Hence, $\sup(A +B) − \sup A$ is an upper bound for any $\color{magenta}{b}$. By the definition of supremum, the previous inequality means: $$\color{magenta}{\sup B} \leq \sup(A + B) − \sup A …

若 \ [f\left ( x \right)\] 是定义在 \ [D\] 上的 有界函数. 记 上确界 为 \ [\mathop {\sup }\limits_ {x \in D} f\left ( x \right)\] ， 下确界 为 \ [\mathop {\inf }\limits_ {x \in D} f\left ( x \right)\] ，则. \ [i)\]\ [ - …

Jul 9, 2019 · Since both A and B are non-empty bounded subsets of R if follows that both A and B have a finite Supremum. Thus sup (A) and sup (B) exist. Obviously sup (A) ≥ a ∀a ∈ A and …

First, we need to show that $\sup(A) + \sup(B)$ is an upper bound for the set $A + B$. Indeed, if $z\in A + B$, then there exists $a\in A$ and $b\in B$ such that $z = a + b$. But by definition of …

Oct 3, 2011 · 令A+B={a+b|a∈A,b∈B} 求证：sup(A+B)=sup A + sup B (sup是上确界)（1）对任意a∈A，有a≤supA ; 对任意b∈B，有b≤supB， 则对任意a+b∈A+B，有a+b≤supA+supB， …

Aug 11, 2020 · sup (X)是取上限函数，inf (X) 是取下限函数。 sup是supremum的简写，意思是：上确界，最小上界。 inf是infimum的简写，意思是：下确界，最大下界。 一、上确界： 上 …

Nov 25, 2023 · 如果 A 和 B 是 R 上的非空子集，我们用 sup (A)，sup (B)，sup (A+B) 分别表示集合 A，集合 B，集合 A+B 的最小上界，本文要证明的就是这三个最小上界的关系。 也就是说 …

Proposition 2.9. Suppose that A, B are nonempty sets of real numbers such that x ≤ y for all x ∈ A and y ∈ B. Then supA ≤ inf B. Proof. Fix y ∈ B. Since x ≤ y for all x ∈ A, it follows that y is an …

supA=a,supB=b,由 上确界 定义,对任给的ε＞0,存在x∈A,y∈B,使得a-ε＜x≤a,b-ε＜y≤b,所以对2ε＞0存在x+y∈A＋B,使得a+b-2ε＜x+y≤a+b,这就是说sup (A＋B)=a+b. 数学分析证明sup …

Oct 14, 2024 · Prove that sup AB = (sup A)(sup B), where AB is the product of the sets and A, B ⊆ R +. Since A, B are bounded above, sup A and sup B exist. Let α = sup A and β = sup B. …