# 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 supB exists and supB=supA.

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

2020-12-01
Given that B ⊂ A. Since sup (A) exist therefore A has an upper bound sya a then a is also an upper bound of B. Therefore sup (B) exist and On the order hand for every x ⊂ A there exist y ⊂ B such that x<=y. Therefore x<=y<=sup(B). Hence sup(A)<=sup(B) Therefore from (1) and (2) sup(A)=sup(B).

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