is the empty set bounded from above?

 

has it got a supremum (l.u.b)?

 

Whether it has or not how can we prove that??

Does not bounded from above refer to a set of real numbers? As the empty set is not a set of numbers I cannot see that question is well-posed.

 

 

Definition

A set [math]S[/math] of real numbers is said to be bounded from above if there exists a real number [math]k[/math], such that [math]k \geq s[/math] for all [math] s \in S[/math].

 

You can also have bounded set as subsets of a topological space, but this is again not what you are talking about here.

http://en.wikipedia.org/wiki/Empty_set#Extended_real_numbers

Right, of course you can consider it as a subset of any set. In particular it can be viewed as a subset of the set of real numbers.

 

Then by definition(?) any real number is an upper and lower bound of the empty set.

Exactly. Given any real x, there is no member of the empty set that is larger than (or smaller than) x. It is a mathematical equivalent of the sound of one hand clapping.

I was hoping some one will produce a solid proof that the formula:

 

there exists aεR such that,for all xεΦ ,[math] x\leq a[/math] , is true

 

and not the vacuously true nonsenses of the wikipedia

I would assume that you simply state that (for any a) there is no x for which x>a.

I think that the empty set is bounded for reasons already stated.

 

I also think that what the bounds are is a matter of convention.

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I was hoping some one will produce a solid proof that the formula:

 

there exists aεR such that,for all xεΦ ,[math] x\leq a[/math] , is true

 

and not the vacuously true nonsenses of the wikipedia

 

Does the statement that "nothing" is greater than (or smaller than) something make any sense?

 

I do not think so and therefore to consider if the empty set is bounded you need to rephrase the definition to take care of this situation.

I was hoping some one will produce a solid proof that the formula:

 

there exists aεR such that,for all xεΦ ,[math] x\leq a[/math]

 

For crying out loud, triclino. Your obsession with "solid proof" has hit a new low (literally). This obsession is severely impeding your ability to learn. I strongly suggest you get over it.

True. Nonetheless, pick an element of R and systematically go through all elements of the empty set to ensure the statement is true.

 

 

 

 

 

Does the statement that "nothing" is greater than (or smaller than) something make any sense?

 

No.

 

But the correct statement is:

 

for all x : if x belongs to "nothing" then this x is smaller or greater than something.

 

And now we wander whether this conditional is true or false.

 

According to wikibedia it is "vacuously" true,meaning that ,since x belongs to 'nothing " is false the conditional is true,irrespectively whether x is smaller or greater than something.

 

However assuming that the conditional is not true then its denial must be true.

 

But its denial is the following:

 

 

There exists an x belonging to "nothing" and this x is greater or smaller to something, ending in a contradiction ,since there dos not exist an x belonging to "nothing".

 

Hence "nothing" is bounded.

 

Now the question is, if "nothing" has a supremum (infemum) or not

I think by convention they are set to [math]\pm \infty[/math].

 

You may be able to make this more rigorous by extending the real numbers to include infinity.

But Patrick Suppes in his book "AXIOMATIC SET THEORY" on page 186 ,ask us to give a proof that the empty set has no least upper bound,meaning that there is a proof that the empty set has no least upper bound.

But Patrick Suppes in his book "AXIOMATIC SET THEORY" on page 186 ,ask us to give a proof that the empty set has no least upper bound,meaning that there is a proof that the empty set has no least upper bound.

 

As any number is an upper bound (and lower bound) it is fine to say that there is no least upper bound.

 

However, as a convention you can set it to [math]- \infty[/math] as (I assume) that can be useful.

 

Remember that [math]\pm \infty[/math] are not real numbers.

But Patrick Suppes in his book "AXIOMATIC SET THEORY" on page 186 ,ask us to give a proof that the empty set has no least upper bound,meaning that there is a proof that the empty set has no least upper bound.

What is the definition used in that book for upper bound? If it is something along the lines of given an ordered structure [math](E,\le)[/math] and a set [math]A\subset E[/math], then an element [math]c\in E[/math] is an upper bound of [math]A[/math] if [math]\{x \in E:x > c\} = \Phi[/math], then it should be obvious that every element of [math]E[/math] is an upper (and lower) bound of the empty set. If the set [math]E[/math] is itself unbounded (e.g., the reals), then there is no least upper bound (greatest lower bound).

What is the definition used in that book for upper bound? If it is something along the lines of given an ordered structure [math](E,\le)[/math] and a set [math]A\subset E[/math], then an element [math]c\in E[/math] is an upper bound of [math]A[/math] if [math]\{x \in E:x > c\} = \Phi[/math], then it should be obvious that every element of [math]E[/math] is an upper (and lower) bound of the empty set. If the set [math]E[/math] is itself unbounded (e.g., the reals), then there is no least upper bound (greatest lower bound).

 

So, in your definition ,let A = (0,1) ,an open interval and E = R ,the real Nos.

 

According to your definition 2 is not an upper bound of (0,1) ,because

 

{xεR : x>2} is not empty.

 

But according to the definition of the book ,page 186, 2 is an upper bound.

 

The definition is:

 

" if A is a set of real Nos then y is an upper bound of A if and only if y is areal No and for every x in A ,[math]x\leq y[/math]"