Extremum

Def'n. Boundedness

Bound A set \(S\subset\mathbb{R}\) is

• bounded above \(\exists M\in\mathbb{R}. \forall s \in S. s\leq M\).
• bounded below \(\exists L\in\mathbb{R}. \forall s \in S. s\geq L\).

Def'n. Extremum

Supremum \(\sup(S)\in\mathbb{R}\) is a upper bound and \(\forall v. v\geq \sup(S)\land v\) is upper bound.
Infimum \(\inf(S)\in\mathbb{R}\) is a lower bound and \(\forall v. v\leq \inf(S)\land v\) is lower bound.

Proposition \(u\) is a supremum of \(S\) IFF \(u\) is a upper bound and \(\forall \epsilon > 0. \exists s_\epsilon \in S. u-\epsilon < s_\epsilon\)

Least upper bound principle

\(\forall S\neq \emptyset, S\) bounded above. \(S\) has a supremum.

proof. Since bounded above, take \(M\in\mathbb{R}\) where \(\forall s \in S. M\geq s\).
Pick \(s'=s_0\),

Find lowest \(a_0\in \{s_0, ..., m_0 + 1\}\) that is an upper bound for \(S\).

Induction \(y_n = \sum_0^n a_i / 10^i\) where \(a_i=\min\{a \in \{0, 1, ..., 9\}\mid y_i \geq S\}\). Take \(x_n \in S\) s.t. \(a_0.a_1...a_n - 10^{-n}\leq x_n \leq y_n\)

WTS \(L = a_0.a_1... = sup(S)\)

Upper bound property: start with any \(s_0.s_1,... \in S\) either
1. \(\forall i. s_i = a_i\Rightarrow S = L\)
2. \(\exists k\) be the first different digit and \(s_k > a_k\) Construct \(y_k^* = a_0.a_1...a_{k-1}s_k\). then \(L < y_k^*\leq s_0.s_1\), while by the ordering this cannot happen.

Subsequence Property: For each \(\epsilon\), you can pick \(n > 0\) s.t. \(L-\epsilon \leq L - 10^{-n} \leq x_n \leq y_n < L\)

Thrm 1. Uniqueness of supremum

proof Assume \(\exists u,v\) be two supremums, \(v < u\). Take \(\epsilon = u -v, \forall \epsilon > 0. \exists s_\epsilon \in S \Rightarrow u - (u-v) < s_\epsilon \Rightarrow v < s_\epsilon\).

Thrm 2.

For all bounded above set \(A\) and \(c\geq 0\). \(\sup(cA) = c \sup(A)\)

proof. Let \(M = \sup(A)\).
Upper bound property: \(\forall s. s\leq M \Rightarrow cs \leq cM\)
Subsequence property: Let \(\epsilon \geq 0\), take \(s_{\epsilon/c}\), then \(cs_{\epsilon/c}\geq u - \epsilon\)