정수에서 유리수로 갈때 정수 ring에 주어진 곱셈에 대한 inverse를 추가하는거로부터 field of quotient(Gallian만 이러는 것 같은.. 나머지는 field of fraction이라고 하는 것 같은데 모르겠다)가 생각되는 것 처럼 Grothendieck group이란게 있더라. 자연수에서 덧셈의 inverse가 추가돼서 abelian group인 정수가 생기는거가 대표적인 예라고.. Commutative monoid마다 대응되는 abelian group이 있는데 자연수의 경우에 그게 정수.
Definition of the Cantor set: see https://en.wikipedia.org/wiki/Cantor_set
Theorem. The Cantor set is uncountable.
proof) Let $\mathcal{C}$ be the Cantor set. Let $J_0=\mathcal{C_0}$. In the process to get the Cantor set, $J_0$ is divided into two line segments($[0,\frac{1}{3}],[\frac{2}{3},1]$). Take one of the two. Let the chosen segment be $J_1$. Of course, $J_1$ also can be divided into two line segments. Again, take one of the two. Let the chosen segment be $J_2$. Repeat this process. Then we get a nested interval $(J_n)_{n=0}^{\infty}$ such that $\bigcap_{n=0}^{\infty}J_n=c$ for some $c\in\mathcal{C}$. Clearly, every element in $\mathcal{C}$ can be represented as in that way by the definition. Now, consider the sequence such that following:
$s_0=0$, $s_n=\begin{cases}
0 \text{ if } J_n=\frac{1}{3}J_{n-1} \\
1 \text{ if } J_n=\frac{1}{3}+\frac{1}{3}J_{n-1}
\end{cases}$.
Then for any $c\in\mathcal{C}$, there exists a nested inverval and for the nested interval, there exists a sequence such that above. And it is easy to know that for distinct $(a_n)$ and $(b_n)$, there exists distinct nested interval $(A_n)_{n=0}^{\infty}$ and $(B_n)_{n=0}^{\infty}$ such that $\bigcap_{n=0}^{\infty}A_n=c_1\in\mathcal{C}$, $\bigcap_{n=0}^{\infty}B_n=c_2\in\mathcal{C}$, and $c_1\neq c_2$. These imply that $\left | \mathcal{C} \right |=\left | 2^{\mathbb{N}} \right |$. $\square $
Not a rigorous proof. 맞으려나 모르겠다.
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