathbb N$ 的上界。但 $\alpha-1<\alpha$，而 $\alpha$ 是最小上界，矛盾。故 $\mathbb N$ 无上界。$\square$

这份证明只有四行，但每一行都有明确作用：第一行用反证假设，第二行用完备性取上确界，第三行用 $\mathbb N$ 对 $+1$ 封闭，第四行用最小性。写正式证明时，可以逐行问自己「这一行用了哪条已知事实」。

数列：把 $\sqrt 2$ 放到实轴上

定理只保证 $\sqrt 2$ 作为一个数存在。若要把它定位在实轴上，一个好办法是看它的十进制截断：

$$1,\quad 1.4,\quad 1.41,\quad 1.414,\quad 1.4142,\ \dots$$

每一项都小于后一项，而 $\sqrt 2$ 是这串数的极限。

定义。 实数列是函数 $a:\mathbb N\to\mathbb R$，常记作 $(a_n)_{n\ge1}$，其中 $a_n=a(n)$。

把下标当自变量，是为了强调「序列是一个函数」；写成 $a_n$ 而不是 $a(n)$ 只是传统。

上面那串数就是一个递增序列：$a_n<a_{n+1}$，且 $\sqrt 2=\lim_{n\to\infty}a_n$。但「极限」目前仍只是直观说法，要等到下一讲才用 $\varepsilon$ 语言定义。

本节结论

• 上界、下界必须取自环境集合 $S$，不必属于子集 $A$。
• $(0,1)$ 在 $S=(0,\infty)$ 中有上界、无下界；换成 $S=\mathbb R$ 则有下界 $0$。
• 完备 = 每个有上界子集都有上确界；$\inf A=-\sup(-A)$。
• 正式证明的结构：写出定理 → 构造集合 → 验证非空与有界 → 取上确界 → 分两部分反证。
• 证明中出现「凭空定义的 $h$」，它来自草稿纸上的估算，正式稿把动机压缩掉了。
• 阿基米德性质由完备性推出：$\sup\mathbb N$ 会导出 $\alpha-1$ 也是上界。
• 数列是 $\mathbb N\to\mathbb R$ 的函数；$\sqrt 2$ 是截断序列的极限。

理解检验

1. 为什么上界 $M$ 不必属于集合 $A$？举一个 $A$ 中根本没有上界的例子。

参考答案：定义只要求 $M$ 属于环境集合 $S$，并且压住 $A$ 的所有元素。取 $A=(0,1)\subset\mathbb R$，$1$ 是上界但 $1\notin A$。而且 $A$ 内没有任何元素是上界：对任意 $x\in A$，令 $y=(x+1)/2$，则 $x<y<1$，所以 $y\in A$ 且 $y>x$。因此上界只能从集合外部取。

2. 设 $S=(0,\infty)$，$A=(0,1)$。把 $A$ 看作 $S$ 的子集，为什么 $A$ 没有下界？这与把 $A$ 看作 $\mathbb R$ 的子集有何不同？

参考答案：下界必须属于 $S$，因此任何候选下界 $m$ 都满足 $m>0$。取 $x=m/2$（若 $m/2\ge1$ 则取 $x=1/2$），则 $x\in A$ 且 $x<m$，所以 $m$ 不是下界；$m$ 任意，故无下界。若把 $A$ 看作 $\mathbb R$ 的子集，$0\in\mathbb R$ 且 $0\le x$ 对一切 $x\in A$ 成立，所以 $0$ 是下界。区别不在于 $A$，而在于允许边界元取自哪个集合。

3. 正式证明里 $h=\dfrac{\alpha^2-2}{4\alpha}$ 看起来凭空出现。它在草稿纸上是怎么被找到的？

参考答案：草稿上的目标是找一个 $h>0$，使 $(\alpha-h)^2>2$。展开得 $\alpha^2-2\alpha h+h^2$，丢掉 $h^2$ 后只需 $\alpha^2-2\alpha h>2$，即 $h<\dfrac{\alpha^2-2}{2\alpha}$。取一个具体的小值，例如把分母放大一倍，得 $h=\dfrac{\alpha^2-2}{4\alpha}$；它既满足所需不等式，又便于在正式稿里算出显式的下界 $\alpha_1>\frac34\alpha$。正式证明把这个摸索过程藏起来，只保留结果。

4. 第一部分为什么必须单独验证 $\alpha_1>0$？如果不验证，后面的哪一步会失效？

参考答案：后面要用「$x^2<2<\alpha_1^2$ 推出 $x<\alpha_1$」，这一步需要 $\alpha_1>0$（开方保序要求两边非负）。若 $\alpha_1\le0$，即使平方大于 $2$ 也不能推出 $x<\alpha_1$，因为负数平方大并不代表它大。验证方式是 $h<\alpha/4$，故 $\alpha_1>\frac34\alpha>0$。

5. 在第二部分中，$\alpha\ge1$ 这个已知条件分别在哪里被用到？

参考答案：两处。第一处用来估计 $h$：$h=\dfrac{2-\alpha^2}{4\alpha}\le\dfrac{2-\alpha^2}{2}$，需要 $4\alpha\ge2$，即 $\alpha\ge1/2$，$\alpha\ge1$ 足够。第二处用来得到 $h<1$，从而 $h^2<h$，把 $(\alpha+h)^2$ 中的 $h^2$ 压成 $h$。若没有这个上界，就无法把二次项吸收，最后的估计 $\alpha_2^2<2$ 不成立。

与前后课的关系

第 1–2 讲建立公理与完备性，本讲把这些材料当作已知，专注训练「写」。下一讲开始正式定义数列收敛：$\varepsilon$-$N$ 语言会把这一讲反复强调的「每一行都要指出依据」变成固定格式。之后的单调收敛定理、柯西准则，都是在这套写法上继续叠加。

课程反复提醒：写证明需要时间，前几周写不顺是正常的。本讲的两份正式证明可以当作模板——先列结构，再填细节，最后检查每一行是否都有出处。

Two different things

The course has two goals, and this lecture is devoted to the second: how to turn "why it ought to be true" into a formal mathematical proof.

These are not the same activity. Lecture 2 argued that $\sqrt2$ exists by imagining the supremum slightly too large or too small and shifting it to create a contradiction. That is scratch-paper reasoning. A proof you hand in is written differently: every quantity (such as $h$) is defined up front, and the argument then follows a fixed structure. A formal proof can look like it comes from nowhere, but it reads far more clearly.

The lecture demonstrates both versions on the same example.

Timeline

Upper bounds, lower bounds, boundedness

Definitions

Let $S$ be an ordered set and $A\subseteq S$.

• $A$ is bounded above if there is $M\in S$ with $a\le M$ for all $a\in A$. Such an $M$ is an upper bound.
• $A$ is bounded below if there is $m\in S$ with $m\le a$ for all $a\in A$. Such an $m$ is a lower bound.
• $A$ is bounded if it is bounded both above and below.

A bound need not belong to the subset

This was the most frequently asked question on the discussion board: which set must $M$ live in?

The answer: $M$ only has to lie in the ambient set $S$; it need not belong to $A$.

Example: $A=(0,1)\subset\mathbb R$. Then $1$ is an upper bound and $0$ is a lower bound, yet neither belongs to $A$. In fact no element of $A$ is an upper bound for $A$: given $x\in A$, the point $(x+1)/2$ lies in $A$ and is larger. Bounds can only come from outside.

In general, suprema and infima may also lie outside $A$. They may belong to $A$, but that is not required.

A subtle example: which set is A a subset of?

One move inside the definition is easy to miss: all bounds must be taken from $S$, not from all of $\mathbb R$.

Take $S=(0,\infty)$ and $A=(0,1)$, viewed as a subset of $S$.

• $A$ is bounded above: $1$ is an upper bound, since $x<1$ for every $x\in A$ and $1\in S$.
• $A$ has no lower bound. Any lower bound $m$ must lie in $S=(0,\infty)$, so $m>0$. Take $x=\min\{m/2,\,1/2\}\in A$; then $x<m$, contradicting that $m$ is a lower bound.

The same $A$, viewed as a subset of $\mathbb R$, does have the lower bound $0$. The difference is not in $A$ but in which set supplies the bounds. Hence one must always specify the ambient set $S$ before discussing bounds.

The lecture also clarifies that $+\infty$ and $-\infty$ are not real numbers and cannot serve as bounds in $\mathbb R$.

Completeness: suprema and infima

Let $S$ be an ordered set and $A\subseteq S$ bounded above. An element $M\in S$ is the least upper bound of $A$, written $\sup A$, if

1. $M$ is an upper bound for $A$;
2. $M\le M_1$ for every upper bound $M_1$ of $A$.

Symmetrically, if $A$ is bounded below, the greatest lower bound $m=\inf A$ satisfies

1. $m$ is a lower bound;
2. $m_1\le m$ for every lower bound $m_1$.

Definition. An ordered set $S$ is complete if every subset bounded above has a least upper bound.

$\mathbb R$ is complete, so every bounded subset has both a supremum and an infimum, related by

$$\inf A=-\sup(-A), \qquad -A=\{-a:a\in A\}.$$

So proving the supremum property suffices; the infimum case follows.

Writing a formal proof: the case of $\sqrt2$

The theorem and the set

Theorem. There is $\alpha\in\mathbb R$ with $\alpha>0$ and $\alpha^2=2$.

Proof. Let

$$A=\{\,x\in\mathbb R : x>0,\ x^2<2\,\}.$$

The set is written out explicitly, including $x\in\mathbb R$ — a formal proof may not skip this.

Step 1: $A$ is nonempty. $1\in A$, since $1>0$ and $1^2=1<2$.

Step 2: $A$ is bounded above. Claim that $2$ is an upper bound. If $x\in A$ and $x\ge2$, then $x^2\ge4>2$, contradicting $x^2<2$. Hence $x<2\le2$.

Step 3: fix $\alpha$. Since $A$ is nonempty and bounded above and $\mathbb R$ is complete, set

$$\alpha=\sup A .$$

Step 4: $\alpha>0$. Since $1\in A$ and $\alpha$ is an upper bound, $1\le\alpha$, so in particular $\alpha>0$. Also $\alpha\le2$, because $2$ is an upper bound and $\alpha$ is the least one.

It remains to prove $\alpha^2=2$, in two parts.

Part one: $\alpha^2\le2$

Suppose not, i.e. $\alpha^2>2$. The goal is to build an upper bound smaller than $\alpha$, contradicting minimality.

Define

$$h=\frac{\alpha^2-2}{4\alpha}, \qquad \alpha_1=\alpha-h .$$

$h>0$: the numerator is positive because $\alpha^2>2$, and the denominator is positive.

$\alpha_1>0$: since

$$h=\frac{\alpha^2-2}{4\alpha}<\frac{\alpha^2}{4\alpha}=\frac\alpha4,$$

we get $\alpha_1=\alpha-h>\frac34\alpha>0$.

$\alpha_1^2>2$: expand and drop the positive term $h^2$:

$$\alpha_1^2=(\alpha-h)^2=\alpha^2-2\alpha h+h^2\ge\alpha^2-2\alpha h .$$

Substituting $h$,

$$2\alpha h=2\alpha\cdot\frac{\alpha^2-2}{4\alpha}=\frac{\alpha^2-2}{2},$$

so

$$\alpha_1^2\ge\alpha^2-\frac{\alpha^2-2}{2}=\frac{\alpha^2+2}{2}>2,$$

where the last inequality uses the hypothesis $\alpha^2>2$.

Now take any $x\in A$. Since $x>0$, $\alpha_1>0$, and $x^2<2<\alpha_1^2$, taking roots gives $x<\alpha_1$. So $\alpha_1$ is an upper bound for $A$, and $\alpha_1<\alpha$, contradicting that $\alpha$ is the least upper bound.

Hence $\alpha^2\le2$.

Part two: $\alpha^2\ge2$

Suppose not, i.e. $\alpha^2<2$. The goal is to find an element of $A$ larger than $\alpha$, contradicting that $\alpha$ is an upper bound.

Define

$$h=\frac{2-\alpha^2}{4\alpha}, \qquad \alpha_2=\alpha+h .$$

$0<h<\frac12$: numerator and denominator are positive, so $h>0$; and since $\alpha\ge1$,

$$h=\frac{2-\alpha^2}{4\alpha}<\frac{2}{4\alpha}\le\frac12<1 .$$

$\alpha_2^2<2$: expand,

$$\alpha_2^2=(\alpha+h)^2=\alpha^2+2\alpha h+h^2 .$$

Because $0<h<1$, we have $h^2<h$, hence

$$\alpha_2^2<\alpha^2+2\alpha h+h .$$

Substituting $2\alpha h=\dfrac{2-\alpha^2}{2}$ and using $h\le\dfrac{2-\alpha^2}{2}$ (which follows from $\alpha\ge1$, i.e. $\frac1{2\alpha}\le1$), we obtain

$$\alpha_2^2<\alpha^2+\frac{2-\alpha^2}{2}+\frac{2-\alpha^2}{2}=2 .$$

Also $\alpha_2>\alpha>0$, so $\alpha_2\in A$. But $\alpha_2>\alpha$, contradicting that $\alpha$ bounds $A$.

Hence $\alpha^2\ge2$.

Together, $\alpha^2=2$, completing the proof. $\square$

From scratch paper to formal proof

Compare Lecture 2 with the argument above; the difference is the order in which information appears:

In the formal version $h$ seems to fall from the sky. Its origin is the scratch calculation: to get $(x-h)^2>2$ it suffices that $2xh<x^2-2$. The formal write-up compresses that calculation into "define $h=\cdots$", sacrificing visible motivation for a gapless logical chain.

One practical rule from the lecture: prove only the properties you need. Establishing every property of the set for the sake of completeness makes the argument harder to follow.

A formal proof of the Archimedean property

Theorem. The natural numbers are not bounded above in $\mathbb R$.

Equivalently, for every real $x$ there is $n\in\mathbb N$ with $n>x$.

Proof (by contradiction). Suppose $\mathbb N$ is bounded above. By completeness, let

$$\alpha=\sup\mathbb N .$$

The contradiction will come from "$\alpha-1$ is also an upper bound".

Since $\alpha$ is an upper bound, $n\le\alpha$ for every $n\in\mathbb N$. Also $n+1\in\mathbb N$, so

$$n+1\le\alpha \implies n\le\alpha-1 \quad\text{for all } n\in\mathbb N .$$

Thus $\alpha-1$ is an upper bound for $\mathbb N$. But $\alpha-1<\alpha$, contradicting that $\alpha$ is the least upper bound. Therefore $\mathbb N$ is unbounded. $\square$

Four lines, each with a job: the first states the contradictory hypothesis, the second invokes completeness, the third uses closure of $\mathbb N$ under $+1$, the fourth uses minimality. When writing a formal proof, one can ask of every line: which known fact does this use?

Sequences: placing $\sqrt2$ on the real line

The theorem guarantees that $\sqrt2$ exists as a number. To locate it on the line, one good device is its decimal truncations:

$$1,\quad 1.4,\quad 1.41,\quad 1.414,\quad 1.4142,\ \dots$$

Each is smaller than the next, and $\sqrt2$ is their limit.

Definition. A real sequence is a function $a:\mathbb N\to\mathbb R$, usually written $(a_n)_{n\ge1}$ with $a_n=a(n)$.

Writing the index as a subscript rather than an argument emphasises that a sequence is a function.

The list above is increasing, $a_n<a_{n+1}$, and $\sqrt2=\lim_{n\to\infty}a_n$. But "limit" is still only intuitive; the $\varepsilon$ definition arrives in the next lecture.

What this lecture establishes

• Upper and lower bounds must come from the ambient set $S$ and need not belong to $A$.
• $(0,1)$ has an upper bound but no lower bound inside $S=(0,\infty)$; inside $\mathbb R$ it has the lower bound $0$.
• Complete = every subset bounded above has a supremum; $\inf A=-\sup(-A)$.
• A formal proof proceeds: state the theorem → build the set → verify nonempty and bounded → take the supremum → argue in two parts by contradiction.
• A quantity such as $h$ that appears "from nowhere" comes from scratch work; the formal draft suppresses the motivation.
• The Archimedean property follows from completeness: $\sup\mathbb N$ would make $\alpha-1$ an upper bound too.
• A sequence is a function $\mathbb N\to\mathbb R$; $\sqrt2$ is the limit of its truncations.

Comprehension check

1. Why need an upper bound $M$ not belong to $A$? Give an example where no element of $A$ is an upper bound.

Reference answer: The definition only requires $M\in S$ and $a\le M$ for all $a\in A$. Take $A=(0,1)\subset\mathbb R$; then $1$ is an upper bound but $1\notin A$. Moreover no element of $A$ bounds $A$: for any $x\in A$, let $y=(x+1)/2$, so $x<y<1$ and $y\in A$ with $y>x$. Bounds must come from outside.

2. Let $S=(0,\infty)$ and $A=(0,1)$. Why does $A$ have no lower bound as a subset of $S$? How does this differ from viewing $A$ as a subset of $\mathbb R$?

Reference answer: A lower bound must lie in $S$, so any candidate $m$ satisfies $m>0$. Take $x=m/2$ (or $x=1/2$ if $m/2\ge1$); then $x\in A$ and $x<m$, so $m$ fails. Since $m$ was arbitrary, there is no lower bound. Viewed inside $\mathbb R$, $0\in\mathbb R$ and $0\le x$ for all $x\in A$, so $0$ is a lower bound. The difference lies in which set is allowed to supply bounds, not in $A$.

3. In the formal proof, $h=(\alpha^2-2)/(4\alpha)$ appears from nowhere. How was it found on scratch paper?

Reference answer: The scratch goal is $h>0$ with $(\alpha-h)^2>2$. Expanding gives $\alpha^2-2\alpha h+h^2$; dropping $h^2$, it suffices that $\alpha^2-2\alpha h>2$, i.e. $h<(\alpha^2-2)/(2\alpha)$. Choosing a concrete small value — for instance doubling the denominator — gives $h=(\alpha^2-2)/(4\alpha)$, which satisfies the needed inequality and also yields the explicit bound $\alpha_1>\frac34\alpha$. The formal proof hides the search and keeps only the result.

4. Why must part one verify $\alpha_1>0$ separately? Which later step would fail without it?

Reference answer: The later step "from $x^2<2<\alpha_1^2$ conclude $x<\alpha_1$" requires $\alpha_1>0$, since taking square roots preserves order only for nonnegative quantities. If $\alpha_1\le0$, a large square would not imply a large value. The verification is $h<\alpha/4$, hence $\alpha_1>\frac34\alpha>0$.

5. In part two, where exactly is the known fact $\alpha\ge1$ used?

Reference answer: Twice. First, to bound $h$: $h=\dfrac{2-\alpha^2}{4\alpha}\le\dfrac{2-\alpha^2}{2}$, which needs $4\alpha\ge2$, i.e. $\alpha\ge1/2$; $\alpha\ge1$ is more than enough. Second, to obtain $h<1$ and hence $h^2<h$, which absorbs the quadratic term in $(\alpha+h)^2$. Without that bound the estimate $\alpha_2^2<2$ fails.

Relation to neighbouring lectures

Lectures 1–2 built the axioms and completeness; this lecture treats them as known and concentrates on writing. The next lecture defines convergence with $\varepsilon$-$N$ language, turning the insistence that "every line names its justification" into a fixed format. The monotone convergence theorem and the Cauchy criterion then build on that format.

The course repeats that proof writing takes time and that early struggles are normal. The two formal proofs here can serve as templates: list the structure, fill in details, then check that every line has a source.