Prove sequence converges to sup

[CAF123]

Homework Statement

Suppose that E is contained in $\mathbb{R}$ is a nonempty bounded set and that $\sup E$ is not in E. Prove that there exists a strictly increasing sequence $\left\{x_n\right\}$ that converges to $\sup E$ such that $x_n \in E$ for all n in $\mathbb{N}$.

Homework Equations

More like Relevant theorems:
Bolzano-Weirstrass,
Completeness,
Monotone convergence.

The Attempt at a Solution

E is contained in R and nonempty so by Completeness, sup E exists. E is bounded so -a < e < a for all e in E and let supE = a which is not in E. I suppose this might be the essence of the proof, but how exactly do we know such a sequence in E exists? If there existed a function $f: \mathbb{N} \rightarrow \mathbb{R}$ with all the $x_n$ in E, then provided I can show such a sequence is strictly increasing (and that it exists) then the rest of the problem becomes trivial. I thought about using nested intervals but I didn't get very far. Any ideas?

Many thanks.

 

[christoff]

Here's an outline/suggestion on how to proceed: let $sup(E)=s_E$.
1) Can you show that given any $\epsilon>0$, there exists an $x\in E$ such that $s_E-\epsilon<x$?
2) Consider the sequence $\epsilon_n=\frac{1}{n}$; apply this sequence of epsilons to part 1 to show that you can generate a sequence in $E$, call it $x_n$.
3) Show that this sequence converges to $sup(E)=s_E$.

Hint for part 1): this is a good exercise in proof by contradiction which teaches you how to negate quantifiers; "for all"'s become "there exists" and vice-versa, and statements get negated.

 

[CAF123]

Here's an outline/suggestion on how to proceed: let $sup(E)=s_E$.
1) Can you show that given any $\epsilon>0$, there exists an $x\in E$ such that $s_E-\epsilon<x$?

Yes, this is the Approximation property for suprema

2) Consider the sequence $\epsilon_n=\frac{1}{n}$; apply this sequence of epsilons to part 1 to show that you can generate a sequence in $E$, call it $x_n$.

I can make the following statements: supE - 1 < x for some x, supE - 1/2 < x for some x, etc.. So supE-1, supE-1/2,...supE-1/n.. would make a sequence and indeed it would be strictly increasing. My only worry is that how can I be sure that the elements would be in E?

 

[christoff]

Yes, this is the Approximation property for supremaI can make the following statements: supE - 1 < x for some x, supE - 1/2 < x for some x, etc.. So supE-1, supE-1/2,...supE-1/n.. would make a sequence and indeed it would be strictly increasing. My only worry is that how can I be sure that the elements would be in E?

The numbers $Sup(E)-1, Sup(E)-1/2, Sup(E)-1/3,...$ aren't the numbers you're interested in; it's the x's that bound them that you want to look at.

Let's start at n=1. You said that there exists an $x\in E$ such that $Sup(E)-1<x$. Let's call this x we found $x_1$.
Now, n=2. We know that there exists an $x\in S$ such that $Sup(E)-1/2<x$. Let's call this second x we've found $x_2$.
Then we could do the same for n=3, n=4, etc.

The important part thing is we can do this for ANY n; so in particular, there exists a sequence $x_n\in E$ with the property that $Sup(E)-1/n<x_n$. Can you show that this sequence converges? Hint below if you can't get it.

Spoiler

You also know that, because $Sup(E)$ is the supremum, that each $x_n$ is bounded by $Sup(E)$. So you have $Sup(E)-1/n<x_n\leq Sup(E)$.

 

[CAF123]

The numbers $Sup(E)-1, Sup(E)-1/2, Sup(E)-1/3,...$ aren't the numbers you're interested in; it's the x's that bound them that you want to look at.

Let's start at n=1. You said that there exists an $x\in E$ such that $Sup(E)-1<x$. Let's call this x we found $x_1$.
Now, n=2. We know that there exists an $x\in S$ such that $Sup(E)-1/2<x$. Let's call this second x we've found $x_2$.
Then we could do the same for n=3, n=4, etc.

The important part thing is we can do this for ANY n; so in particular, there exists a sequence $x_n\in E$ with the property that $Sup(E)-1/n<x_n$. Can you show that this sequence converges? Hint below if you can't get it.

Spoiler

You also know that, because $Sup(E)$ is the supremum, that each $x_n$ is bounded by $Sup(E)$. So you have $Sup(E)-1/n<x_n\leq Sup(E)$.

Ok so the last statement implies $$\lim_{n \rightarrow \infty} \sup E - \lim_{n \rightarrow \infty} \frac{1}{n} \leq \lim_{n \rightarrow \infty} x_n$$ From this $x_n$ must converge to $\sup E$. It caanot be greater because the $x_n$ are bounded above by $\sup E$. Is this all I need to do? I haven't checked your hint yet, I prefer to have feedback first.

 

[christoff]

Ok so the last statement implies $$\lim_{n \rightarrow \infty} \sup E - \lim_{n \rightarrow \infty} \frac{1}{n} \leq \lim_{n \rightarrow \infty} x_n$$ From this $x_n$ must converge to $\sup E$. It caanot be greater because the $x_n$ are bounded above by $\sup E$. Is this all I need to do? I haven't checked your hint yet, I prefer to have feedback first.

Yes, that is basically the idea. Since the sequence $sup(E)-1/n$ converges to $sup(E)$, and $x_n$ is bounded above by the $sup(R)$ and below by this $sup(E)-1/n$, it must converge to $sup(E)$.