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I know the statement above is true because of the Archimedian Theory. Would the following proof make sense to prove it?

This is a proof by contradiction.

If the set of natural numbers does have an upper bound, then it has a least upper bound. Let $x = \sup \mathbb{N}$, supposing that each does exist as a finite real number. Then there is not a natural number $n > x$. Therefore $n \leq x$. Then, $n \leq x - 1$ cannot be true for all natural numbers.

There is some natural number $m$ such that $m > x - 1$. Since $m$ is a natural number, $m + 1$ must also exist, so $m + 1 > x$. But this cannot be so since we defined $x$ as the largest number.

Therefore, by contradiction, the above statement stands true.

Ken
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knerd
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    Why can't $n \le x - 1$ be true for all natural numbers? – Lucas Mann Sep 10 '14 at 19:13
  • It can't be true for all natural numbers since it is less than or equal to x. If n was equal to x to begin with, then it would be greater than x - 1 which isn't true since we supposed that there is no number n > x. – knerd Sep 10 '14 at 19:16
  • Ok, suppose all natural numbers are $\le x$. Now take $x' := x + 1$. Then obviously all natural numbers are $\le x'$. If we apply your proof to $x'$, it claims that $n \le x' - 1$ cannot be true for all natural numbers. But $x' - 1 = x$. So aren't you just saying that the assumption is true? – Lucas Mann Sep 10 '14 at 19:21
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    Suppose there is an $x$ such that $x\ge n$ for every natural number. Then the set of natural numbers is bounded above, so has a least upper bound $b$. Now you can produce an argument roughly along the lines described. But as it stands what you have is not a proof. – André Nicolas Sep 10 '14 at 19:22
  • The statement you're trying to prove is usually taken as an axiom, by the way. – Greg Martin Sep 10 '14 at 19:35
  • Please look at what I changed above and see if it now makes sense. Thanks all for your help! – knerd Sep 13 '14 at 23:28

2 Answers2

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You need to use some specific properties of the real numbers. Otherwise, your proof might work for non-standard number systems which are not Archimedean.

If you construct your reals as Dedekind cuts of rationals then you only need the Archimedean property of the rationals: just take any rational in the set of upper bounds. If you use Cauchy's construction, pick a point beyond which all elements are at distance at most one (say), and again use the Archimedean property of the rationals.

It remains to prove the Archimedean property of the rationals. Assuming the construction as equals class of fractions $p/q$ with $p$ positive, we know that $p/q \leq p$.

Yuval Filmus
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  • The OP's proof does use a specific property of the real numbers. Namely, it uses the least upper bound property of the real numbers, which does imply the Archimedean property, as the OP showed. – bof Jun 04 '15 at 04:30
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I am taking a crack at this!

Now, what ever a large real number is, arbitrarily select one but keep it fixed throughout the proof. Let us call it $x$.

Next, for a contradiction, let us suppose there are no natural numbers $n$ satisfying the inequality $n > x$. Thus, every natural number $n$ satisfies the inequality $n\leq x$. In other words, the set $\Bbb{N}$ of all natural numbers as a subset of the set $\Bbb{R}$ of real numbers is bounded above by $x$ (an upper bound). The set $\Bbb{R}$ of all real numbers is complete--- This means Every nonempty subset of real numbers which possesses an upper bound also possesses a least upper bound or supremum. So let $b = \hbox{sup}\Bbb{N}$ (it exists by the completeness of $\Bbb{R}$).

Now observe!

  • $b\leq u$ for any upper bound $u$ for $\Bbb{N}$.

  • $n\leq b$ for every $n\in\Bbb{N}$ and

  • For any $\varepsilon > 0$ (no matter how small), the real number $b - \varepsilon$ is not an upper bound for $\Bbb{N}$ because $b -\varepsilon < b$.

In particular, $b - 1$ (with $\varepsilon = 1 > 0$) is not an upper bound for $\Bbb{N}$. Consequently, there must be some $n_0\in\Bbb{N}$ such that $b - 1 < n_0\leq b$. This last inequality can be written as $$b < n_0 + 1\leq b + 1.$$But we know that if $n_0$ is a natural number, then $n_0+1$---the immediate successor of $n_0$ is also a natural number. In the last inequality above we have $b < n_0 + 1$ which contradicts the definition of $b$. Hence, our assumption that $\Bbb{N}$ is bounded above is false. Hence, it must be that $\Bbb{N}$ is unbounded above and so whatever real number $x$ we choose there must be a natural number $n$ such that $n > x$ as we wanted to show.

Foster Antony
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