8
$\begingroup$

I've written this proof and I'd like to ask if this is a valid proof of the Archimedean Property:

$\underline{\text{Claim}:}$

$\forall x \in \mathbb{R}\exists n_x \in \mathbb{N}:x \leq n_x$.

$\underline{\text{Proof}:}$

We know that $1 \in \mathbb{N}$ and $1 \in \mathbb{R}$.

Thus, if $x \leq 1$ then $n_x = 1$ and we are done.

Now assume, $x > 1$. Let M = $\{k \in \mathbb{N} : k < x\}$. Then $1 \in M$.

Thus, $M\neq \varnothing$ and is bounded above by $x$.

By the completeness property, we can then infer that sup$M$ exists. If we let $u = \text{sup}M -1 $ then we know that $\exists k \in M$ such that $k > u$.

$\therefore\text{sup}M < k + 1$, which is a natural number by the inductive property of natural numbers.

Since, $k + 1 > \text{sup}M$, we have that $k + 1 \not \in M$. By setting $n_x = k + 1$, we have proved the Archimedean Property. $\square$

$\endgroup$
3
  • 3
    $\begingroup$ It looks good. Clever way around the fact that we don’t know that $u$ is a natural number. $\endgroup$ Jul 4 '21 at 17:32
  • $\begingroup$ Since your proof is more or less the same as this in Direct proof of Archimedean Property (not by contradiction), I suggest we close this question. Any alternative proof/idea can be posted there. $\endgroup$ Jul 4 '21 at 17:33
  • 2
    $\begingroup$ That answer there is very similar, but goes out of its way to proves the equivalent of $u$ is a natural number, which this proof cleverly skips right by. $\endgroup$ Jul 4 '21 at 17:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.