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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. 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.2012-03-12
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    use dollar signs \$ to indicate your latex... I fixed it, but you might want to go in a pretty it up a bit more for readability.2012-03-12
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    @BenjaminLim I think the proof is about "If a set of real numbers is bounded above, then the set has a least upper bound" - I think the author is trying to prove the axiom itself (and thus treat it as a theorem)2012-03-12
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    @alittledon thanks :D2012-03-12
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    If there is an upper bound, then there is a least upper bound. If there is a lower bound, then there is a greatest lower bound. If there is a bound (upper or lower) on a subset of the reals, then it acts as a boundary or border, past which, you can be sure the subset never ventures. But that "point of no return" is unique.2012-03-12
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    About Q1: no, the supremum and infimum of $X$ need not be in $X$. Consider $X=\left\{\frac1n\mid 0< n\in N\right\}$ and $Y=\left\{1-\frac1n\mid 0< n\in N\right\}$. Then $\inf X=0\not\in X$ and $\sup Y=1\not\in Y$.2012-03-12
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    @user269334 What is the definition of an axiom?2012-03-12
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    @BenjaminLim: the completeness axiom is an axiom if you want it to be. But one doesn't usually go about real numbers claiming their existence abstractly. Rather, they are constructed via several different methods, and each construction requires a *proof* of the completeness "axiom".2012-03-12

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