# Thread: Relation between gaps and supremum properties

1. ## Relation between gaps and supremum properties

Hi so I am studying and there is a sentence saying that

There is (S,<) an ordered set,
then S is without gaps if and only if S has the properties of Supremum,

Can someone give me an intuitive explanation and give me a formal proof?
Thank you.

2. ## Re: Relation between gaps and supremum properties

Originally Posted by davidciprut
There is (S,<) an ordered set,
then S is without gaps if and only if S has the properties of Supremum,
Can someone give me an intuitive explanation and give me a formal proof?
As you can see there are many forms of ordered sets.
You also define what a gap means.

3. ## Re: Relation between gaps and supremum properties

Also, what do you mean by "properties of Supremum". Do you mean has a supremum?

4. ## Re: Relation between gaps and supremum properties

1-)So the ordered set they defined has two properties
O1. x,y belongs to set S then on of the three situation exists, x<y, x=y, x>y (Trichotomous)
O2.x,y,z belongs to set S, if x<y and y<z then x<z (Transitivity)
And it's written (S,<), which means S is an ordered set which has the order with <.

2-)Gap,
It is said that (S,<) is a a sat without gaps if and only if:
L,U are subsets of S (They are not empty sets) and L<U (L is equal or less than U)
In other words for every u that belongs to U and for every l belongs to L, l<u (l is equal or less than u)
Then there exists c that belongs to S such that l<c<u (equal or less signs)

3-)So the property of supremum is,
Definition:
There is an ordered set S
X is a subset of S
s (belongs to S) is called the Supremum of set X if and only if the two properties exist,
1.x<s (x is equal or less than s) , x belongs to X (s is upper bound of X)
2.If x<u (x is equal or less than u) for every x that belongs to X then s<u (s is equal or less than u)

So these are the information that is given as an ordered set and the third is the properties of supremum.

By the way I don't know how to write the mathematical signs in the forum, so that is why I use more words,
And I hope the information I gave is enough. I try to be as precise as I can because I am translating it from Hebrew to English.

5. ## Re: Relation between gaps and supremum properties

Originally Posted by davidciprut
1-)So the ordered set they defined has two properties
O1. x,y belongs to set S then on of the three situation exists, x<y, x=y, x>y (Trichotomous)
O2.x,y,z belongs to set S, if x<y and y<z then x<z (Transitivity)
And it's written (S,<), which means S is an ordered set which has the order with <.

2-)Gap,
It is said that (S,<) is a a sat without gaps if and only if:
L,U are subsets of S (They are not empty sets) and L<U (L is equal or less than U)
In other words for every u that belongs to U and for every l belongs to L, l<u (l is equal or less than u)
Then there exists c that belongs to S such that l<c<u (equal or less signs)

3-)So the property of supremum is,
Definition:
There is an ordered set S
X is a subset of S
s (belongs to S) is called the Supremum of set X if and only if the two properties exist,
1.x<s (x is equal or less than s) , x belongs to X (s is upper bound of X)
2.If x<u (x is equal or less than u) for every x that belongs to X then s<u (s is equal or less than u)

So these are the information that is given as an ordered set and the third is the properties of supremum.

By the way I don't know how to write the mathematical signs in the forum, so that is why I use more words,
And I hope the information I gave is enough. I try to be as precise as I can because I am translating it from Hebrew to English.
It appears that by ordered your text means what is called total order or linear order
That is standard.

The definition of supremun is a bit garbled but understandable.

But I am sorry to say that I find the description of gaps unintelligible. It sounds like an idea related to cuts.
But if gaps are something similar to cuts then your translation is wanting.

Can you try again?

You can learn to use LaTeX coding.
$$a \prec b$$ gives $a \prec b$ read $a\text{ preceeds }b$ which is usually a reflexive and anti-symmetric relation, $a \prec a$

6. ## Re: Relation between gaps and supremum properties

Hmmm I see, I wish I could give another definition for gaps. But what the lecturer meant by gap he meant, for example, rational numbers has gaps, because it has numbers like square root 2 or the number e, and many numbers that rational numbers doesn't contain, so it it hes gaps.

By the way, I learned to use the program called LyX , I heard that it uses latex coding, so will it work here if I write like I write in LyX?

7. ## Re: Relation between gaps and supremum properties

It might have to do something with dedekind cuts but not sure

8. ## Re: Relation between gaps and supremum properties

For "without gaps", I think the OP means:

A totally ordered set $(S,<)$ is said to be "without gaps" if for any pair of nonempty subsets $L\subseteq S, U\subseteq S$ with $l for all $l\in L, u \in U$ there exists $c\in S$ with $l\le c \le u$ for all $l \in L, u \in U$.

Next, from the statement that $S$ "has the supremum property", I think the OP means that every subset $X \subseteq S$, the supremum $\sup X \in S$.

9. ## Re: Relation between gaps and supremum properties

Originally Posted by SlipEternal
For "without gaps", I think the OP means:
A totally ordered set $(S,<)$ is said to be "without gaps" if for any pair of nonempty subsets $L\subseteq S, U\subseteq S$ with $l for all $l\in L, u \in U$ there exists $c\in S$ with $l\le c \le u$ for all $l \in L, u \in U$.
Next, from the statement that $S$ "has the supremum property", I think the OP means that every subset $X \subseteq S$, the supremum $\sup X \in S$.
I really doubt that is what is meant by those definitions.
Look at this about the rationals.
As a metric space the rationals have no gaps. But it is not complete, thus lacks the supremun property.

10. ## Re: Relation between gaps and supremum properties

Let $S = \mathbb{Q}$. Let $L = (-\infty,\sqrt{2})\cap \mathbb{Q}, U = (\sqrt{2},\infty)\cap \mathbb{Q}$. Both $L$ and $U$ are subsets of $S$, but there does not exist any $c\in S$ such that $l\le c\le u$ for all $l\in L, u \in U$, so $\mathbb{Q}$ definitely has gaps by those definitions.