Linear Continuum - Non-examples

Non-examples

• The set of rational numbers is not a linear continuum. Even though property b) is satisfied, property a) is not. Consider the subset:

A = { x | x < √2 }

of the set of rational numbers. Even though this set is bounded above by any rational number greater than √2 (for instance 3), it has no least upper bound in the rational numbers.

• The set of non-negative integers with its usual order is not a linear continuum. Property a) is satisfied (let A be a subset of the set of non-negative integers that is bounded above. Then A is finite so that it has a maximum. This maximum is the desired least upper bound of A). On the other hand, property b) is not. Indeed, 5 is a non-negative integer and so is 6, but there exists no non-negative integer that lies strictly between them.

• The ordered set A of nonzero real numbers:

A = (-∞, 0) ∪ (0, +∞)

is not a linear continuum. Property b) is trivially satisfied. However, if B is the set of negative real numbers:

B = (-∞, 0)

then B is a subset of A which is bounded above (by any element of A greater than 0; for instance 1), but has no least upper bound in B. Notice that 0 is not a bound for B since 0 is not an element of A.

• Let Z_- denote the set of negative integers and let A = (0,5) ∪ (5,+∞). Let:

S = Z_- ∪ A

Then S satisfies neither property a) nor property b). The proof is similar to the previous examples.