# How do we prove that the real numbers have no upper bound

I want to use proof by contradiction.

Suppose that real numbers are bounded, then according to the axiom of continuity, there exists a least upper bound $b$.

But if $x\in \Bbb R$, then $x+1\in \Bbb R$ because of the inclusion property of real numbers.

But $x+1\in \Bbb R\Longrightarrow x+1\leq b\Longrightarrow x\leq b-1$, hence $b-1$ is an upper bound for $\Bbb R$.

However since $b$ is a least upper bound we must have: $b\leq b-1\Longrightarrow 1\leq 0$, a contradiction, since $1>0$

Thus $\Bbb R$ is not bounded.

Is that proof correct?

• If $r$ is that upper bound, can you find a way to make a bigger number than $r$? Jan 6, 2016 at 13:01
• hint: consider $r+\pi^{10000000000}$ Jan 6, 2016 at 13:01
• yes $r+1>r$ ,so what Jan 6, 2016 at 13:11
• You should clarify: Is it required that the "upper bound" also be a real number? (The negative reals have an upper bound that is not a negative real, so this is a valid question.) In the affine extended real number system, $+\infty$ is an upper bound on the real numbers: it just is not itself a real number. In any system that combines the reals with the transfinite ordinal numbers, any transfinite ordinal number is an upper bound on the reals. Jan 6, 2016 at 13:26
• @vonbrand, don't go overboard. Just $r+\frac12$ is enough.
– mrf
Jan 6, 2016 at 14:43

From Calculus by Apostol:

Theorem #1: The set P of positive integers $$1, 2, 3,...$$ is unbounded above.

Proof #1: Assume P is bounded above. We shall show that this leads to a contradiction. Since P is nonempty, P has a least upper bound, say $$b$$. The number $$b-1$$, being less than $$b$$, cannot be an upper bound for P. Hence, there is at least one positive integer $$n$$ such that $$n>b-1$$. For this $$n$$ we have $$n+1>b$$. Since $$n+1$$ is in P, this contradicts the fact that $$b$$ is an upper bound for P.

Theorem #2: For every real $$x$$ there exists a positive integer $$n$$ such that $$n>x$$.

Proof #2: If this were not so, some $$x$$ would be an upper bound for P, contradicting Theorem #1.

$$\therefore$$ The set of real numbers has no upper bound.

• It seems that theorem 2 is used in theorem 1 when we say $n \gt b-1$! Isn't it? :) Jan 6, 2016 at 14:01
• No, it is just a basic deduction that if there was not such a number $n$ then $b-1$ would be an upper bound which is impossible. Jan 6, 2016 at 14:04
• Ah! It seems that I confused myself! You are right. :) Jan 6, 2016 at 14:05
• I read the whole first volume about 6 month ago. It was fantastic. I enjoyed reading it. :) When I saw this answer I recalled those days! :) Jan 6, 2016 at 14:06
• I don't know the context of Apostol's proof, but it's clearly overkill here: assume $r = \sup \Bbb R$; since $\Bbb R$ is closed under addition, then $r + 1 \in \Bbb R$, but $r + 1 > r$, contradiction. Jan 6, 2016 at 14:26

Suppose there is such upper bound $r\in\mathbb{R}$.

$r+1>r$ and $r+1\in \mathbb{R}$ since $\mathbb{R}$ is closed under addition.

• So you showed if $r$ is an upper bound then $r+1$ is also an upper bound. But you will have to do a little more to get a contradiction. Jan 6, 2016 at 14:48

The Archimedian axiom states:

For every $x, y \in \mathbb{R}$ with $0 < x < y$ there exists a $n \in \mathbb{N}$ such that $y < n x$.

Of course $n x \in \mathbb{R}$, so there is no largest real number $y$.