 Georg Ferdinand Ludwig Philipp Cantor (/ˈkæntɔːr/ KAN-tor, German: [ˈɡeːɔʁk ˈfɛʁdinant ˈluːtvɪç ˈfiːlɪp ˈkantɔʁ]; March 3 [O.S. February 19] 1845 – January 6, 1918) was a German mathematician. He created set theory, which has become a fundamental theory in mathematics. Cantor established the importance of one-to-one correspondence between the members of two sets, defined infinite and well-ordered sets, and proved that the real numbers are more numerous than the natural numbers. In fact, Cantor’s method of proof of this theorem implies the existence of an infinity of infinities. He defined the cardinal and ordinal numbers and their arithmetic. Cantor’s work is of great philosophical interest, a fact he was well aware of.

# Understanding these basic definition and statement~

• Mathematical induction is a method of proof used to establish that a given statement is true for all natural numbers. Let $S(n)$ be a statement about the positive integer $n .$ If
1. $S(1)$ is true and
2. for all $k \geq 1$, the truth of $S(k)$ implies the truth of $S(k+1)$,
then $S(n)$ is true for all $n \geq 1$ Verifying $S(1)$ is true is called the basis step. The assumption that $S(k)$ is true for some $k \geq 1$ is called the induction hypothesis. Using the induction hypothesis to prove $S(k+1)$ is true is called the induction step. There are variants of mathematical induction used in practice, for example if one wants to prove a statement not for all natural numbers but only for all numbers greater than or equal to a certain number $b$, then
3. Show $S(b)$ is true.
4. Assume $S(m)$ is true for $m \geq b$ and show that truth of $S(m)$ implies the truth of $S(m+1)$
Another generalization, called strong induction, says that in the inductive step we may assume not only the statement holds for $S(k+1)$ but also that it is true for $S(m)$ for all $m \leq k+1$. In strong induction it is not necessary to list the basis step, it is clearly true that the statement holds for all previous cases. The inductive step of a strong induction in this case corresponds to the basis step in ordinary induction.
• Let $A$ be a nonempty subset of $\mathbb{R}$. The number $b$ is called an upper bound for $A$ if for all $x \in A$, we have $x \leq b$. A number $b$ is called a least upper bound of $A$ if, first, $b$ is an upper bound for $A$ and, second, $b$ is less than or equal to every upper bound for $A$. The supremum of $A$ (also called least upper bound of $A$ ) is denoted by $\sup (A), \sup A$ or $\operatorname{lub}(A)$. If $A \subset \mathbb{R}$ is not bounded above, we say that $\sup A$ is infinite and write $\sup A=+\infty$.
• A lower bound for a set $A \subset \mathbb{R}$ is a number $b$ such that $b \leq x$ for all $x \in A$. Also $b$ is called a greatest lower bound if and only if it is a lower bound and for any lower bound $c$ of $A$, $c \leq b .$ The infimum of $A$ (also called greatest lower bound of $A$ ) is denoted by inf $(A), \inf A$ or $g l b(A)$. If $A \subset \mathbb{R}$ is not bounded below, we set $\inf A=-\infty$.
• Well-Ordering Property: If $A$ is a nonempty subset of $\mathbb{N}$, then there is a smallest element in $A$, i.e., there is an $a \in A$ such that $a \leq x$ for every $x \in A$.
• Archimedean Property: If $x \in \mathbb{Q}$, then there is an integer $n$ with $x<n .$
• For each $n \in \mathbb{N}$, let $I_{n}$ be a nonempty closed interval in $\mathbb{R} .$ The family $\left{I_{n}: n \in \mathbb{N}\right}$ is called a nest of intervals if the following conditions hold:
• $I_{n+1} \subset I_{n}$ for all $n \in \mathbb{N}$
• For each $\varepsilon>0$, there is some $n \in \mathbb{N}$ such that $\left|I_{n}\right|<\varepsilon$.

# Try the following problem to test yourself！

Problem 1.

Prove that $\sqrt{2}$ is not rational.

Proof .

Assume not. Let $r \in \mathbb{Q}$ such that $r=\sqrt{2}$ or $r^{2}=2 .$ Without loss of generality, we may assume $r \geq 0$. And since $r^{2}=2$, we have $r>0$. Since $r$ is rational, there exist two natural numbers $n \geq 1$ and $m \geq 1$ such that
$$r=\frac{n}{m} .$$
Moreover one may assume that $n$ and $m$ are relatively prime, i.e., the only common divisor is $1 .$ Since $r^{2}=2$, we get
$$\left(\frac{n}{m}\right)^{2}=\frac{n^{2}}{m^{2}}=2$$
which implies $2 m^{2}=n^{2}$. Therefore, $n^{2}$ is even, so it is a multiple of $2 .$ Assume that $n$ is not even, then $n=2 k+1$ for some natural number $k$. Hence $n^{2}=4 k^{2}+4 k+1=2\left(2 k^{2}+2 k\right)+1$. In other words, if $n$ is not even, then $n^{2}$ will not be even. Therefore, $n$ is also even. Set $n=2 k$ for some natural number $k$. Then $n^{2}=4 k^{2}$ and since $n^{2}=2 m^{2}$, we deduce that $m^{2}=2 k^{2}$. The same previous argument will imply that $m$ is also even. So both $n$ and $m$ are even so both are multiples of 2 . This is a contradiction with our assumption that both are relatively prime. Therefore, such a rational number $r$ does not exist which completes the proof of our statement.

Problem 2.

Show that two real numbers $x$ and $y$ are equal if and only if $\forall \varepsilon>0$ it follows that $|x-y|<\varepsilon$

Proof .

This is an if and only if statement, and we need to prove the implications in both directions. $(\Rightarrow):$ If $x=y$, then $|x-y|=0$ and thus $|x-y|<\varepsilon$ no matter what $\varepsilon>0$ is chosen. $:(\Leftarrow)$ We give a proof by contradiction. Assume $x \neq y$, then $\varepsilon_{0}=|x-y|>0$. However, the statements
$$|x-y|=\varepsilon_{0} \text { and }|x-y|<\varepsilon_{0}$$
cannot be both true, our assumption is wrong, thus $x=y$.

Problem 3.

Use the induction argument to prove that
$$1^{2}+2^{2}+\cdots+n^{2}=\frac{2 n^{3}+3 n^{2}+n}{6}$$
for all natural numbers $n \geq 1$.

Proof .

Since
$$1^{2}=\frac{2+3+1}{6},$$
the above identity holds in the case $n=1$. Assume that the identity holds for $n$ and let us prove that it also holds for $n+1$. Since
$$1^{2}+2^{2}+\cdots+(n+1)^{2}=1^{2}+2^{2}+\cdots+n^{2}+(n+1)^{2},$$
the induction assumption implies
$$1^{2}+2^{2}+\cdots+(n+1)^{2}=\frac{2 n^{3}+3 n^{2}+n}{6}+(n+1)^{2} .$$
Algebraic manipulations imply
\begin{aligned} \frac{2 n^{3}+3 n^{2}+n}{6}+(n+1)^{2} &=\frac{2 n^{3}+3 n^{2}+n+6(n+1)^{2}}{6} \ &=\frac{2 n^{3}+3 n^{2}+n+3(n+1)^{2}+3(n+1)^{2}}{6} \ &=\frac{2 n^{3}+6 n^{2}+7 n+3+3(n+1)^{2}}{6} . \end{aligned}
On the other hand, we have
\begin{aligned} \frac{2(n+1)^{3}+3(n+1)^{2}+(n+1)}{6} &=\frac{2 n^{3}+6 n^{2}+6 n+2+3(n+1)^{2}+(n+1)}{6} \ &=\frac{2 n^{3}+6 n^{2}+7 n+3+3(n+1)^{2}}{6} \end{aligned}
which implies
$$1^{2}+2^{2}+\cdots+(n+1)^{2}=\frac{2(n+1)^{3}+3(n+1)^{2}+(n+1)}{6} .$$
So our identity is also valid for $n+1$. By induction we clearly showed that the identity is valid for any natural number $n \geq 1$.