Let S be an ordered set and A is a nonempty subset such that sup A exists. Suppose there is a B⊂A such that whenever x∈A there is a y∈B such that x≤y. Show that ⊃B exists and ⊃B=⊃A.

Given that B⊂A. Since ⊃(A) exist therefore A has an upper bound sya a then a is also an upper bound of B. Therefore ⊃(B) exist and On the order hand for every x⊂A there exist y⊂B such that x≤y. Therefore x≤y≤⊃(B). Hence ⊃(A)≤⊃(B) Therefore from (1) and (2) ⊃(A)=⊃(B).

Answered: "Let S be an ordered set and A..."

College
Calculus and Analysis
Differential Calculus
Analyzing functions

Lipossig

Answered question

2020-11-30

Let S be an ordered set and A is a nonempty subset such that sup A exists. Suppose there is a

B \subset A

such that whenever

x \in A

there is a

y \in B

such that

x \leq y

. Show that

\supset B

exists and

\supset B =\supset A

Answer & Explanation

timbalemX

Skilled2020-12-01Added 108 answers

Given that $B \subset A$ . Since $\supset (A)$ exist therefore $A$ has an upper bound sya a then a is also an upper bound of $B$ . Therefore $\supset (B)$ exist and On the order hand for every $x \subset A$ there exist $y \subset B$ such that $x \leq y$ . Therefore $x \leq y \leq\supset (B)$ . Hence $\supset (A) \leq\supset (B)$ Therefore from (1) and (2) $\supset (A) =\supset (B)$ .

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