Prove that this is no real number such that $x \leq a$ for all real $x$.

I want to know if the way I proved it is valid or not.

Proof. We first prove that there is no real number $a$ such that $x=a$ for all real $x$. If this were true, then all real numbers would be equal to $a$. This is not possible, because the axiom that garantees the existence of identity elements tells us that the set of real numbers has the numbers $0$ and $1$. Therefore, the set of real numbers has more than only one element.

If $x<a$ for all real $x$, then, of course $a>0$. It can be proven that $1>0$. There is an axiom that tells that if two real numbers are in $R^{+}$, the sum of the two is in $R^{+}$. So, $a+1>0$ is in in $R^{+}$ and in $R$. But we assumed that $a$ is greater than all the real numbers. Thus, $a>a+1$, which is an absurd.

Is this proof correct? Can I improve it? Thank you.

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    $\begingroup$ Why not just say that if such $a$ exist then since $a+1$ is also a real number (by axiom, since both $a$ and $1$ are real) and if $a+1\leq a$ it would imply $1\leq 0$ which you said you can prove otherwise. (the first part is not relavent as far as I can tell) $\endgroup$
    – Belgi
    Aug 8, 2012 at 4:19
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    $\begingroup$ Your proof is no good at all. "$x\le a$ for all $x$" means "$(x < a \hbox{ or } x = a)$ for all $x$". This is not the same as "$(x<a$ for all $x)$ or $(x=a$ for all $x)$", as you seem to think. $\endgroup$
    – MJD
    Aug 8, 2012 at 5:20
  • $\begingroup$ @MJD You are of course correct, but this makes little change to the proof. OP needs to change "more than only one element" to "at least one element different from a", and "for all real x" to "for all real x different from a". $\endgroup$
    – Taemyr
    Jul 7, 2014 at 6:46
  • $\begingroup$ I think you are missing a step. Why is a>a+1 absurd? This should be proven since it's the heart of your proof. How to prove it depends on what axioms you can use. $\endgroup$
    – Taemyr
    Jul 7, 2014 at 6:52

3 Answers 3


Let me propose an alternative proof.

Proof: By contradiction, suppose such an $a \in \mathbb{R}$ exists. Then $a + 567 \le a$, which implies $567 \le 0$, a contradiction.

  • 2
    $\begingroup$ What do you even mean by "ordinary reals"? I have repeatedly said that $a$ is a real number. And yes, for any $a, b, c \in \mathbb{R}$, $a \le b \implies a + c \le b + c$ is a valid consequnce of the field and order axioms of the real numbers. $\endgroup$
    – Ink
    Aug 8, 2012 at 7:34
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    $\begingroup$ As long as $a \in \mathbb{R}$, all the field axioms and order axioms hold for $a$. The reason $a$ cannot exist is why I got a contradiction. And what you wrote isn't even a field axiom. $\endgroup$
    – Ink
    Aug 8, 2012 at 8:16
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    $\begingroup$ The desired property has nothing to do with the field being Archimedean and everything to do with the field being ordered, seeing as it holds for any ordered field, not just the Archimedean ones. $\endgroup$ Aug 8, 2012 at 8:36
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    $\begingroup$ Googling for "ordered field" smallest element gives, for example, proof in Elements of Real Analysis By Charles Denlinger p.14 of the fact that an ordered field does not have smallest/largest element; the proof goes by shifting the element by one; i.e., similarly as in this answer. $\endgroup$ Aug 8, 2012 at 10:29
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    $\begingroup$ @user1296727: The fact that there is no greatest element in the field does not imply that the field is Archimedean. In fact, every ordered field has no greatest element, no matter whether it is Archimedean or not. $\endgroup$
    – user5501
    Aug 8, 2012 at 10:54

This property is a consequence of the construction of $\mathbb{R}$. In particular, it follows from the fact that $\mathbb{R}$ has no upper bound: indeed, the existence of your $a$ would imply that $\sup \mathbb{R} \in \mathbb{R}$. But this is false, since $\mathbb{N} \subset \mathbb{R}$ and $\mathbb{N}$ has no upper bound, because $n+1 > n$ for any $n \in \mathbb{N}$.

  • 1
    $\begingroup$ The closed interval $[0,1]$ has an upper bound although $[0,1)\subset[0,1]$ and $[0,1)$ has no upper bound, because $\frac{x+1}2\gt x$ for every $x$ in $[0,1)$. $\endgroup$
    – Did
    Aug 8, 2012 at 11:13
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    $\begingroup$ I don't understand your comment, sorry. If $x \leq a$ for every $x \in \mathbb{R}$, then $\sup \mathbb{R} \leq a$. Moreover, please remark that $1$ is an upper bound for $[0,1)$. It seems you confuse upper bound, least upper bound, and maximum. $\endgroup$
    – Siminore
    Aug 8, 2012 at 12:26
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    $\begingroup$ Surely you noted that my comment follows exactly your answer: replace $\mathbb R$ by $[0,1]$ and $\mathbb N$ by $[0,1)$ and everything you state about $\mathbb R$ and $\mathbb N$ holds about $[0,1]$ and $[0,1)$. Ergo, your proof is bogus. More constructively, you might want to wonder what makes you accept that $[0,1)\subset[0,1]$ has an upper bound, but not $\mathbb N\subset\mathbb R$. $\endgroup$
    – Did
    Aug 8, 2012 at 12:42
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    $\begingroup$ Well, if $x \in [0,1)$, then $x+1 \notin [0,1)$, and this is rather important, isnt'it? My "proof" used the fact that $n+1 \in \mathbb{N}$ whenever $n \in \mathbb{N}$. Now, if $a$ is as in the statement, by definition of $\mathbb{R}$ there exists $\bar{n} \in \mathbb{N}$ such that $\bar{n} \leq a < \bar{n}+1$, and this is a contradiction. However, in my opinion, it is hard to write a proof if we do not agree on what we know about $\mathbb{R}$. $\endgroup$
    – Siminore
    Aug 8, 2012 at 14:18
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    $\begingroup$ I still do not understand. My comment is not off-topic at all! Who cares if $\frac{x+1}{2}>x$ for $x \in [0,1)$? The point is that $\mathbb{N}$ is closed under the addition of a fixed positive quantity, viz. $1$. Essentially, my "proof" boils down to Brian's proof: just remark that $a+1 \leq a$ implies $1 \leq 0$. Again: why do you say that $[0,1)$ has no upper bound? $\endgroup$
    – Siminore
    Aug 8, 2012 at 17:18

This is a consequence of the Archimedean property of the reals. (I assume that you have some familiarity with real analysis; in particular, I assume that you know that the reals have the supremum property.)

Theorem: For every real $\alpha > 0$ and real $\beta$, there exists an integer $n$ (in the naive sense of integers) such that

$$n \alpha > \beta$$

Proof: Suppose that there were to exist counterexamples $\alpha_0$ and $\beta_0$ (i.e., $\alpha_0$ such that for every integer $n$, $n\alpha_0 < \beta_0$). If we form $S = \{s \in \mathbb{R}: s = n\alpha \text{ for } n \in \mathbb{N}\}$, then this would say that every member of S is smaller than $\beta_0$ by hypothesis.

But clearly $S$ is not empty if $n\alpha_0 < \beta_0$ for every $n \in \mathbb{N}$; since it is bounded above (and since these are real numbers!) there exists $\lambda = \mathrm{sup}(S)$.

Now, for every $\epsilon > 0$, we must have some $s' \in S$ such that $s' > \lambda - \epsilon$. (Otherwise, every $s \in S$ is smaller than $\lambda - \epsilon$, so $\lambda - \epsilon$ would be an upper bound of $S$ strictly smaller than $\lambda$, which was by definition the smallest upper bound. This would be absurd.)

But this means that for some $s' \in S$, $s' > \lambda - \alpha$ (since we put $\alpha > 0$). Since $s' \in S$ if, and only if, $s'$ is an integer multiple of $\alpha$, this says

$$ s' = n\alpha > \lambda - \alpha $$

But then we get

$$ (n+1)\alpha > \lambda $$

so that some member of $S$ is greater than $\lambda$. This is absurd; so we conclude that there are no counterexamples to the theorem in $\mathbb{R}$.


Now, the result you want is an immediate corollary.

Theorem: There is no $\gamma \in \mathbb{R}$ such that $\delta < \gamma$ for all $\delta \in \mathbb{R}$.

Proof: Suppose there were a counterexample $\gamma_0$. Pick any real $\delta_0$. By the Archimedean property,

$$\exists n: n\delta_0 > \gamma_0.$$

This directly contradicts the hypothesis, since $n\delta_0$ is a real greater than $\gamma_0$.


  • 5
    $\begingroup$ Why do it this needlessly complicated way, instead of just using that $\mathbb{R}$ is an ordered field? $\endgroup$ Aug 8, 2012 at 9:01
  • $\begingroup$ @Tobias Nothing about $\mathbb{R}$ being an ordered field necessarily guarantees the absence of infinite, or infinitesimal, elements. There are non-Archimedean fields in which infinite elements do indeed exist. You cannot derive a contradiction simply from having a totally ordered field. $\endgroup$
    – Chris
    Aug 8, 2012 at 9:06
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    $\begingroup$ Your comments there are not actually correct. In any ordered field $F$, $a < b$ and $c < d$ implies that $a+c < b+d$, so even though the field can contain elements greater than any real, it cannot contain an element larger than all others, which is the issue here. $\endgroup$ Aug 8, 2012 at 9:17
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    $\begingroup$ @user1296727: This has nothing to do with whether an ordered field is Archimedian or not. Take "the" hyperreals for example. There is no hyperreal larger than every hyperreal, and the hyperreals are non-Archimedian and contain infinitesimals and infinite elements. $\endgroup$ Aug 8, 2012 at 9:42
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    $\begingroup$ @user1296727 The question is not about the existence of infinite elements but, rather, the existence of a maximal element. Ordered fields can have infinite elements, but they cannot have a maximal element, since $\rm\:x+1 > x\:$ for all $\rm\,x.\ \ $ $\endgroup$ Aug 8, 2012 at 20:23

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