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Archimedean Property. For any $x$, $\varepsilon$ in $\mathbb{R}$ with $\varepsilon >0$, there is an $n$ in $\mathbb{N}$ such that $n\varepsilon >x$.

Proof. If $x\leq 0$, just take $n=1$. So, assume $x>0$.

Consider $A=${$n\varepsilon: n$ in $\mathbb{N}$}. We can do show that $x$ is less than ar least one element of $A$. Suppose not.

Then, $x$ is an upper bound for $A$. So, by the Dedekind axiom, $A$ has a least upper bound, say $s$. By the definition, $s-1$ is not an upper bound because $s-1<s$. Then, there is an element in $A$ strictly greater than $s-1$. In other words, there is an $n$ in $\mathbb{N}$ such that $n\varepsilon>s-\varepsilon$. But the implies $(n+1)\varepsilon >s$. This is a contruduction because $(n+1)\varepsilon$ is in $A$ and $s$ is an upper bound of $A$. We are done.

I couldn't understand boldface sentences.How '' there is an element in $A$ strictly greater than $s-1$''? So, how ''there is an $n$ in $\mathbb{N}$ such that $n\varepsilon>s-\varepsilon$''? Finally, how did we get a contradiction? Can you explain clearly?

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    $\begingroup$ It has to read "$s-ε$ is not an upper bound because ..." The "$s-1$" is a mis-print. -- And think about what it means that "$s$ is the least upper bound". $\endgroup$ Commented Nov 13, 2016 at 17:35
  • $\begingroup$ @LutzL Yes, thanks. $\endgroup$
    – user295645
    Commented Nov 13, 2016 at 17:50
  • $\begingroup$ @LutzL I think the "s−1" is NOT a mis-print, the definition of the least upper bound says ε could be any positive real, thus permissible to choose ε=1 $\endgroup$
    – iMath
    Commented Nov 20, 2016 at 14:23
  • $\begingroup$ @imath: And what if $ ε=\frac13$? Then $nε>s-1$ does not automatically imply $(n+1)ε>s$. $\endgroup$ Commented Nov 20, 2016 at 14:30
  • $\begingroup$ @LutzL Yes , it is a bit hard to get that result , but just divide ε from both sides of nε>x, and you get n>x/ε, then see proof here en.wikibooks.org/wiki/Real_Analysis/… $\endgroup$
    – iMath
    Commented Nov 20, 2016 at 15:08

2 Answers 2

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Claim: Suppose that $s$ is a least upper bound of $S$. Then for every $\varepsilon>0$, there exists some $y\in S$ such that $s-\varepsilon<y\le s$

Suppose there were no such $y$. Then for all $y\in S$, we must have $y\le s-\epsilon$. This means that $s-\varepsilon$ is an upper bound for $S$, and as $\varepsilon>0$ we have $s-\varepsilon<s$. But we declared that $s$ is the least upper bound, so this doesn't make sense - $s-\varepsilon$ is smaller than $s$ but is still an upper bound.

So by contradiction there must be such $y$.

Note: This property is often known as supremum property, so you should make yourself very familiar with this property - it pops up almost everytime you deal with least upper bounds and usually no detailed explanation is given.


Now apply above to your $A$. Then it follows that there is some $y=n\varepsilon$ such that $n\varepsilon>s-\varepsilon$. Clearly this implies that $(n+1)\varepsilon>s$.

However, this is a nonsense. Since $n+1 \in\mathbb{N}$, by definition of $A$, $(n+1)\varepsilon$ is a member of $A$ and is still larger than $s$. But we said $s$ was an upper bound for $S$, so $s$ must be larger than (or equal to) everymember of $A$. So this contradicts the fact that $s$ is an upper bound.

Hence, we can conclude that $A$ is not bounded above (in particular by $x$). So there exist some element $n\varepsilon \in A$ such that $n\varepsilon>x$

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$s=\sup A$ is the least upper bound. As $ε>0$, $s-ε<s$ is not an upper bound. There are elements of $A$ that are above $s-ε$. However, all elements of $A$ are of the form $nε$, $n\in\Bbb N$. Thus $nε>s-ε$ for some $n\in \Bbb N$.

Now that implies $(n+1)ε>s$, but $(n+1)$ is also an element of $A$ and thus has to be below all upper bounds, i.e., $(n+1)ε\le s$. As both inequalities can not be true at the same time, one gets a contradiction to the assumption of an upper bound.

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