Is the Empty Set Bounded? Proof and Contradiction

In summary, the conversation discusses whether or not the empty set has a least upper bound, and if so, what that bound would be. It is determined that, by convention, the empty set is bounded by -\infty in the extended real numbers. However, there is no proof provided for this convention. The completeness axiom is mentioned, which states that non-empty subsets of R with upper bounds have least upper bounds, but it does not apply to the empty set. It is argued that the empty set has no real least upper bound, but it could have -\infty as an extended real upper bound. The conversation also delves into different types of proofs and logic, but no concrete proof for the empty set's upper bound is
  • #1
evagelos
315
0
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?
 
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  • #2
By convention it is [tex] -\infty [/tex]. No proof :)
 
  • #3
Focus said:
By convention it is [tex] -\infty [/tex]. No proof :)

IS that new mathematics??
 
  • #4
Since the set is empty, it really has no bounds. Therefore one can prescribe bounds by convention.
 
  • #5
mathman said:
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
 
  • #6
The empty set is bounded
 
  • #7
The question is asking for the least upper bound not just for bounds
 
  • #8
peos69 said:
IS that new mathematics??

No

peos69 said:
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.

peos69 said:
The empty set is bounded

Please would you tell us what the bound is?

peos69 said:
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.
 
  • #9
YOU want to go and sleep or shall we Curry on on this thread?
 
  • #10
Focus said:
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.
 
  • #11
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
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
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
LukeD said:
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 [itex]-\infty.[/itex]
 
  • #15
Oh, LukeD, I see you addressed my above point in your second post. Sorry about that.

peos69 said:
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) [itex]\top[/itex] and [itex]\bot[/itex] 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 [itex]\forall x x=x[/itex] and its negation. I'm not sure what complications would result here from using a paraconsistent framework, though.
 
  • #16
LukeD said:
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.
 
  • #17
Focus said:
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.
 
  • #18
morphism said:
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.
 
  • #19
Focus said:
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)"
 
  • #20
Give me a definition of the 'Vacuously true' expression please
 
  • #21
LukeD said:
.

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
 
  • #23
evagelos said:
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.
 
  • #24
  • #25
LukeD said:
.

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
 
  • #26
I am sorry to say nobody yet has given me the definition of the 'Vacuously true' expression
 
  • #27
evagelos said:
I am sorry to say nobody yet has given me the definition of the 'Vacuously true' expression

Er...?
 
  • #28
evagelos said:
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.
 
  • #29
LukeD said:
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.

Focus said:
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.
 
  • #30
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.
 
  • #31
GR you discovered foofootos;;;;;;;;;
 
  • #32
evagelos said:
Give me a definition of the 'Vacuously true' expression please

evagelos said:
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".
 
  • #33
Thanks:
can this 'Vacuously true' expression' be considered as an axiom, a theorem, or what?
 
  • #34
A definition?
 
  • #35
evagelos said:
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.
 
<h2>1. What is the empty set supremum proof?</h2><p>The empty set supremum proof is a mathematical proof that shows that the empty set, denoted by ∅, has a supremum (least upper bound) of -∞. This proof is important in understanding the properties of sets and their elements.</p><h2>2. Why is the empty set supremum proof significant?</h2><p>The empty set supremum proof is significant because it helps establish the properties of the empty set, which is a fundamental concept in mathematics. It also allows for the application of set theory in various fields, such as computer science and statistics.</p><h2>3. How is the empty set supremum proof derived?</h2><p>The empty set supremum proof is derived using the definition of supremum, which states that a supremum of a set S is the smallest upper bound of S. Since the empty set has no elements, it is vacuously true that all real numbers are upper bounds of the empty set. Therefore, the supremum of the empty set must be the smallest of these upper bounds, which is -∞.</p><h2>4. What are the implications of the empty set supremum proof?</h2><p>The implications of the empty set supremum proof are vast and far-reaching. It helps establish the foundations of set theory and provides a basis for understanding more complex mathematical concepts. It also has practical applications in fields such as computer science, where the empty set is used to represent an absence of data.</p><h2>5. Are there any misconceptions about the empty set supremum proof?</h2><p>One common misconception about the empty set supremum proof is that it implies that the empty set is equal to -∞. However, this is not the case. The proof simply shows that -∞ is the supremum of the empty set, but the two are not equivalent. Another misconception is that the empty set has no properties, but the empty set supremum proof shows that it does have the property of having -∞ as its supremum.</p>

1. What is the empty set supremum proof?

The empty set supremum proof is a mathematical proof that shows that the empty set, denoted by ∅, has a supremum (least upper bound) of -∞. This proof is important in understanding the properties of sets and their elements.

2. Why is the empty set supremum proof significant?

The empty set supremum proof is significant because it helps establish the properties of the empty set, which is a fundamental concept in mathematics. It also allows for the application of set theory in various fields, such as computer science and statistics.

3. How is the empty set supremum proof derived?

The empty set supremum proof is derived using the definition of supremum, which states that a supremum of a set S is the smallest upper bound of S. Since the empty set has no elements, it is vacuously true that all real numbers are upper bounds of the empty set. Therefore, the supremum of the empty set must be the smallest of these upper bounds, which is -∞.

4. What are the implications of the empty set supremum proof?

The implications of the empty set supremum proof are vast and far-reaching. It helps establish the foundations of set theory and provides a basis for understanding more complex mathematical concepts. It also has practical applications in fields such as computer science, where the empty set is used to represent an absence of data.

5. Are there any misconceptions about the empty set supremum proof?

One common misconception about the empty set supremum proof is that it implies that the empty set is equal to -∞. However, this is not the case. The proof simply shows that -∞ is the supremum of the empty set, but the two are not equivalent. Another misconception is that the empty set has no properties, but the empty set supremum proof shows that it does have the property of having -∞ as its supremum.

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