Recall that (a,b)⊆R means an open interval on the real number line: (a,b)={x∈R|a<x<b}. Let ≤ be...

Recall that (a,b)⊆R means an open interval on the real number line:

(a,b)={x∈R|a<x<b}.

Let ≤ be the usual “less than or equal to” total order on the set

A=(−2,0)∪(0,2)

Consider the subset

B={−1/n | n∈N, n≥1}⊆A.

Determine an upper bound for B in A. Then formally prove that B has no least upper bound in A by arguing that every element of A fails the criteria in the definition of least upper bound.

Note:

least upper bound is an upper bound for B⊆A that is less than every other upper bound

Expert Solution

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