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