# Show that if a ∈ R, then: sup{r ∈ Q : r < a} = a

• cooljosh2k2
In summary, the conversation discusses how to prove that if a is an element of the set of real numbers, then the supremum of the set {r ∈ Q : r < a} is equal to a. The conversation covers various approaches and proofs, including using the density of rationals in the reals and starting from the definition of supremum. It is ultimately determined that starting from the definition is the most effective approach.
cooljosh2k2

## Homework Statement

Show that if a ∈ R, then:
sup{r ∈ Q : r < a} = a .

## The Attempt at a Solution

Seems like a very simple question, but the r is throwing me off. Do you prove it the same way as a natural number using an epsilon? For example:

If a is an upper bound of S and if v < a, then we put ε = a - v... etc.

or is there a different way to prove a supremum with a rational number.

The definition of supremum is invariant under any choice of poset.

So yes. You prove it the same way. You'll be wanting to use some equivalency or consequence of the fact that the rationals are dense in the reals.

Like for example:

we know that ε = sup {r∈Q : r < a} exists.

Let ε ≤ a because a is an upper bound of S. Then say that ε < a, then ε < (a + ε)/2 < a

Which gives a contradiction and therefore proves that a is the supremum of S.

How do you know the supremum exists? Important question.

It might help to solve the problem in reals first and then realize how you need to modify the proof to make it work for the rationals. That is, can you prove that $$a=\sup \{x\in \bold{R} : x < a \}$$?

Isnt it obvious that a supremum exists? The conditions state that r < a, therefore, r has an upper bound which therefore means that it has a supremum. a is the max, we're just trying to prove that its the surpremum.

These things are not trivial. Existence of supremums and infimums need to be justified. For the real numbers with standard ordering it is sufficient for a set to be bounded above in order to have a supremum. I realize that you may already know this but it is a very important concept.

Do note that neither $$\{x \in \bold{R} : x < a \}$$ and $$\{x \in \bold{R} : x < a \}$$ have maxima.

I strongly suggest you prove $$\sup \{x \in \bold{R} : x < a \}=a$$.

Would it work if we proved:

$$\sup \{x \in \bold{R} : x < a \}=a$$

and then just state that since

$$\{r \in \bold{Q} : r < a \}$$ is a subset of $$\{x \in \bold{R} : x < a \}$$

The proof is complete?

xento said:
Would it work if we proved:

$$\sup \{x \in \bold{R} : x < a \}=a$$

and then just state that since

$$\{r \in \bold{Q} : r < a \}$$ is a subset of $$\{x \in \bold{R} : x < a \}$$

The proof is complete?

No. The set {a-1} is a subset of {x: x < a} but it has a different supremum.

You guys lost me. Can somebody please explain? I am really trying to learn this, my textbook sucks!

K i received help from a friend who took the course, he's not sure if he's right though.

He says that since Q is dense in R, every nonempty interval of R contains a rational.

From the definition of the set S = sup{r ∈ Q : r < a} it follows a is an upper bound of the set S. For every ε >0, the density of the rationals in R shows there is a rational r in (a - ε, a). So, a - ε < r < a, which shows a = sup S is the least upper bound of S.

Is this approach correct?

I'm pretty lost myself. Here's another idea I had.

Density Theorem: if $$x$$ and $$y$$ are any real numbers with $$x < y$$ , then there exists a rational number $$r \in \bold{Q}$$ such that $$x < r < y$$.

Let $$X=\{x \in \bold{R} : x < a \}$$
Let $$Y=\{r \in \bold{Q} : r < a \}$$

If $$a$$ is an upper bound of $$X$$ such that $$x \le a$$ for all $$x \in X$$ and if $$u < a$$ then we have $$\epsilon := a-u$$. Then $$\epsilon > 0$$ so there exists $$x' \in X$$ such that $$u = a-\epsilon < x'$$. Therefore, $$u$$ is not an upper bound of $$X$$ and we conclude that $$\sup (X)=a$$. But, since $$x'$$ and $$a$$ are both real numbers, the density theorem says that there exists a rational number $$r \in \bold{Q}$$ such that $$x'<r<a$$ and since $$r<a$$ then this r is also $$r \in Y$$. Thus, $$\sup (Y)=a$$.

Would this be sufficient to conclude the proof?

You need to start proving things from the definition. When you do that, you will not have to ask whether a proof is correct or not.

The case for R follows directly from the definition of the supremum.

Suppose a is the supremum of $$\{x \in \bold{R} : x < a\}$$. Further, suppose b is another upper bound for this set. From the definition of an upper bound $$x \le b \forall\, x \in \{x \in \bold{R} : x < a\}$$. Therefore, $$b \ge a$$ as otherwise (a<b) would imply that b is no longer an upper bound.

Last edited:
What about the proof i posted above, also no good?

cooljosh2k2 said:
K i received help from a friend who took the course, he's not sure if he's right though.

He says that since Q is dense in R, every nonempty interval of R contains a rational.

From the definition of the set S = sup{r ∈ Q : r < a} it follows a is an upper bound of the set S. For every ε >0, the density of the rationals in R shows there is a rational r in (a - ε, a). So, a - ε < r < a, which shows a = sup S is the least upper bound of S.

Is this approach correct?

Where have you shown that any upper bound for S must be greater than or equal to a?

Havent i showed that? I show that a is the least upper bound of S, so any other upper bound has be greater or equal to a.

cooljosh2k2 said:
Havent i showed that? I show that a is the least upper bound of S, so any other upper bound has be greater or equal to a.

I don't know, did you?

You said something that certainly gives that consequence but when you prove things, especially when you're doing graded work, you should show the full inferences to arrive at the final result. If you don't explicitly make them conform to the definitions you should at the very least give some remark towards noting that you have accomplished what was required to be shown.

Also note the different routes we took for our proofs. You and xento showed that there are no upper bounds below a. I showed any upper bound must be greater than or equal to a. These are, of course, equivalent approaches. I used the fact that there are no upper bounds below a contradicts the assumption that b is an upper bound.

Last edited:
So what am i missing, it seems right to me. I am starting to get frustrated with the question :S

What do you find yourself not believing in your proof?

Nothing, but you put a doubt in my mind, and now I am not sure if I am missing something.

Only you can put doubt in your mind.

Go over the definitions and the steps and try to convince yourself it's right.

Ok i will, but if i find that i think its right, am i right, lol. I mean there must be a right answer, if i think mine is right, it doesn't mean I am right. Ok, now I am just confusing myself... Can you please just tell me if I am right or wrong, and if I am wrong ill go back and see what's wrong?

What I find hard from studying real analysis is that the type of questions can vary tremendously, and there are only limited examples in textbooks. Also, it is not like doing calculus for example where you do problems, check the answers and build up your confidence as you get more and more right answers. In analysis, getting validation for your solutions is hard so I feel the learning curve is steeper than it should be since you don't really know where you stand.

xento said:
What I find hard from studying real analysis is that the type of questions can vary tremendously, and there are only limited examples in textbooks. Also, it is not like doing calculus for example where you do problems, check the answers and build up your confidence as you get more and more right answers. In analysis, getting validation for your solutions is hard so I feel the learning curve is steeper than it should be since you don't really know where you stand.

Couldnt have said it better myself, the way i usually study math is by trying some examples, looking at the answer, and then learning from it. But there are no examples to learn from. So I am having a lot of trouble understanding this class, as apposed to other math classes.

## 1. What is the definition of the supremum of a set?

The supremum of a set is the least upper bound of the set, meaning it is the smallest number that is greater than or equal to all the elements in the set.

## 2. How does this statement relate to real numbers and rational numbers?

This statement is specifically referring to real numbers and rational numbers. The set in question is the set of all rational numbers that are less than the given real number, and the supremum of this set is equal to the given real number.

## 3. Why is it important to prove this statement?

Proving this statement is important because it helps to establish the relationship between real numbers and rational numbers. It also helps to solidify the concept of supremum and its definition.

## 4. Can you provide an example to illustrate this statement?

Yes, for example, if a = 2, then the set of all rational numbers less than 2 would be {1, 1.5, 1.9, 1.99, 1.999, ...}. The supremum of this set would be 2, which is the given value of a.

## 5. Is this statement always true for any real number a?

Yes, this statement is always true for any real number a. This is because the supremum of a set is always a real number, and in this case, the set in question only contains rational numbers that are less than the given real number a. Therefore, the supremum of this set will always be equal to a.

Replies
7
Views
2K
Replies
8
Views
2K
Replies
14
Views
2K
Replies
3
Views
1K
Replies
9
Views
2K
Replies
7
Views
2K
Replies
1
Views
1K
Replies
12
Views
2K
Replies
1
Views
5K
Replies
1
Views
1K