[Calculus] 1. Dedekind Cut

2023-01-04

Definition 1.1 Let \(S\sube\R\) and \(r\in\R\). We say that:

\(r\) is an upper(lower) bound of \(S\) if \(\forall s\in S[r\ge(\le)s]\)
\(r\) is the greatest(least) element of \(S\) if:
1. \(r\) is an upper(lower) bound of \(s\)
2. \(r\in S\)
\(r\) is the least upper(greatest lower) bound of \(S\) if \(r=\min\{u\in\R|u\text{ is an upper(lower) bound of }S\}\). In this case, \(r\) is also denoted as \(\sup S(\inf S)\).

Definition 1.2 (Dedekind cut) Let \(A,B\sube\R\). We say that \((A,B)\) is a Dedekind cut of \(\R\) if:

\(A\ne B\ne\varnothing\)
\(A\cup B=\R\)
\(\forall a\in A,b\in B[a < b]\)

Proposition 1.3 (Dedekind's gapless property) If \((A,B)\) is a Dedekind cut of \(\R\), then exactly one of the following happens:

\(\max A\) exists, but \(\min B\) doesn't
\(\min B\) exists, but \(\max A\) doesn't

Proof (齐神说要用区间套定理和阿基米德性质，但是我没用上。麻烦高人帮忙看看这个证明有没有问题)

Let \(P_1\): '\(\max A\) exists', \(P_2\): '\(\min B\) exists'. There may be four cases:

\(P_1 \land P_2\)
\(P_1 \land\lnot P_2\)
\(\lnot P_1 \land P_2\)
\(\lnot P_1 \land\lnot P_2\)

The proposition is proved if case 1 and 4 can not be true since case 2 and 3 excludes mutually.

Case 1. \(\max A\) exists, and \(\min B\) exists too. Write them as \(x,y\) respectively.

Then \(\forall a\in A, b\in B[a \le x < y \le b]\). Consider \(z=\frac{x+y}{2}\), then \(x < z < y \Rarr z\notin A \land z\notin B\), a contradiction.

Case 4. None of \(\max A\) and \(\min B\) exists.

Since every \(b\in B\) is an upper bound of \(A\), by Weierstrass' theorem, \(\sup A\) exists.

(a) \(\sup A \in A\).

Then \(\sup A = \max A\). Otherwise there exists an \(a_0 \in A\) such that \(\sup A < a_0\), which is a contradiction.

However, by hypothesis, A has no maximal.

(b) \(\sup A \in B\).

Then \(\sup A = \min B\), by the definition of sup.

However, by hypothesis again, B has no minimal.

In summary, \(\sup A \notin A \land \sup A \notin B\), i.e., \(\sup A \notin \R\). This is also a contradiction. □

Theorem 1.4 (Weierstrass) Let \(S(\ne\varnothing)\sube\R\). If S has an upper bound, then \(\sup S\) exists.

Proof Let \(B:=\{b\in\R|b\text{ is an upper(lower) bound of }S\}\), and \(A:=\R\backslash B\).

Since:

\(S\ne\varnothing\Rarr A\ne\varnothing\), and '\(S\) has an upper bound' \(\rArr B\ne\varnothing\).
\(A\cup B\) by definition.
\(\forall a\in A,b\in B[a < b]\). Were this false, \(a\ge b\Rarr a\text{ is an upper bound of }S\Rarr a\in B\), a contradiction.

Therefore, \((A,B)\) is a Dedekind cut.

Suppose \(\max A\) exists, denoted as \(a_0\). Then \(a_0\in A\Rarr a_0\notin B\Rarr a_0\text{ is not an upper bound of }S \Rarr \exists s_0 \in S [a_0 < s_0]\), a contradiction. i.e..

Thus, \(\max A\) doesn't exists.

By Dedekind's gapless property, \(\min B\) must exist, i.e., \(\sup S\) exists. □

Proposition 1.5 (Archimedean property) \(\forall r\in\R \left[r > 0 \Rarr \exists n\in\N\left[\frac{1}{n} < r\right]\right]\)

Proof It is equvalent to prove that \(\N\) has no upper bound.

Suppose \(\N\) has an upper bound. By Weierstrass's theorem, \(\sup\N\) exists, denoted as \(s_0\).

There must exist an \(N\in\N\), such that \(s_0 - 1 < N\). Were this false, i.e, \(\forall n \in\N(s_0 - 1 > n)\). Then we have \(s_0 - 1\) is an upper bound of \(\N\), and hence \(s_0 - 1 \ge s_0\) since \(s_0\) is the least upper bound. This is a contradiction.

Then we obtain that \(N + 1\) is an upper bound of \(\N\), since it is greater than \(s_0\).

But, apparently, \(N + 1 < N + 2\), which means \(N + 1\) cannot be the upper bound of \(\N\).

Thus, \(\N\) has no upper bounds, i.e., Archimedean property is proved. □