# Proof Verification - Archimedean Property

I am self-learning real analysis and learning to write proofs. I am trying to prove the Archimedean Property and would like to check if my attempt at a proof is correct and how to improve my proof writing skills.

Given any number $$x\in R$$, there exists an $$n \in N$$ satisfying $$n>x$$.

My understanding of this statement is that the set of natural numbers $$N$$ is not bounded above.

(Proof): By contradiction, there exists an $$x\in R$$,such that $$\forall n \in N$$,$$n \leq x$$. $$x$$ is an upper bound for $$N$$, so by the Axiom of Completeness $$N$$ has a least upper bound $$\alpha = sup (N)$$.

By the approximation property, if $$\alpha = sup (N)$$ then $$\forall \epsilon >0$$ ,$$\exists n \in N$$ such that $$\alpha - \epsilon < n \leq \alpha$$ $$\implies \alpha - \epsilon < n$$

$$\implies \alpha < n +\epsilon$$

$$\implies \alpha \leq n$$

$$\implies n \geq \alpha$$ which contradicts that $$\alpha$$ is the least upper bound.

• (There exists some $x\in\mathbb R$ such that $n\leq x$ for all $n\in\mathbb N$.)
– cqfd
Aug 21, 2019 at 18:08
• Thank you I made the correction.
– TQFT
Aug 21, 2019 at 19:00
• Still not correct.
– cqfd
Aug 22, 2019 at 1:53
• Why is $n \ge \alpha =\sup \mathbb N$ a contradiction that $\alpha$ is the least upper bound of $N$? You haven't found an upper bound that is smaller than $\alpha$. And you havent found an integer that is larger than $\alpha$. So nothing has been contradicted. Aug 22, 2019 at 5:46
• So either try to find a $\beta < \alpha$ were $\beta$ is an upper bound. (That would be a contradiction). Or try to find an $m\in \mathbb N$ so that $m > \alpha$. (That would be a contradiction.) Aug 22, 2019 at 5:47

There's a mistake.

Let $$\alpha = \sup(\mathbb{N})$$ which exists by the reasons you mentiond. It is true that this means that for all $$\varepsilon>0$$ there exists $$n\in\mathbb{N}$$ such that $$\alpha-\varepsilon .

From this you conclude that $$\alpha which is fine.

However this does not mean that $$\alpha\leq n$$.

You can't deduce that because $$n$$ depends on $$\varepsilon$$, therefore the usual trick of letting $$\varepsilon = (n-\alpha)/2$$ isn't possible (you can't define $$\varepsilon$$ using a variable $$n$$ which depends on $$\varepsilon$$).

Instead you should fix $$\varepsilon$$. If you choose $$\varepsilon=\frac{1}{2}$$, then $$\alpha for some natural number $$n\in\mathbb{N}$$ which corresponds to $$\varepsilon=\frac{1}{2}$$. From this you can conclude that $$\alpha, since $$n+1$$ is a natural number we get a contradiction to the fact that $$\alpha$$ is an upper bound.

Two mistakes.

For all $$\epsilon > 0$$ the will indeed exist an $$n_\epsilon\in \mathbb N$$ so that $$\alpha -\epsilon < n_\epsilon \le \alpha$$ and $$n_\epsilon < \alpha +\epsilon$$ but that does not mean $$n_\epsilon < \alpha + \epsilon$$ for all $$\epsilon$$.

$$n_\epsilon < \alpha + \epsilon$$ is only true for that $$n_\epsilon$$ and that $$\epsilon$$. For a different value of $$\delta > 0$$ it will follow that that is an $$n_\delta$$ so that $$n_\delta < \alpha + \delta$$ but $$n_\delta$$ could be a completely different value than $$n_\epsilon$$.

Second.

$$n\ge \alpha$$ does not contradict that $$\alpha$$ is a least upper bound. $$\alpha$$ is a least upper bound, and $$n \in \mathbb N$$ will mean that $$\alpha \ge n$$ and we have $$n \ge \alpha$$. That's not a contradiction.

......

So here's a hint.

let $$0 < \epsilon <1$$.

Let $$n_\epsilon$$ but the natural number where $$\alpha - \epsilon < n_\epsilon \le \alpha$$.

Now I'll tell you right off the bat, you will never find a contradiction with $$n_\epsilon$$. You can note that $$n_\epsilon < \alpha+\epsilon$$ if you want but that won't be a contradiction nor will it help you.

You will find nothing wrong with $$n_\epsilon$$.

Try to find a different natural number that does cause a contradiction.

Second hint. Don't bother trying to find a different $$\delta > 0$$ and a different $$n_\delta$$ so that $$\alpha - \delta < n_\delta \le \alpha$$. If you do that you will find something very important about $$n_\epsilon$$ vs. $$n_\delta$$ but it won't be a contradiction.

Third hint: You have $$\alpha -\epsilon < n_\epsilon \le \alpha$$. Try to find an $$m\in \mathbb N$$ so that $$m > \alpha$$. That was your original goal after all. How does knowing $$\alpha - \epsilon < n_\epsilon \le \alpha$$ help you find $$m$$ so that $$m > \alpha$$?

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Fourth Hint: FORGET ANALYSIS! How would a five year old answer answer this?

Try it. Go up to a five year old and ask her "I'm thinking of a real big number. How do you know that there is a bigger one?" I bet you she will say the answer that is the utter key to this proof!