1. ## Completeness Property

Do any of the following sets have the completeness property:

1. Q
2. closed interval from 0 to 1
3. open interval from 0 to 1
4. 0 to 1 including 0 but not 1
5. 0 to infinity including 0

I dont think number 1 does because you could have the set of rationals where the sqaure of the element in the set is less than two. This has upper bounds but does not have a least upper bound

3. what is the least rational number greater than root 3

4. You are right about question 1.

what is the least rational number greater than root 3
It does not exist. For every rational $\displaystyle r > \sqrt{3}$ you may consider $\displaystyle \sqrt{3} + (r - \sqrt{3}) / 2$, which is less than $\displaystyle r$.

I think you get the idea: for a set A, whenever there is a sequence of point in A that converges to a point outside of A, A is not complete.

5. o okay that makes perfect since now what about the other four sets in the first post.

6. Well, for (3), can't you think of a sequence of numbers $\displaystyle x_n, n=1, 2,\ldots$ such that $\displaystyle 0 < x_n < 1$ but $\displaystyle x_n$ converge to 0?

Think of an interval from 0 to 1 as a iron footpath. If 0 is included, then the left end of the path has a nice round place to stand on (and similarly for 1). If 0 is not included, imagine that close to 0 the path bends down until it becomes vertical, so it's very easy to slip down. Now the question is, can you safely walk on such path?

Evgeny

7. So you're saying there is no greatest lower bound or least upper bound for 3

8. ## Greatest lower bound

I only now noticed this reading your first and last posts: If a nonempty set (of real numbers) has a lower bound, then it necessarily has a greatest lower bound (also called infimum), and similarly for the least upper bound (supremum). What may not exist is minimum and maximum, also called the least and greatest elements of the set. (Minimum is not the same as the least element on partially ordered sets, but this is a different story. Since any two reals can be compared, minimum and the least element are the same.)

The existence of inf for sets bounded from below and of sup for sets bounded from above follows (I believe, is equivalent) to one of the axioms of real numbers. In fact, this is a very important axiom that sets reals apart from rational numbers: see
Real number - Wikipedia, the free encyclopedia

The least element, on the other hand, is first and foremost an element of the set in question, while the set's infimum does not have to be an element. So an inteval (0, 1) without endpoints has neither least nor greatest element, but its inf is 0 and its sup is 1.

9. ok so 2-5 all have the completeness property

but one point confuses me. Any set containing real numbers will have a lub and glb given that it has upper bounds and lower bounds but arent sets containing rational numbers also sets containing real numbers since every rational number is a real number. So shouldn't it follow that all sets containing only rationals would follow the completeness property but I have already shown an example where a set of rationals with an upper bound does not have a lub

thank you

10. ok so 2-5 all have the completeness property
Sorry, no.

Completeness is when you can't accidentally wander outside the set (when you make ever smaller steps). Since 0 is not in the interval (0, 1), it's like a bent down path, as I said. If you approach 0, you can't hold on to it and will eventually (after infinitely many steps ) leave the interval. More precisely, consider the sequence $\displaystyle 1/n$ for $\displaystyle n=1, 2\ldots$. Every single element of the sequence is also an element of (0, 1). However, the sequence tends to 0, which is not in (0, 1); hence, no completeness.

Let $\displaystyle A=\{x\in\mathbb{R}\mid0<x<1\}$ and $\displaystyle S=\{1/n\mid n=1, 2,\ldots\}$. Then $\displaystyle \inf A = \inf S = 0$, but $\displaystyle 0\notin A$ and $\displaystyle 0\notin S$. Also, $\displaystyle A$ and $\displaystyle S$ do not have least elements.

Any set containing real numbers will have a lub and glb given that it has upper bounds and lower bounds.
Yes.
but arent sets containing rational numbers also sets containing real numbers since every rational number is a real number.
Yes.
So shouldn't it follow that all sets containing only rationals would follow the completeness property but I have already shown an example where a set of rationals with an upper bound does not have a lub
What you showed in the original post is that the set $\displaystyle \{x\in\mathbb{Q}\mid x^2<2\}$ does not have the greatest element, not that it does not have a supremum. Supremum, which is $\displaystyle \sqrt{2}$, exists, but it is not in the set.

Here is some explanation on the relationship between completeness and inf/sup. Let $\displaystyle A$ be a nonempty set of reals that has a lower bound. $\displaystyle A$ being complete and the existence of $\displaystyle \inf A$ are different things. In one direction: If $\displaystyle A$ is complete , then $\displaystyle \inf A\in A$. This is also equivalent to: If $\displaystyle \inf A\notin A$, then $\displaystyle A$ is not complete.

However, in the opposite direction nothing can be said about completeness of $\displaystyle A$ from the fact that $\displaystyle \inf A$ exists. For $\displaystyle A$ as described, $\displaystyle \inf A$ always exists, but $\displaystyle A$ may not be complete.

The reason that $\displaystyle A$ complete implies $\displaystyle \inf A\in A$ is this. There is always a sequence inside $\displaystyle A$ that converges to $\displaystyle \inf A$, regardless of whether $\displaystyle \inf A\in A$ or not. If $\displaystyle A$ is complete, then by definition the limit of every sequence is in $\displaystyle A$, so $\displaystyle \inf A\in A$. However, $\displaystyle \inf A\in A$ means that only sequences that tend to $\displaystyle \inf A$ have their limit in $\displaystyle A$. Completeness is a much stricter condition because it requires that for all sequences, whatever their limit may be, this limit (if it exists) in in $\displaystyle A$.

In fact, in that Wikipedia article, the condition that $\displaystyle \inf A\in A$ is called Dedekind-completeness. Nevertheless, it is very different from the regular completeness, which is defined via sequences and limits.

I hope I this makes it clearer.