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

Question

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.

Answer 1

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.

Answer 2

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 $n \in \mathbb{N}$.

Answer 3

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.)

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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 $\mathbb{R}$.

Q.E.D.

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Now, the result you want is an immediate corollary.

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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 $\gamma_0$.

Q.E.D.