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formalization

solakis

Active member
Dec 9, 2012
326
formalize the following definition:

We define the supremum of a non empty subset of the real Nos (S) bounded from above ,denoted by Sup(S), to be a real No a ,which is the smallest of all its upper bounds
 

Deveno

Well-known member
MHB Math Scholar
Feb 15, 2012
1,967
oh gosh i might have this wrong but i think it is:

$\{y \in \Bbb R : \forall x \in S\ \forall b \in \Bbb R: (x < b) \implies (x \leq y)\}$
 

Evgeny.Makarov

Well-known member
MHB Math Scholar
Jan 30, 2012
2,503
formalize the following definition:
What do you mean by formalizing?

oh gosh i might have this wrong but i think it is:

$\{y \in \Bbb R : \forall x \in S\ \forall b \in \Bbb R: (x < b) \implies (x \leq y)\}$
It's not true that \(y=\sup S\) iff \(\forall x \in S\ \forall b \in \Bbb R: (x < b) \implies (x \leq y)\). The latter formula is true if $y$ is any upper bound of $S$.
 

solakis

Active member
Dec 9, 2012
326
What do you mean by formalizing?

.
What did you do when you transformed the Problem:

if ,for all,x $ a|x|+bx\geq 0$ then $ a\geq |b|$

into:

\[
\forall a,b,\,[(\forall x,\,a|x|+bx\ge0)\to a\ge|b|]
\]
 

Plato

Well-known member
MHB Math Helper
Jan 27, 2012
196
What did you do when you transformed the Problem:
if ,for all,x $ a|x|+bx\geq 0$ then $ a\geq |b|$
into:
$\forall a,b,\,[(\forall x,\,a|x|+bx\ge0)\to a\ge|b|]$
There is an interesting textbook Symbolic Logic and the Real Number System by AH Lightstone. He symbolizes almost every property of real numbers except the one you have asked about. It is known as the completeness theorem in his book.

We can define some terms.
If $A\ne\emptyset$ then $\mathcal{UB}(A)=\{x: (\forall a\in A)[a\le x]\}$

If $\mathcal{UB}(A)\ne \emptyset$ then $(\exists t\in\mathcal{UB}(A))(\forall x\in\mathcal{UB}(A))[t\le x]$. We say $t=\sup(A)$. p.192
 

Deveno

Well-known member
MHB Math Scholar
Feb 15, 2012
1,967
What do you mean by formalizing?

It's not true that \(y=\sup S\) iff \(\forall x \in S\ \forall b \in \Bbb R: (x < b) \implies (x \leq y)\). The latter formula is true if $y$ is any upper bound of $S$.
yup, left out an y ≤ b somewhere. danke.
 

solakis

Active member
Dec 9, 2012
326
There is an interesting textbook Symbolic Logic and the Real Number System by AH Lightstone. He symbolizes almost every property of real numbers except the one you have asked about. It is known as the completeness theorem in his book.

We can define some terms.
If $A\ne\emptyset$ then $\mathcal{UB}(A)=\{x: (\forall a\in A)[a\le x]\}$

If $\mathcal{UB}(A)\ne \emptyset$ then $(\exists t\in\mathcal{UB}(A))(\forall x\in\mathcal{UB}(A))[t\le x]$. We say $t=\sup(A)$. p.192
The above is a semi formalization of the following sentence:

We say that a non empty set A , whose all the unbounded sets ($ \mathcal{UB}(A) $) are non empty has a supremum ,t, if the unbounded sets have a minimum which is t
 

Evgeny.Makarov

Well-known member
MHB Math Scholar
Jan 30, 2012
2,503
The above is a semi formalization of the following sentence:

We say that a non empty set A , whose all the unbounded sets ($ \mathcal{UB}(A) $) are non empty has a supremum ,t, if the unbounded sets have a minimum which is t
No. The quote above have several errors. First, the phrase "set A , whose all the unbounded sets" does not make sense because one can't say, for some set B, "whose" set B is. For example, the phrase "John, whose all unmarried sons" makes sense because, given person Bill, we can say whose son Bill is and whether he is a son of John. However, I don't know when some set B is a set of some other set A. You probably mean "A, whose all the unbounded elements" or possible "A, whose all the unbounded subsets," because the relations "an element of" and "a subset of," just like "a son of," are well-defined. But the elements of A are numbers, not sets. Indeed, in the definition of UB(A) (i.e., upper bounds of A), x is compared using <=. Only numbers, not sets, can be compared using <=. Further, note that UB(A) does not consist of elements of A but of numbers that exceed all elements of A, i.e., the upper bounds of A.

Second, the definition does not say that "all the unbounded sets [rather, elements] ($ \mathcal{UB}(A) $) are non empty," but that the set UB(A) itself is nonempty. The definition cannot refer to "supremum" because this is a definition of supremum. Finally, again, t <= x cannot mean that the minimum of x is t because <= is defined only for numbers, not sets.

I recommend you start by understanding why UB(A) is the set of upper bounds of A.
 

solakis

Active member
Dec 9, 2012
326
No. The quote above have several errors. First, the phrase "set A , whose all the unbounded sets" does not make sense because one can't say, for some set B, "whose" set B is. For example, the phrase "John, whose all unmarried sons" makes sense because, given person Bill, we can say whose son Bill is and whether he is a son of John. However, I don't know when some set B is a set of some other set A. You probably mean "A, whose all the unbounded elements" or possible "A, whose all the unbounded subsets," because the relations "an element of" and "a subset of," just like "a son of," are well-defined. But the elements of A are numbers, not sets. Indeed, in the definition of UB(A) (i.e., upper bounds of A), x is compared using <=. Only numbers, not sets, can be compared using <=. Further, note that UB(A) does not consist of elements of A but of numbers that exceed all elements of A, i.e., the upper bounds of A.

Second, the definition does not say that "all the unbounded sets [rather, elements] ($ \mathcal{UB}(A) $) are non empty," but that the set UB(A) itself is nonempty. The definition cannot refer to "supremum" because this is a definition of supremum. Finally, again, t <= x cannot mean that the minimum of x is t because <= is defined only for numbers, not sets.

I recommend you start by understanding why UB(A) is the set of upper bounds of A.
I do not understand why so much fuss for a typo .

Surely one can easily see that i mean the upper bounds sets and NOT the unbounded sets

Now whether is one upper bound set or many is not very clear of the definition,but this is not of great importance,they all have the same minimum.

Finally this semi formalization is not a formalization of the definition of the OP
 

Plato

Well-known member
MHB Math Helper
Jan 27, 2012
196
I do not understand why so much fuss for a typo .
That was clearly not a typo. You used the term unbounded too many times for it to have been a typo.

Finally this semi formalization is not a formalization of the definition of the OP
OK. Why don't you offer a formalization of the definition of the OP?

But remember definitions vary in this area of logic.
 

Evgeny.Makarov

Well-known member
MHB Math Scholar
Jan 30, 2012
2,503
Surely one can easily see that i mean the upper bounds sets and NOT the unbounded sets
I am not sure what "the upper bounds sets" means. An upper bound of A is a number, not a set. And UB(A) is not "the upper bounds sets," but the set of upper bounds.

Now whether is one upper bound set or many is not very clear of the definition,but this is not of great importance,they all have the same minimum.
There is a single set of upper bounds. It may be either empty (if A is unbounded) or infinite (if A is bounded).

Finally this semi formalization is not a formalization of the definition of the OP
Why not? To get a single formula, you can substitute the definition of the set UB(A) into the second line and replace "If ..., then ..." with $\to$. This gives a theorem, which gives rise to a definition. If you need a bare definition, it is a part of the second line: we say that $t$ is $\mathrm{Sup}(A)$ if \(t\in\mathcal{UB}(A)\land (\forall x\in\mathcal{UB}(A), t\le x)\). In any case, the formula says exactly what the English text says in the OP (except S is replaced by A).
 

solakis

Active member
Dec 9, 2012
326
That was clearly not a typo. You used the term unbounded too many times for it to have been a typo..

We say that a non empty set A , whose all the unbounded sets ($ \mathcal{UB}(A) $) are non empty has a supremum ,t, if the unbounded sets have a minimum which is t
As you can observe clearly my definition ends with the phrase:

"if the unbounded sets have a minimum which is t"


Can un unbounded set have a minimum??
 

Evgeny.Makarov

Well-known member
MHB Math Scholar
Jan 30, 2012
2,503
Can un unbounded set have a minimum??
Yes, if it is unbounded from above.

To summarize the thread: In post #5, Plato offered a formula that expresses what the English text in post #1 says. In post #7, you offered an alternative English text that supposedly says the same thing as Plato's formula. However, even though your English text makes some sense, it contains many errors (e.g., "the unbounded sets (UB(A)) are non empty" instead of "the set of upper bounds is nonempty").

Is your original question answered?
 

solakis

Active member
Dec 9, 2012
326
I am not sure what "the upper bounds sets" means. An upper bound of A is a number, not a set. And UB(A) is not "the upper bounds sets," but the set of upper bounds.

There is a single set of upper bounds. It may be either empty (if A is unbounded) or infinite (if A is bounded).

Why not? To get a single formula, you can substitute the definition of the set UB(A) into the second line and replace "If ..., then ..." with $\to$. This gives a theorem, which gives rise to a definition. If you need a bare definition, it is a part of the second line: we say that $t$ is $\mathrm{Sup}(A)$ if \(t\in\mathcal{UB}(A)\land (\forall x\in\mathcal{UB}(A), t\le x)\). In any case, the formula says exactly what the English text says in the OP (except S is replaced by A).
The concept of the upper bounds set(s) is not mentioned at all in the definition of the OP.

Now if you want to baptize the concept of the "set bounded from above" to :" the set of the upper bounds of a set" it is your decision.

The definition of the 1st is:

$\exists y\forall x [x\in S\Longrightarrow x\leq y]$

While the definition of the 2nd is:

$UB(S)$ ={$x: \forall s(s\in S\Longrightarrow s\leq x)$}

The 1st one defines a Real No y.

The 2nd one a set .
 

solakis

Active member
Dec 9, 2012
326
Yes, if it is unbounded from above.
I did not write :

Unbounded from above ( in my translation to Plato's semi formalized definition).

But ,simply:

Unbounded sets

There is a great difference between the two.


To summarize the thread: In post #5, Plato offered a formula that expresses what the English text in post #1 says. In post #7, you offered an alternative English text that supposedly says the same thing as Plato's formula. However, even though your English text makes some sense, it contains many errors (e.g., "the unbounded sets (UB(A)) are non empty" instead of "the set of upper bounds is nonempty").

Is your original question answered?
No, thank you for your help, i am working on it.