Home page for accesible maths 2 Sequences 2.2 Cauchy-sequences and the Bolzano-Weierstrass Theorem 2.4 Bounded and unbounded sequences

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2.3 The Least Upper Bound Principle. Maximum vs. Supremum

The Least Upper Bound Principle is a kind of a first principle in the theory of real analyis, so it deserves its own section. You surely understand the “principle” part, but the notion of “upper bound” needs to be explained. If $S$ is a set of real numbers, then $x$ is an upper bound of $S$ , if for any $s\in S$ $s\leq x$ . (similarly, $y$ is a lower bound of $S$ if for any $s\in S$ , $s\geq y$ .) E.g. if a sequence is bounded by the real positive number $M$ , then $M$ is an upper bound of the sequence and $-M$ is a lower bound of the sequence. The sequence $\{n^{3}\}^{\infty}_{n=1}$ has a lower bound (say $1$ or $-12$ ) but has no upper bound. On the other hand, the sequence $\{-n^{3}\}^{\infty}_{n=1}$ has an upper bound and has no lower bound. The following definition is more English than Math.

Definition 2.3.1

If $T$ is a set of real numbers, then $x\in T$ is the minimal element of $T$ if for any $s\in T$ , $x\leq s$ . Similarly, $y$ is the maximal element of $T$ if for any $s\in T$ , $y\geq s$ .

Of course, the minimal element of a set is a lower bound and clearly the best possible lower bound of a set. This means that if a set $T$ has no lower bound than it has no minimal element.

Nitpicking 2.3.1

Why is it not possible that a certain set $S$ has more than one minimal elements?

Note that there are sets having lower bound without having a minimal element. Say, the set of positive real numbers. $0$ is a lower bound of this set, but the zero is not IN the set. Obviously there is no such thing as the smallest positive number, since for any $x>0$ , $0<\frac{x}{2}<x$ . However, we have the following theorem.

Theorem 2.3.1 (Plain English Version of the Least Upper Bound Principle)

If
you have a set $S$ that has an upper bound, then there is a special guy $x$ that is greater or equal than all the elements of $S$ , but if $y$ is just a tad bit smaller than $x$ , then some elements of $S$ will be greater than $y$ .

Theorem 2.3.2 (The Least Upper Bound Principle)

If a non-empty set $S$ has an upper bound then the set $S$ has a least upper bound. That is, the set of all upper bounds of $S$ has a minimal element. Clearly, there can be only one such element. It is the only upper bound that is a limit point of a sequence from $S$ .

Proof: Let $S$ be our set having upper bound.

Lemma 2.3.1

Suppose that $x$ is an upper bound of $S$ . Then either $x$ is the least upper bound of $S$ or for any $\varepsilon>0$ there is an upper bound of $S$ , $y_{\varepsilon}<x$ , so that for some $s_{\varepsilon}\in S$ , $(y_{\varepsilon}-s_{\varepsilon})\leq\varepsilon$ .

Proof: Suppose that $x$ is not the least upper bound of $S$ . Take the sequence $x_{0}=x,x_{n}=x-n\varepsilon$ . Clearly, it is not possible that all of these $x_{n}$ ’s are upper bounds(this is obvious, but it is formally follows from the Law of Archimedes!!). Let $n$ be the largest integer such that $x_{n}$ is still an upper bound and set $x_{n}=y_{\varepsilon}$ . Obviously, $y_{\varepsilon}$ is an upper bound and there is an element of $S$ in between $y_{\varepsilon}-\varepsilon$ and $y$ . $\square$

By the lemma above, we have a sequence of upper bounds of $S$ $u_{1}\geq u_{2}\geq u_{3}\dots$ such that

•

There exists $s_{n}\in S$ such that $u_{n}-s_{n}\leq\frac{1}{10^{n}}$ .
•

$u_{n}-u_{n+1}\leq 10^{-n}$ .

Indeed, for some upper bound $x$ pick $y_{1/10}$ using the Lemma. This will be $u_{1}$ . Then, using the lemma again we pick $u_{2}$ (using $u_{1}$ as the” $x$ ”) and so on we construct the sequence $\{u_{n}\}^{\infty}$ . In the Non-Mathematical World this is the algorithm the get close to a wall in a dark room. You face the wall and makes a step of $\frac{1}{10}$ meter long. You make another step and another step. If you hit the wall, then you step back to the last position. Then you make steps $1/{100}$ -meter long and so on. Clearly, $\{u_{n}\}^{\infty}_{n=1}$ is a Cauchy-sequence and for its limit $u$ , $u\leq u_{n}$ for any $n\geq 1$ . Observe that $u$ is a least upper bound for $S$ !! First, it is really an upper bound, since if $u<s$ for some $s\in S$ then $u_{n}<s$ for some $n$ that is impossible. Also, there cannot be a smaller upper bound than $u$ . Indeed, $\{s_{n}\}^{\infty}_{n=1}$ is again a Cauchy-sequence by the Sum Rule having $u$ as its limit. $\square$

Note that there are two possibilites.

•

Either the least upper bound $u$ is in $S$ itself. Example: $S=[0,1]$ , where $1$ is the least upper bound.
•

Or $u$ is not in $S$ , but there is a convergent sequence in $S$ converging to $u$ . Example: $S=[0,1)$ , where $1$ is still the least upper bound.

Observe that the Bolzano-Weierstrass Theorem can be viewed as a trivial consequence of the Least Upper Bound Principle. If you have a bounded sequence $\{x_{n}\}^{\infty}_{n=1}$ then it has a smaller upper bound and this number is surely a limit of a subsequence of $\{x_{n}\}^{\infty}_{n=1}$ (yes, the $s_{n}$ guys will form the subsequence). We can also use the Least Upper Bound Principle to introduce the notion of “supremum”.

Definition 2.3.2 (Supremum)

Let $S$ be a set of real numbers having an upper bound. Then the least upper bound of $S$ $u$ is called the supremum of $S$ . If $u$ is in the set $S$ , then we call $u$ the maximum of $S$ .

A bounded set always have a supremum, but sometimes it has no maximum. Finite sets of course, always has a maximum, the greatest number of the set. Similarly, one can define the Largest Lower Bound of a set.

Theorem 2.3.3 (The Largest Lower Bound Principle)

If a non-empty set $S$ has a lower bound then the set $S$ has a largest lower bound. That is, the set of all lower bounds of $S$ has a largest element. Clearly, there can be only one such element.

Proof: The set of lower bounds has an upper bound (the elements of the set $S$ !!). So the lower bounds have a supremum. Clearly, this supremum is the largest lower bound. Alternatively, let $x$ be the Least Upper Bound of the set $-S$ ( $s\in-S$ if and only if $-s\in S$ ). Then $-x$ is the Largest Lower Bound of the set $S$ . $\square$

Definition 2.3.3 (Infinum)

Let $S$ be a set of real numbers having a lower bound. Then the largest lower bound of $S$ $l$ is called the infinum of $S$ . If $l$ is in the set $S$ , then we call $l$ the minimum of $S$ .

You can prove the Bolzano-Weierstrass Theorem from the Largest Lower Bound Principle as well. It is interesting to see that if the bounded sequence $\{x_{n}\}^{\infty}_{n=1}$ is not convergent. Then the two convergent subsequences obtained by the two principles are converging to different limits!

Example 2.3.1

Let $x_{n}=0$ if $n$ is even and let $x_{n}=1$ if $n$ is odd. So, we have the bounded sequence $1,0,1,0,1,\dots$ . Then the Least Upper Bound Principle gives us the subsequence $\{x_{2n+1}\}_{n=1}^{\infty}$ converging to $1$ and the Largest Lower Bound Principle gives us the subsequence $\{x_{2n}\}_{n=1}^{\infty}$ converging to $0$ .

Remark 2.3.1

(The Thing You Should Always Remember About The Least Upper Bound Stuff) The whole point of introducing infinite decimals was to have a number system with the Least Upper Bound Principle. Let $x\in\mathbb{R}$ be an arbitrary real number. Let $S_{x}$ be the set of rational numbers that are smaller than $x$ . It is important to see that $x$ is the least upper bound of $S_{x}$ . Clearly, $x$ is an upper bound for $S_{x}$ and if $y<x$ , then there exists $s\in S_{x}$ such that $y<s<x$ so $y$ is not an upper bound for $S_{x}$ .