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

[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.

 

[ZioX]

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.

 

[cooljosh2k2]

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.

 

[ZioX]

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

 

[cooljosh2k2]

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.

 

[ZioX]

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

 

[xento]

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?

 

[jgens]

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.

 

[cooljosh2k2]

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

 

[cooljosh2k2]

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?

 

[xento]

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?

 

[ZioX]

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.

 

[cooljosh2k2]

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

 

[ZioX]

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?

 

[cooljosh2k2]

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.

 

[ZioX]

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.

 

[cooljosh2k2]

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

 

[ZioX]

What do you find yourself not believing in your proof?

 

[cooljosh2k2]

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

 

[ZioX]

Only you can put doubt in your mind.

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

 

[cooljosh2k2]

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?

 

[xento]

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.

 

[cooljosh2k2]

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.