For a set S, there is always a sequence converging to sup(S)

[Mr Davis 97]

Homework Statement

[/B]
Let $S\subset \mathbb{R}$ be nonempty and bounded above. Show that there must exist a sequence $\{a_n\}_{n=1}^\infty\subseteq S$ such that $\lim_{n\to\infty}a_n=\sup(S)$.

Homework Equations

The Attempt at a Solution

Here is my idea. Let $\epsilon >0$. Then there must exist an $a \in S$ such that $\sup(S) - a< \epsilon$, by nature of $\sup(S)$ being the least upper bound.

So clearly we can always find an element $a$ that is arbitrarily close to $\sup(S)$. My problem is that I don't know how to make this formal. How would I construct the sequence necessary?

What about this: for each $n \in \mathbb{N}$ there must must exist an $a \in (\sup(S) - 1/n,~ \sup(S))$. And so we construct the sequence out of the $a$'s that must exist, so that $a_1 \in (\sup(S) - 1,~ \sup(S))$, $a_2 \in (\sup(S) - 1/2,~ \sup(S))$, and so on.

 

[LCKurtz]

That's the idea. So you have $a_n \in ( \sup S - \frac 1 n,\sup S)$. Now just finish:
To show $a_n \to \sup S$, suppose $\epsilon >0$. Then if $n > ~...$

 

[Mr Davis 97]

Homework Statement

[/B]
Let $S\subset \mathbb{R}$ be nonempty and bounded above. Show that there must exist a sequence $\{a_n\}_{n=1}^\infty\subseteq S$ such that $\lim_{n\to\infty}a_n=\sup(S)$.

Homework Equations

The Attempt at a Solution

Here is my idea. Let $\epsilon >0$. Then there must exist an $a \in S$ such that $\sup(S) - a< \epsilon$, by nature of $\sup(S)$ being the least upper bound.

So clearly we can always find an element $a$ that is arbitrarily close to $\sup(S)$. My problem is that I don't know how to make this formal. How would I construct the sequence necessary?

That's the idea. So you have $a_n \in ( \sup S - \frac 1 n,\sup S)$. Now just finish:
To show $a_n \to \sup S$, suppose $\epsilon >0$. Then if $n > ~...$

Well... Suppose that we construct the sequence $(a_n)$ by letting $a_n$ be the element in $(\sup(S) - \frac{1}{n}, \sup(S))$ that is guaranteed to exist by nature of the supremum. Then $\forall n \in \mathbb{N}$, $\sup(S) - a_n < \frac{1}{n}$.

Now, let $\epsilon > 0$. Then by the Archimedean principle there exists an $N \in \mathbb{N}$ such that $\frac{1}{N} < \epsilon$. If $n \ge N$, then $|a_n - \sup(S)| = |\sup(S) - a_n| = \sup(S) - a_n < \frac{1}{n} \le \frac{1}{N} < \epsilon$. This proves that $\lim a_n = \sup(S)$.

Is this a correct argument?

 

[ConfusedMonkey]

Why not just say that if there is no sequence convering to sup(S) then there is an epsilon nbd about sup(S) which contains no members of S, and this contradicts the fact that sup(S) is the least upper bound.

 

[Eclair_de_XII]

Why not just say that if there is no sequence convering to sup(S) then there is an epsilon nbd about sup(S) which contains no members of S, and this contradicts the fact that sup(S) is the least upper bound.

When I was doing this problem, I learned that if $sup(S)$ isn't an accumulation point of the sequence, then it is contained in the sequence.

 

[ConfusedMonkey]

When I was doing this problem, I learned that if $sup(S)$ isn't an accumulation point of the sequence, then it is contained in the sequence.

If sup(S) belongs to S then the proof is trivial, so assume sup(S) doesn't belong to S.

 

[member 587159]

Well... Suppose that we construct the sequence $(a_n)$ by letting $a_n$ be the element in $(\sup(S) - \frac{1}{n}, \sup(S))$ that is guaranteed to exist by nature of the supremum. Then $\forall n \in \mathbb{N}$, $\sup(S) - a_n < \frac{1}{n}$.

Now, let $\epsilon > 0$. Then by the Archimedean principle there exists an $N \in \mathbb{N}$ such that $\frac{1}{N} < \epsilon$. If $n \ge N$, then $|a_n - \sup(S)| = |\sup(S) - a_n| = \sup(S) - a_n < \frac{1}{n} \le \frac{1}{N} < \epsilon$. This proves that $\lim a_n = \sup(S)$.

Is this a correct argument?

Totally correct. You could have also used squeeze theorem to conclude.

 

[Mr Davis 97]

Totally correct. You could have also used squeeze theorem to conclude.

Awesome. Two questions: First, how can I justify that we can let $a_n$ be the element in $(\sup(S) - \frac{1}{n},~ \sup(S))$ that must exist? I feel like such a construction is very non-constructive, so I don't see why it's valid. Second, why in particular did the the value $\frac{1}{n}$ work? Is the only criteria that it converges to zero? Could I have just as easily replaced $\frac{1}{n}$ with $\frac{1}{n^2}$?

 

[LCKurtz]

Awesome. Two questions: First, how can I justify that we can let $a_n$ be the element in $(\sup(S) - \frac{1}{n},~ \sup(S))$ that must exist?

Not "the" element, "any" such element. You can pick one and call it $a_n$.

I feel like such a construction is very non-constructive, so I don't see why it's valid. Second, why in particular did the the value $\frac{1}{n}$ work? Is the only criteria that it converges to zero? Could I have just as easily replaced $\frac{1}{n}$ with $\frac{1}{n^2}$?

Yes.

 

[Mr Davis 97]

Not "the" element, "any" such element. You can pick one and call it $a_n$.Yes.

Why are we allowed to pick one and call it $a_n$?

Also, I feel like we could also show that there is always a nondecreasing sequence that converges to $\sup(S)$. I can visualize the process of assigning values in the same way as above, but instead of using $\frac{1}{n}$ we assign each element $a_n$ to be in the interval $(\sup(S) - a_{n-1}, \sup(S))$ so that $a_n \ge a_{n-1}$ and $a_n$ converges to $\sup(S)$. Would this work?

 

[member 587159]

Awesome. Two questions: First, how can I justify that we can let $a_n$ be the element in $(\sup(S) - \frac{1}{n},~ \sup(S))$ that must exist? I feel like such a construction is very non-constructive, so I don't see why it's valid. Second, why in particular did the the value $\frac{1}{n}$ work? Is the only criteria that it converges to zero? Could I have just as easily replaced $\frac{1}{n}$ with $\frac{1}{n^2}$?

(1) This is a very good question. Consider this lemma.

Let $A \subseteq\mathbb{R}$ be a non empty set with a LUB $c:= \sup A$.

For every $r >0$, there is $a \in A$ with $c \leq a + r$.

Proof: If not, there is r>0 s.t. for all $a \in A$ we would have $c > a + r$ or equivalently $c-r>a$, contradicting the definition of $c$.

Now, you can apply this lemma for $r = 1/n$ and $a = a_n$.

(2) Yes, any sequence $(b_n)$ converging to 0 works, because we have:

$$0\leq |\sup A - a_n| \leq b_n$$

By the squeeze theorem:

$$\lim_n |\sup A -a_n| = 0$$

Which clearly is equivalent with

$$\lim_n a_n = \sup A$$

 

[LCKurtz]

Why are we allowed to pick one and call it $a_n$?

You know there is an element in the interval so there may be many, but at least one. So pick one. There's no law against naming it, so call it something sensible like $a_n$.

Also, I feel like we could also show that there is always a nondecreasing sequence that converges to $\sup(S)$. I can visualize the process of assigning values in the same way as above, but instead of using $\frac{1}{n}$ we assign each element $a_n$ to be in the interval $(\sup(S) - a_{n-1}, \sup(S))$ so that $a_n \ge a_{n-1}$ and $a_n$ converges to $\sup(S)$. Would this work?

Yes, it could work, but you need to be careful. Just requiring $a_{n+1} > a_n$ might not be enough after you pick $a_1$. Do you see how you need to be careful if the set S has a cluster point $p < sup(S)$?

 

[Mr Davis 97]

You know there is an element in the interval so there may be many, but at least one. So pick one. There's no law against naming it, so call it something sensible like $a_n$.

Yes, it could work, but you need to be careful. Just requiring $a_{n+1} > a_n$ might not be enough after you pick $a_1$. Do you see how you need to be careful if the set S has a cluster point $p < sup(S)$?

We haven't defined cluster points yet, but I think I might see what you mean. Defined this way maybe $a_n$ will converge to some $p < \sup(S)$. I'm not seeing how to get $a_n$ to both converge and be nondecreasing using my idea...

 

[LCKurtz]

Think about choosing $a_{n+1} \ge \max\{a_n,\sup S - \frac 1 n\}$

 

[Mr Davis 97]

Think about choosing $a_{n+1} \ge \max\{a_n,\sup S - \frac 1 n\}$

So do we have essentially exactly what we had before, except we have $a_{n+1} \in ( \operatorname{max} \{ a_{n}, \sup(S) - \frac{1}{n} \} , \sup(S))$ instead of just $a_n \in (\sup(S) - \frac{1}{n}, \sup(S))$? For the latter part we have an inductive definition. What would the initial case be?