# Need help understanding a proof on the least upper bound property of the reals.

I'm trying to learn from a book called "Vector Calculus, Linear Algebra, and differential forms".

In chapter 0.5, there's a proof on the least upper bound property of reals. I'm learning the material by myself, so I don't have anyone to ask about what the author is saying on the proof.

The Proof goes on like this: (the bold texts are written by the author, and the normal-texts are my questions)

Need to show: Every nonempty subset $X\subset \mathbb{R}$ that has an upper bound has a least lower bound $\sup X$ (*Q1. It doesn't mean $\sup X \in X$ in all cases, does it? I need to be sure).

Proof. We'll construct successive decimals of supX

Suppose $x \in X$ (since $X \neq \varnothing$ , $\exists x \in X$), and $a$ is an upper bound of $X$. We'll assume that $x >0$.

If $x = a$, then supX = a. If $x \neq a$ (which means $x < a$), there is a largest $j \in \mathbb{N}$ such that $[x]_{j} < [a]_{j}$

(* I understood this with this example: Suppose x = 3.127898... and a = 3.129876... Then, $[x]_{-2}$ = 3.12, $[a]_{-2}$ = 3.12 BUT $[x]_{-3}$ = 3.127 < 3.129 = $[a]_{-3}$ Thus, in this case, $j = -3$

But when $x_{2} = 35.054345$ and $a_{2} = 100.34523$, then, $[a_{2}]_{2} = 100 > 35 = [x_{2}]_{2}$, and in this case, $j = 2$

)

There are 10 numbers that have the same $k_{2}$th digit as x for $k_{1} > j$ and that have 0 as the $k_{2}$ th digit for $k_{2} < j$

(* Following the example above with x = 3.127898... and j=-3, I understood those 10 numbers to be: $n_{0}$ = 3.12000000 $n_{1}$ = 3.12100000... $n_{2}$ = 3.12200000... ... $n_{9}$ = 3.1290000...)

Consider those that are in $[[x]_{j},a]$ (* I thought the author meant "Among $n_{0},...,n_{9}$,consider those ...)

This set is not empty, since $[x]_{j}$ is one of them. (* I guessed "this set" to be: {$n_{0},n_{1},n_{2},...,n_{9}$}$\cap$ $[[x]_{j},a]$, and since $[x]_{j} \in \left \{n_{0},n_{1},...,n_{9}\right \}$ "this set" is not empty.)

Let $b_{j}$ be the largest such that $X \cap [b_{j},a] \neq \varnothing$; such a $b_{j}$ exists, since $x \in X \cap [[x]_{j},a]$. (*I'm confused about what this is saying...is $b_{j}$ one of $n_{1},...,n_{9}$ or is it just another real number?)

Now, consider the set of numbers in $[b_{j},a]$ that have the same kth digit as $b_{j}$ for $k>j-1$ and 0 for $k<j-1$. This is a non-empty set with at most 10 elements, and $b_{j}$ is one of them (the smallest). Call $b_{j-1}$ the largest such that $X \cap [b_{j-1},a] \neq \varnothing$. Such a $b_{j-1}$ exists, since if necessary we can choose $b_{j}$. Keep going this way, defining $b_{j-2},b_{j-3},$ and so on.

Let b be the number whose nth decimal digit (for all n) is the same as the nth decimal digit of $b_{k}$.

We claim that b = supX. Indeed, if $\exists y \in X$ with $y > b$, then there is a first digit k of y that differs from the kth digit of b, and $b_{k}$ was not the largest number with k digits that is not an upper bound, since using the kth digit of y would give a bigger one. (* ok...I'm not 100% getting what the author meant here...)

So b is an upper bound. Now, suppose that b' < b is also an upper bound. Again, there is a first digit k of b that differs from the kth digit of b'. This contradicts the fact that $b_{k}$ was not an upper bound, since then $b_{k} > b$.

gosh that was long...anyway, help me understand this proof! Thanks :D

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The fact that every set that is bounded above (below) has a least upper bound (resp. greatest lower bound) is an axiom, what are you trying to prove? Perhaps if you say you want to prove that $X$ implies the least upper bound then it is different. – user38268 Mar 12 '12 at 7:28

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