---- > [!definition] Definition. ([[supremum]]) > The **supremum** of a [[poset]] $P$, denoted $\sup P$ is the least element in $P$ that is greater than or equal to every element of $P$, if such an element exists. ^definition > [!equivalence] Equivalent definition for $\mathbb{R}$ > If $S \subset \mathbb{R}$ is nonempty, then $\alpha=\sup S$ if and only if $(1)$ $\alpha \geq s$ for all $s \in S$ and $(2)$ for all $\varepsilon>0$, there exists $s \in S$ such that $\alpha-\varepsilon<s \leq \alpha$. ^equivalence > [!proof]- Proof. > Suppose $\alpha=\sup S$. By definition $\alpha$ satisfies $(1)$. For $(2)$, suppose there exists $\varepsilon>0$ such that $\alpha-\varepsilon \geq s$ for all $s \in S$. Then clearly $\alpha$ must not be a *least* upper bound (for $\alpha-\varepsilon$ is also an upper bound): $\alpha \neq \sup S$. > > > > Conversely suppose $\alpha \in \mathbb{R}$ is such that $\alpha \geq s$ for all $s \in S$ and for all $\varepsilon>0$, there exists $s \in S$ such that $\alpha-\varepsilon < s \leq \alpha$. Suppose $a$ is another upper bound for $S$: $a \geq s$ for all $s \in S$. If $a<\alpha$, set $\varepsilon=\alpha-a$. Then there exists $s \in S$ such that $\alpha-(\alpha-a)<s \leq \alpha$ hence $a<s$ for some $s \in S$ and so $a$ must not be an upper bound. > [!basicproperties] > - If $S \subset \mathbb{R}$ is nonempty and bounded above (so $\sup S \in \mathbb{R}$) then there always exists a [[monotonic map|monotone]] sequence $s_{1} \leq s_{2} \leq \dots$ in $S$ [[sequence|converging]] to $\sup S$. > > > [!proof]- Proof. > > Put $\alpha := \sup S \in \mathbb{R}$. If we just need convergence, we can immediately define $(s_{n})$ using the equivalence above: for $n \in \mathbb{N}$, let $s_{n}$ be such that $\alpha-\frac{1}{n}<s_{n} \leq \alpha$. Fix $\varepsilon>0$. Choose $N>1/\varepsilon$. Then for all $n \geq N$, $|\alpha-s_{n}| \leq |\alpha-\left( \alpha- \frac{1}{n} \right)|=\frac{1}{n} \leq \frac{1}{N} < \varepsilon.$ > We can obtain from this construction a *monotone* sequence $(\hat{s}_{n})$ by taking $\hat{s}_{n}:=\max \{ s_{1},\dots,s_{n} \}$. Then $|\alpha-\hat{s}_{n}| \leq \frac{1}{n}$ for all $n$ and so $(\hat{s}_{n}) \to \alpha$. > ^proof > > > > > > - For any $R \in \mathbb{R}_{>0}$ one has $R= \sup_{0 < c < 1}c R$. > > > [!proof]- Proof. > > We appeal to the equivalence. Clearly $R$ is an upper bound for the set $\{ cR:0<c<1 \}$. Fix $\varepsilon>0$. We just have to find $c \in (0,1)$ such that $R-\varepsilon< cR \leq R$. This is intuitive; a valid explicit choice would be $c:= \begin{cases} > 1-\frac{\varepsilon}{2R} & 0 < \varepsilon \leq R \\ > \frac{1}{59} & \varepsilon \geq R. > \end{cases}$ > ---- #### ---- #### References > [!backlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM [[]] > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ``` > [!frontlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM outgoing([[]]) > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ```