Is the Empty Set Bounded? Proof and Contradiction

[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?

 

[Focus]

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

 

[peos69]

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

IS that new mathematics??

 

[mathman]

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

 

[peos69]

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

Sure when we cannot prove something we use the convention stuff

 

[peos69]

The empty set is bounded

 

[peos69]

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

 

[Focus]

IS that new mathematics??

No

Sure when we cannot prove something we use the convention stuff

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.

The empty set is bounded

Please would you tell us what the bound is?

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

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.

 

[peos69]

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

 

[peos69]

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.

READ your completeness Axiom more carefully,it says:
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.

 

[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.

 

[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).

 

[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

 

[CRGreathouse]

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.

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

 

[CRGreathouse]

Oh, LukeD, I see you addressed my above point in your second post. Sorry about that.

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

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.

 

[Focus]

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.

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.

 

[morphism]

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.

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.

 

[Focus]

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.

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.

 

[LukeD]

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.

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)"

 

[evagelos]

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

 

[evagelos]

.

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.

since u is the least upper bound we have u<u-1 and not u-1<u so where is the contradiction

 

[LukeD]

Skimmed the Wikipedia article a bit. It seems pretty good.
http://en.wikipedia.org/wiki/Vacuous_truth

 

[LukeD]

since u is the least upper bound we have u<u-1 and not u-1<u so where is the contradiction

For any real number it's the case that x - 1 < x (you're subtracting 1, so you get a smaller number). So we have both u < u - 1 (since u is the least upper bound) and u - 1 < u. This is a contradiction.

 

[evagelos]

Skimmed the Wikipedia article a bit. It seems pretty good.
http://en.wikipedia.org/wiki/Vacuous_truth

From your reading on the Wikipedia how would you define the 'Vacuously true' expression!

 

[evagelos]

.

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.

But you do not mention that u<u-1 in your proof

 

[evagelos]

I am sorry to say nobody yet has given me the definition of the 'Vacuously true' expression

 

[asdfggfdsa]

I am sorry to say nobody yet has given me the definition of the 'Vacuously true' expression

Er...?

 

[LukeD]

I am sorry to say nobody yet has given me the definition of the 'Vacuously true' expression

I thought the Wikipedia article was pretty clear. If I had to define it, I'd give you something pretty similar to the intro in the Wiki article, probably pretty similar wording too.

 

[HallsofIvy]

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.

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.

Yes, but wording it "If x is in the empty set then |x|<= M" gives a valid, vacuously true statement.

 

[CRGreathouse]

To me, "For all foo in bar, phi" is exactly the same as "For all foo, foo is in bar implies phi". Not just that they have equal truth values, but that the latter is the definition for the former.

 

[evagelos]

GR you discovered foofootos;;;;;;;;;

 

[HallsofIvy]

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

I am sorry to say nobody yet has given me the definition of the 'Vacuously true' expression

In logic the statement A=> B or "If A then B"is true if A is false no matter whether B is true or false. That is called "vacuously true".

 

[evagelos]

Thanks:
can this 'Vacuously true' expression' be considered as an axiom, a theorem, or what?

 

[morphism]

A definition?

 

[LukeD]

Thanks:
can this 'Vacuously true' expression' be considered as an axiom, a theorem, or what?

Depending on the axioms that you use for logic, that A => B is true whenever A is false can be either an axiom or a theorem.

Vacuous Truth is just what it's called when A is false because it doesn't at all matter what B is.

 

[evagelos]

Depending on the axioms that you use for logic, that A => B is true whenever A is false can be either an axiom or a theorem.

Vacuous Truth is just what it's called when A is false because it doesn't at all matter what B is.

up to now i was informed of three possibilities:

morphism said a definition with a question mark, you said a theorem or an axiom depending on the set of axioms. You succeeded in making me to open the books of logic to find out
myself.
so wait i get back to you

 

[Dragonfall]

Your logic book will only tell you the formalism preferred by its author.

 

[evagelos]

The laws and the facts about logic are eternal and no author and his preferred formalism
can change that

 

[CRGreathouse]

The laws and the facts about logic are eternal and no author and his preferred formalism can change that

Some logic books define \phi \vee \psi as \neg\phi\rightarrow\psi while others define it as its own privative with 3-4 axioms (for example, in paraconsistent logics). What would you say to that?

 

[Dragonfall]

The laws and the facts about logic are eternal and no author and his preferred formalism
can change that

It's as if you're talking about a religion.

 

[evagelos]

It's as if you're talking about a religion.

If you subtract religion from our life we will live in a better world,but if you take away

logic can you live in para consistent logic as GR put it

 

[Dragonfall]

You sound just like a postmodernist mathematician! You wouldn't happen to be a disciple of Jacques Lacan, would you?

 

[evagelos]

pardon me but i like to postmortem mathematics.

Who the hell is that Lucas guy anyway

 

[LukeD]

The laws and the facts about logic are eternal and no author and his preferred formalism
can change that

Why do you think that the laws of mathematics are such? Why are they "true" and "eternal"? Did you prove this in some way? Did some higher power tell you that this must be true? Is it just intuitive?

I'm not trying to goad you into an argument. I'm just curious about your answer, and I wanted to illustrate something.

 

[HallsofIvy]

The laws and the facts about logic are eternal and no author and his preferred formalism
can change that

And this is true because you SAY so?

 

[evagelos]

The F----->T, F----->F implications are just part of the conditional definition of two statements p and q......p----->q.

So the ' vacuously true' expression is not an axiom is not a theorem in the axiomatic
foundation of the predicate calculus.

It can be used in a semantical type of proof and not in a syntactical type where all the mathematical proofs are.

Anyway to get rid of the 'vacuously true' myth here is a proof that the empty set is bounded

We want to prove that:...Ea(x)[ xεΦ====>absvalue(a)>=x]........

Let...xεΦ====>( xεΦ v absvalue(a)>=x) <=====> ( ~xεΦ===> absvalue(a)>=x) and
since ~xεΦ ( because no object belongs to the empty set) we have...absvalue(a)>=x.

Thus......xεΦ====>absvalue(a)>=x ..........

And......Ea(x)[ xεΦ====>absvalue(a)>=x].........

Where (x) means for all x, Ea means there exists an a, ~xεΦ means x does not belong to the empty set.

Perhaps an easier and shorter proof is the following:

We know...(x)[ ~xεΦ]====> ~xεΦ====>(~xεΦ v absvalue(a)>=x) <=====>
........xεΦ====>absvalue(a)>=x .........

And.......Ea(x)[ xεΦ====>absvalue(a)>=x]........

 

[Focus]

p -> q is a shorthand way of writing \neg p \vee q. The other operators (such as "not" and "or") are defined, you can call them axioms but its a bit pointless because they really do not tell you anything. As for the rest of the post with the proof, I can't read it, please use http://en.wikibooks.org/wiki/LaTeX/Mathematics" to write it, ascii is not made for maths, it makes my eyes bleed.

 

[HallsofIvy]

pardon me but i like to postmortem mathematics.

Who the hell is that Lucas guy anyway

So you think mathematics is dead? That's explains a lot about your posts!

 

[evagelos]

the law that i used in my proof is the tautology or theorem of the propositional calculus :

[(pvq)<----->(~p---->q)].............1

If we give values to p and q , (T,F) all the values of 1 will be true,thus 1 is a tautology,

meaning that pvq and ~p--->q are logically equivalent ,meaning that pvq logically implies
~p--->q and ~p---->q logically implies pvq,written as pvq <====> ~p---->q,notice the double line in the double implication.

If we now put p= ~xεΦ and q= absvalue(x)=<a, we have:

we have p......a fact not an assumption
but p===>pvq ( because of the law called disjunction introduction) <=====> ~p---->q (because of the above law).

hence...~p=====>q and if we substitute p and q we we have that the empty set is

bounded

 

[evagelos]

Another maybe understandable proof is to reason by contradiction:

Assume that the empty set is not bounded ,then we must have : Αn x belonging to the empty set, xεΦ,and an absolute x bigger than a, absvalue(x)> a.

But since no x belongs to the empty set ,~xεΦ, we have a contradiction :

xεΦ & ~xεΦ

Thus,,,,,,,,,,,,,, the empty set is bounded