Introductory Real Analysis Problem

[zeronem]

Homework Statement

If x is any real number, prove that

$$x=sup\{ r \in Q: r<x\}$$ $$=inf\{ s \in Q: x<s\}$$

Relevant Equations

$$r<x<s$$
$$s-r>0$$
$$n(s-r)>1$$
$$nr\in[m,m+1)$$

$$r<x<s$$
$$s-r>0$$

We enploy the Archimedean principle where

$$n(s-r)>1$$

We employ density of rationals where

$$\exists [m,m+1] \in Q$$

Such that

$$nr\in [m,m+1)$$
Therefore
$$m\leq nr \lt m+1$$$$ \frac m n \leq r \lt \frac m n + \frac 1 n $$

Since

$$ \frac m n \leq r $$

Then

$$ \frac m n \leq r \lt r + \frac 1 n \lt s $$

Since $$r \lt x \lt s$$

two possibilities:

First possibility:

$$ r < x \leq \frac m n + \frac 1 n \lt s $$

Where

$$ \forall x \in R $$

$$ \exists\{ \frac m n + \frac1 n\} \in Q$$ such that $$x \leq \frac m n + \frac 1 n $$

Or

Second possibility:

$$ r < \frac m n + \frac 1 n \leq x \lt s $$

Where

$$ \forall x \in R$$

There $$ \exists \{ \frac m n + \frac 1 n \} \in Q $$ such that $$ \frac m n + \frac 1 n \leq x $$

Therefore

$$x=sup\{ r \in Q : r<x\}= inf\{ s \in Q : x<s\}$$

QED

Is this a valid argument?

 

[PeroK]

I find your working is impossible to follow. You don't state your assumptions, how the variables you use are defined, nor what theorems or results you are invoking.

My guess is that you have shown:

$\forall \ x \in \mathbb{R}, \ \exists \ q \in \mathbb{Q}: \ q \le x$

In which case, no that is not a proof.

 

[zeronem]

I find your working is impossible to follow. You don't state your assumptions, how the variables you use are defined, nor what theorems or results you are invoking.

My guess is that you have shown:

$\forall \ x \in \mathbb{R}, \ \exists \ q \in \mathbb{Q}: \ q \le x$

In which case, no that is not a proof.

I’m using the Archimedean principle with and the density of rationals for my proof. Not sure where you are getting “q”

 

[PeroK]

I’m using the Archimedean principle with and the density of rationals for my proof. Not sure where you are getting “q”

Can you annotate your working?

 

[WWGD]

I’m using the Archimedean principle with and the density of rationals for my proof. Not sure where you are getting “q”

What is the m you're referring to? Do you mean there is an m with that property or you have a specific m in mind? Edit: can you write the gist of your argument?

 

[zeronem]

Can you annotate your working?

Yea hold on. Took me forever to type this out haha. Let me show you where Archimedean principle and density of rationals come into employment. I’ll do it though edit

 

[PeroK]

Yea hold on. Took me forever to type this out haha. Let me show you where Archimedean principle and density of rationals come into employment. I’ll do it though edit

The formulas are the time-consuming things to type. Saying what you are doing shouldn't take long.

 

[zeronem]

The formulas are the time-consuming things to type. Saying what you are doing shouldn't take long.

Ok I put it in the edit

 

[PeroK]

Homework Statement: If x is any real number, prove that

$$x=sup\{ r \in Q: r<x\}$$ $$=inf\{ s \in Q: x<s\}$$
Homework Equations: $$r<x<s$$
$$s-r>0$$
$$n(s-r)>1$$
$$nr\in[m,m+1)$$

$$r<x<s$$
$$s-r>0$$

We enploy the Archimedean principle where

$$n(s-r)>1$$

We employ density of rationals where

$$\exists [m,m+1] \in Q$$

Such that

$$nr\in [m,m+1)$$
Therefore
$$m\leq nr \lt m+1$$$$ \frac m n \leq r \lt \frac m n + \frac 1 n $$

Since

$$ \frac m n \leq r $$

Then

$$ \frac m n \leq r \lt r + \frac 1 n \lt s $$

Since $$r \lt x \lt s$$

two possibilities:

First possibility:

$$ r < x \leq \frac m n + \frac 1 n \lt s $$

Where

$$ \forall x \in R $$

$$ \exists\{ \frac m n + \frac1 n\} \in Q$$ such that $$x \leq \frac m n + \frac 1 n $$

Or

Second possibility:

$$ r < \frac m n + \frac 1 n \leq x \lt s $$

Where

$$ \forall x \in R$$

There $$ \exists \{ \frac m n + \frac 1 n \} \in Q $$ such that $$ \frac m n + \frac 1 n \leq x $$

Therefore

$$x=sup\{ r \in Q : r<x\}= inf\{ s \in Q : x<s\}$$

QED

Is this a valid argument?

No, that's not valid. What I think you've shown is:

$\forall r, s \$ where $r < x < s \$ then $\exists q \in \mathbb{Q}$ such that either:

$r < q \le x < s$ or $r < x \le q < s$

I think the proof you have of that is valid. But, that's not what you are trying to prove.

Anyway, I'll make the following suggestions:

1) I don't think it's a good idea to try to prove both at the same time. Prove the sup and inf cases separately.

2) It might be necessary to show that both sets are non-empty to start with.

3) Explicitly use the definition of sup and inf. Hint: sup is the least upper bound. You can see your proof is not right, because you never really used the definition of sup and inf.

4) You should start with let $x \in \mathbb{R}$. Logically that is the starting point. The sets are not defined until you have chosen $x$.

5) It's easier to give the sets names. E.g. $A$ and $B$.

6) Once you have something like $\frac m n + \frac 1 n$, it's better to call that $q$.

 

[zeronem]

No, that's not valid. What I think you've shown is:

$\forall r, s \$ where $r < x < s \$ then $\exists q \in \mathbb{Q}$ such that either:

$r < q \le x < s$ or $r < x \le q < s$

I think the proof you have of that is valid. But, that's not what you are trying to prove.

Anyway, I'll make the following suggestions:

1) I don't think it's a good idea to try to prove both at the same time. Prove the sup and inf cases separately.

2) It might be necessary to show that both sets are non-empty to start with.

3) Explicitly use the definition of sup and inf. Hint: sup is the least upper bound. You can see your proof is not right, because you never really used the definition of sup and inf.

4) You should start with let $x \in \mathbb{R}$. Logically that is the starting point. The sets are not defined until you have chosen $x$.

5) It's easier to give the sets names. E.g. $A$ and $B$.

6) Once you have something like $\frac m n + \frac 1 n$, it's better to call that $q$.

10-4 thanks for the suggestions. I’m self learning and I appreciate anyone who would be equivalently critical as if they were grading the answer in school.

Edit message: Now if I employed the definitions of LUB and GLB in the middle of the proof to give the professor the understanding of where I’m coming from would that make proof more valid above? As well as slip in the $$\forall x \in R $$ at the beginning?

 

[zeronem]

I ascertained two separate nonempty sets A and B and I was able to prove inf B = x but having trouble with sup A.

Where $$ (r,x) \subset A $$ and $$ (x,s) \subset B $$

I also decided to employ Archimedian principle and density of rationals to both sets separately to try to arrive to the conclusive equivalence. So only was capable of showing inf B = x.

My next attempt was to do $$ A \cup B \subset Q = r \lt x \lt s $$

Then employing archimedian principle with density of rationals as done previously in that manner while ascertaining the definitions of least upper bound and greatest least bound to show the equivalence. I’m not sure why I couldn’t get the equivalence when defining as separate sets. I know there has got to be multiple routes and the conclusions should arrive the same.

 

[PeroK]

I ascertained two separate nonempty sets A and B and I was able to prove inf B = x but having trouble with sup A.

Where $$ (r,x) \subset A $$ and $$ (x,s) \subset B $$

I also decided to employ Archimedian principle and density of rationals to both sets separately to try to arrive to the conclusive equivalence. So only was capable of showing inf B = x.

My next attempt was to do $$ A \cup B \subset Q = r \lt x \lt s $$

Then employing archimedian principle with density of rationals as done previously in that manner while ascertaining the definitions of least upper bound and greatest least bound to show the equivalence. I’m not sure why I couldn’t get the equivalence when defining as separate sets. I know there has got to be multiple routes and the conclusions should arrive the same.

None of that makes any sense to me. The proof should be very similar for sup and inf. What was your proof for the inf?

I'll show you how my proof would start.

Let $x \in \mathbb{R}$ and $A = \{ r \in \mathbb{Q}: r < x \}$

By definition, $x$ is an upper bound for $A$.

Let $y < x$ ...

 

[zeronem]

None of that makes any sense to me. The proof should be very similar for sup and inf. What was your proof for the inf?

I'll show you how my proof would start.

Let $x \in \mathbb{R}$ and $A = \{ r \in \mathbb{Q}: r < x \}$

By definition, $x$ is an upper bound for $A$.

Let $y < x$ ...

I actually had a proof started out like that and ascertained that x is an upper bound. So do I employ density of rationals to show that x is a least upperbound given there exists an element greater than x that belongs to set B? But is not in A.

I was able to show inf B = x, where $$ B = \{s \in Q : x \lt s \} $$ separately, but with sup A = x, I ran into a road block since there is no element in A that is greater than x unless you include set B. If x is an element in A and B then I guess I can show that there exists elements greater and less than x that show x is both a least upper bound in A and greatest least bound in B.

 

[PeroK]

I actually had a proof started out like that and ascertained that x is an upper bound. So do I employ density of rationals to show that x is a least upperbound given there exists an element greater than x that belongs to set B? But is not in A.

I was able to show inf B = x, where $$ B = \{s \in Q : x \lt s \} $$ separately, but with sup A = x, I ran into a road block since there is no element in A that is greater than x unless you include set B. If x is an element in A and B then I guess I can show that there exists elements greater and less than x that show x is both a least upper bound in A and greatest least bound in B.

$B$ is irrelevant to the proof that $x = sup A$ and vice versa.

From what you say here I suspect your proof that $x = inf B$ may be faulty.

A common approach to showing that $x = sup A$, where $A$ is any set is:

1) Confirm that $A$ is non-empty. Often trivial.

2) Show that $x$ is an upper bound for $A$.

3) Take any number $y < x$ and show that $y$ is not an upper bound for $A$.

These three steps constitute a proof that $x = sup A$.

With a similar process to show that $x = inf B$.

How you complete these steps depends, of course, on the sets in question.

 

[zeronem]

A least upper bound can be considered an upper bound correct? Or in the set of upper bounds. So if I show that x is an upper bound, then of course I have to show it is a least upper bound which seems easier than said.

If I show x is an upper bound in A then does that show that x is a least upper bound in Q? it sounds like my proof for it as a least upper bound is finished once I show that any element y is not an upper bound. Is the element y where I employ density of rationals to show that there exists an element $\gt x$ that is an upper bound, which then shows x can be a least upper bound?

What if i state: given

$$Q \subset R$$

$\forall x \in Q$ $\exists x \in (r,s)$

Let $x \in R$, where a nonempty subset A is bounded below in Q $\forall x \in A ; \{ r \lt x \}$

Then let r be a lower bound in A, then by definition x is greatest least bound in Q?

Now I’m just going off in a tangent with ideas. I’ll stick to individual set A and start off from your position see if can I come up with it.

Thanks for your continued help.

 

[PeroK]

What if i state: given

$$Q \subset R$$

$\forall x \in Q$ $\exists x \in (r,s)$

There are some basic concepts in real analysis that are not that easy to master on your own. In fact, they are not easy for students at university to master, despite having lectures and tutorials.

What you write generally is not precise mathematics. In particular, this construction is logically meaningless. You need to sort this out. Easier said than done, I know.

For this problem it's relatively easy to see that in some sense the sets A and B almost touch and $x$ is the point in the middle, being the supremum of A and the infimum of B. But, proving that rigorously is a different matter.

That said, go back to post #14 and try to follow the argument through.

 

[zeronem]

There are some basic concepts in real analysis that are not that easy to master on your own. In fact, they are not easy for students at university to master, despite having lectures and tutorials.

What you write generally is not precise mathematics. In particular, this construction is logically meaningless. You need to sort this out. Easier said than done, I know.

For this problem it's relatively easy to see that in some sense the sets A and B almost touch and $x$ is the point in the middle, being the supremum of A and the infimum of B. But, proving that rigorously is a different matter.

That said, go back to post #14 and try to follow the argument through.

I’ll admit that statement about Q does not convey the information I am trying to come across with. Basically trying to state that Q is a nonempty subset of R that is a bounded interval (r,s). I wrote it rather quickly without thinking much on it. I admit it needs revision. Basically I’m saying Q = (r,s) is contained in R if that’s legit. Where x is a real number that is a also element in the interval (r.s).

I’ll admit it isn’t easy self learning without a weekly professor critique and an everyday class to go to; but I’m intrigued by the subject and am determined to master it and contribute to it if I can.