Bounded sets do have a least upper bound

I think this is very fundamental property of real numbers, as it allows us to discuss about limits. I do not know, how the limits are playing an important role to complete the following one. Please explain.

" Any non-empty set of real numbers which is bounded above has a supremum"

This is available in standard books. But, how to complete the proof of the cited above statment by introducing LIMITS, I don't know. Advanced thanks to MSE members, who can help me.

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3 Answers

The fact that a nonempty set of real numbers bounded above has a supremum is often taken as an axiom in analysis courses. One can construct the set of real numbers from the set of rational numbers via Dedekind cuts, and then prove this as a theorem instead. This is done for example in Rudin's Principles of Mathematical Analysis, and the proof does not have anything to do with limits at all!

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!We need a good notation for a real number given by its decimal representation. A real number has the form a =$a_0$ $a_1$$a_2..., where a_0 is an integer a_1$$a_2$... belongs to {1, 2, 3,...,9}. To eliminate ambiguity in defining real numbers by their decimal representation,let us decide that if the sequence of decimals ends up with nines:a =$a_0$ $a_1$$a_2...a_n9999... (where a_n < 9). then we choose this number’s decimal representation as a =a_0 a_1$$a_2$...($a_n$ +1)0000...(For example, instead of 0.4999999.. we write 0.5.). – gama Dec 3 '12 at 4:46
Let S be a nonempty set of real numbers, bounded above. Let us construct the least upper bound of S. Consider first all the approximations by integers of the numbers a of S: if a = $a_0$$a_1$...collect the $a_0$'s. This is a collection of integer numbers. It is bounded above (by assumption). Then there is a largest one among them, call it $B_0$. Then what to do I don't know.If nay one can help me in this regard, I am so thankful to them. – gama Dec 3 '12 at 4:48

The proof of existence of supremums that I saw relied on the completeness of the real numbers, and not much else. Basically, we constructed a Cauchy (and thus convergent) sequence of real numbers that was always an upper bound of the set, and got infinitely close to the set.

Here's an outline:

Theorem. Let $U \subset \mathbb{R}$ be a bounded set of real numbers. Then there exists $s \in \mathbb{R}$ such that for all $u \in U$, $s \geq u$, and for any $\varepsilon > 0$, $s - \varepsilon$ is not an upper bound of $U$.

Let $M > 0$ be the upper bound for $U$ (for example, if $U = (0,1)$, we could say $M = 1000000000$, and it would be fine).

Take $T_1 := M$, and $B_1$ to be some number that is not an upper bound for $U$.

Now, we given $T_i$ and $B_i$, we will define $T_{i+1}$ and $B_{i+1}$ as follows:

We take the midpoint of $T_i$ and $B_i$ to be called $m_i$, and see if $m_i$ is an upper bound for $U$. If it is, then we'll take $T_{i+1} := m_i$ and $B_{i+1} := B_i$. We define $a_i$ to be $m_i$

If $m_i$ is not an upper bound for $U$, then we define $a_i := a_{i-1}$, and $T_{i+1} := T_i$, and $B_{i+1} := B_i$.

The distance between $T_i$ and $B_i$ halves for each iteration, so since $a_i$ is contained in the interval, it is squeezed into convergence. Its limit $a$ must be an upper bound, and the claim (I will leave for you to play with) is that $a$ is the least upper bound of the set $U$. $\Box$

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The hypotheses of your theorem are not clear to me. What are you taking as the definition/axioms of the real number system? $\mathbb R$ is an Archimedean ordered field such that every Cauchy sequence converges? – Jonas Meyer Dec 3 '12 at 5:29
What is used here is that $\mathbb{R}$ is a complete (every Cauchy sequence converges), ordered field. The Archimedian property is not relevant to this proof, but yes, I do take that as a part of the definition. – andybenji Dec 3 '12 at 5:41
The Archimedean property is used in your proof, and it must be for the proof to be correct, because an ordered field in which Cauchy sequences converge need not have the least upper bound property. (At least one example can be found in the book Counterexamples in Analysis.) Where you used it is in concluding that $(a_i)$ is a Cauchy sequence from the fact that the intervals get cut in half. $\lim\limits_{n\to \infty}\dfrac{1}{2^n}=0$ is equivalent to the Archimedean property. – Jonas Meyer Dec 3 '12 at 5:46
Oh, my bad. So yes, that is my definition of the real numbers. – andybenji Dec 3 '12 at 6:58
@andybenji! My question is quite different. Please see my comments, which are given above. – gama Dec 3 '12 at 6:58

Let $U\subset\mathbb{R}$ be bounded above. Then there exists $x'\in\mathbb{R}$ such that $u\leq x$, for all $u\in U$. Therefore, $x'$ is the upper bound of $U$. Let, $X=\{x\in \mathbb{R}:u\leq y\,for\,u\in U\}$ be the set of all upper bounds of $U$. Then obviously $x'\in X$ and $x^*=\min X$ is the least upper bound of $U$. As $u\leq x^*$ for all $u\in U$ and $x^*\leq x$, for all upper bounds $x$ of $U$.

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