# Direct Proof of Archimedian Property

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. $$\square$$

• It looks good. Clever way around the fact that we don’t know that $u$ is a natural number. Jul 4 '21 at 17:32
• 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. Jul 4 '21 at 17:33
• 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. Jul 4 '21 at 17:39