About definition of 'Bounded above' and 'Least Upper Bound Property'

[jwqwerty]

The definition of 'Bounded above' states that:

If E⊂S and S is an ordered set, there exists a β∈S such that x≤β for all x∈E. Then E is bounded above.

The 'Least Upper Bound Property' states that:
If E⊂S, S be an ordered set, E≠Φ (empty set) and E is bounded above, then supE (Least Upper Bound) exists in S.

My question is that why doesn't the definition of 'Bounded Above' include E≠Φ? Is there a problem when E=Φ? If not, then why does it matter when E=Φ for the 'Least Upper Bound Property'?

 

[tiny-tim]

hi jwqwerty!

The 'Least Upper Bound Property' states that:
If E⊂S, S be an ordered set, E≠Φ (empty set) and E is bounded above, then supE (Least Upper Bound) exists in S.

Because Φ has no least upper bound (unless S is finite).

 

[micromass]

If we're working in [itex]\mathbb{R}[/itex], then we sometimes use the convention

[tex]\sup \emptyset =-\infty[/tex]

But we should be careful because [itex]-\infty[/itex] is NOT a real number. The above is not an equality of real numbers, but merely a notation.

But don't ever use that notation in class unless your instructor uses it.

 

[mathman]

The definition of 'Bounded above' states that:

If E⊂S and S is an ordered set, there exists a β∈S such that x≤β for all x∈E. Then E is bounded above.

The 'Least Upper Bound Property' states that:
If E⊂S, S be an ordered set, E≠Φ (empty set) and E is bounded above, then supE (Least Upper Bound) exists in S.

My question is that why doesn't the definition of 'Bounded Above' include E≠Φ? Is there a problem when E=Φ? If not, then why does it matter when E=Φ for the 'Least Upper Bound Property'?

There is something missing in the statement of the existence of least upper bound. Example: Let S = set of all rational numbers and let E = set of all rational numbers < √2, then E is bounded, but supE is not in S.