The empty set supremum proof

1. Aug 4, 2008

evagelos

Does the empty set have a supremum ( least upper bound)? if yes, can anybody give me a proof please? if no, again a proof please?

2. Aug 4, 2008

Focus

By convention it is $$-\infty$$. No proof :)

3. Aug 4, 2008

peos69

IS that new mathematics??

4. Aug 4, 2008

mathman

Since the set is empty, it really has no bounds. Therefore one can prescribe bounds by convention.

5. Aug 4, 2008

peos69

Sure when we cannot prove something we use the convention stuff

6. Aug 4, 2008

peos69

The empty set is bounded

7. Aug 4, 2008

peos69

The question is asking for the least upper bound not just for bounds

8. Aug 4, 2008

Focus

No

If you want to attack mathematics please publish your papers and stop posting on these forums unless you have something useful or relevant to say. Its defined this way, if you don't like it, write an article why and publish it.

Please would you tell us what the bound is?

If a subset of R is bounded then a supremum exists by the completeness of R. There is no bound and a supremum does not exist for the empty set.

9. Aug 4, 2008

peos69

YOU want to go and sleep or shall we Curry on on this thread???

10. Aug 4, 2008

peos69

If a NON EMPTY subset of the real Nos is bounded from above then it has a least upper bound
YOU FORGOT the NON EMPTY part,another fatal mistake.

11. Aug 6, 2008

LukeD

Focus: The definition of a bound (at least the one I've been taught) is that M, a real number, bounds S, a subset of the real numbers, if for all x in S, |x| <= M (and you can also define upper bound and lower bound. Clearly, a set that is bounded has both upper and lower bounds)
Since the empty set has no elements, the statement "for all x in the empty set, |x| <= M" is vacuously true no matter what M is. Therefore, every real number is a bound of the empty set.

Anyway, the completeness axiom only says that non-empty subsets of R with upper bounds have least upper bounds. It doesn't say anything about the empty set, and it's easy to prove that it does not have a least upper bound.

Proof:
Assume for sake of contradiction that the empty set has a least upper bound, we'll call it u. u-1 also bounds the empty set (since every real number bounds the empty set), so it is an upper bound. However, u-1 < u, which is the least upper bound. This is a contradiction, and therefore, the empty set has no least upper bound.

12. Aug 6, 2008

LukeD

Another comment, if we consider the extended real numbers (real numbers as well as positive and negative infinity), then every subset of the extended reals is trivially bounded by infinity.
Furthermore, we get that every subset of the extended reals (including the empty set) has a least upper bound in extended reals.

We can see that the proof that the empty set has no real least upper bound fails for the extended reals because it is not the case that u-1 < u when u is positive or negative infinity. However, it does show that the least upper bound cannot be a real number. Therefore, the only numbers left to check are positive and negative infinity. -infinity < infinity, and -infinity bounds the empty set above.
Since -infinity is less than or equal to every extended real number, it is true that -infinity is the least upper bound of the empty set (we cannot find an upper bound less than -infinity).

13. Aug 6, 2008

peos69

In a mathematical proof we have a sequence of theorems,axioms ,definitions,logical conclusions due to the laws of logic it is so simple and powerfully.
When you say vacuously true you violate the above definition
That short of proof is used many times where people are unable to give a solid proof
like proving that the empty set is closed e.t.c e.t.c
Besides that is a semantical proof based simply on the F----->T truthfulness
In a real proof which is syntactical the words true false are not used.
hence the proof that the empty set is bounded from above
is...........................pending

14. Aug 6, 2008

CRGreathouse

Yes, it has no real upper bound, but typically the extended reals are used for bounds. In that case it's $-\infty.$

15. Aug 6, 2008

CRGreathouse

Are you saying that (1) you don't like RAA proofs, (2) that you're a constructivist, (3) that you prefer paraconsistent to classical logic, or that (4) $\top$ and $\bot$ are not technically valid symbols in 'official' proofs?

LukeD's first proof combines with his remark in the second to form a constructive proof, addressing (1) and (2). For (4), proofs can be rewritten to avoid these symbols, using expressions known to be true or false: say $\forall x x=x$ and its negation. I'm not sure what complications would result here from using a paraconsistent framework, though.

16. Aug 6, 2008

Focus

There is no x an element of the empty set. Thats a bit of a contradictory statement to make. I am not worried about the M part, its the bit that says for all x in empty set.

17. Aug 6, 2008

morphism

But that's the point -- the statement "for all x in the empty set <blah blah blah>" is vacuously true no matter what, since there is no x in the empty set.

18. Aug 6, 2008

Focus

Hmm sorry my bad. Might be more useful to define it like for all x, x in empty set implies x is less or equal than M.

19. Aug 6, 2008

LukeD

i'm pretty sure that "for all x in S, P(x)" is equivalent to (if not in fact defined to be) "for all x, x in S => P(x)"

20. Aug 16, 2008

evagelos

Give me a definition of the 'Vacuously true' expression please