Question

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:

  • make sure you are addressing the technical definitions of upper bound and least upper bound, not just your conceptions of these terms.
  • Please note that the set A does not contain the number 0, and we are talking about upper bounds and the (potential for a) least upper bound for B in A. So your argument should not have anything to do with the number 0. Just pretend that number doesn't exist.
  • For the least upper bound part, you are proving a "for every" statement. You need to provide a piece of evidence for each and every element in A that demonstrates that that element cannot be the least upper bound of B in A.
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Answer #1

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