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.

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- Nov 8th 2013, 11:39 AMdavidciprutRelation 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. - Nov 8th 2013, 12:12 PMPlatoRe: Relation between gaps and supremum properties
You must give us some more information.

As you can see there are many forms of ordered sets.

You also define what ameans.**gap** - Nov 8th 2013, 12:17 PMHallsofIvyRe: Relation between gaps and supremum properties
Also, what do you mean by "properties of Supremum". Do you mean

**has**a supremum? - Nov 8th 2013, 12:50 PMdavidciprutRe: 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. - Nov 8th 2013, 03:58 PMPlatoRe: Relation between gaps and supremum properties
It appears that by

your text means what is called total order or*ordered**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 ofunintelligible. It sounds like an idea related to cuts.*gaps*

But if gaps are something similar to cuts then your translation is wanting.

Can you try again?

You can learn to use LaTeX coding.

[tex]a \prec b[/tex] gives $\displaystyle a \prec b$ read $\displaystyle a\text{ preceeds }b$ which is usually a reflexive and anti-symmetric relation, $\displaystyle a \prec a$ - Nov 10th 2013, 10:40 AMdavidciprutRe: 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? - Nov 10th 2013, 10:42 AMdavidciprutRe: Relation between gaps and supremum properties
It might have to do something with dedekind cuts but not sure

- Nov 10th 2013, 12:40 PMSlipEternalRe: Relation between gaps and supremum properties
For "without gaps", I think the OP means:

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

Next, from the statement that $\displaystyle S$ "has the supremum property", I think the OP means that every subset $\displaystyle X \subseteq S$, the supremum $\displaystyle \sup X \in S$. - Nov 10th 2013, 03:59 PMPlatoRe: Relation between gaps and supremum properties
I really doubt that is what is meant by those definitions.

Look at this about the rationals.

As a metric space the rationals have. But it is not complete, thus lacks the supremun property.*no gaps* - Nov 10th 2013, 05:17 PMSlipEternalRe: Relation between gaps and supremum properties
Let $\displaystyle S = \mathbb{Q}$. Let $\displaystyle L = (-\infty,\sqrt{2})\cap \mathbb{Q}, U = (\sqrt{2},\infty)\cap \mathbb{Q}$. Both $\displaystyle L$ and $\displaystyle U$ are subsets of $\displaystyle S$, but there does not exist any $\displaystyle c\in S$ such that $\displaystyle l\le c\le u$ for all $\displaystyle l\in L, u \in U$, so $\displaystyle \mathbb{Q}$ definitely has gaps by those definitions.