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While reading this SE thread, I saw in the comments someone say "the proof [that the completeness axiom implies the well ordering principle] will take some work".

However, this other thread shows that by using the definition of infimum we can establish the above-mentioned result rather easily.

An even easier way of proving the result would be to say:

Let $A\subseteq \mathbb N$ be non-empty. Then, as $A\subseteq \mathbb R$ and $0\le x, \forall x\in A$, the completeness axiom gives us the existence of $s=\inf A$. Then, by topological arguments one can show that $\mathbb N$ has no limit points hence is closed. Therefore, $s\in A$ and $A$ has a minimum.

Is this easier way of proving the result valid ?

Thanks in advance

PS: the topological arguments are that if $p$ doesn't coincide with the natural numbers, $p$ can't be a limit point (there is void around it). And if $p$ coincide with one natural number, it can't be a limit point either (since the definition of limit point excludes the point of consideration).

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    $\begingroup$ You can dispense with topological concepts (beyond the idea of the infimum and supremum, which one could argue is topological in nature), but the key steps remain the same. You need to show that if $p$ is not integer, there is some $q > p$ such that any integer $> p$ is also $> q$. This sounds easy, but if you examine the obvious proof, it assumes that there is a least integer $> p$, which is effectively what you are trying to prove. You need a deeper dive into the nature of real numbers to accomplish it (which is also true of the topological argument), $\endgroup$
    – Paul Sinclair
    Commented Sep 9, 2023 at 14:52
  • $\begingroup$ Yes, I was suspecting that my proof was fallacious. Concerning your argument, I apology if it was obvious, but what does $p$ stand for in your argument? Is it the $\inf$, is it the studied limit point, or is it something else? $\endgroup$
    – niobium
    Commented Sep 9, 2023 at 15:10
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    $\begingroup$ Sorry - I mixed up the $p$ in your postscript with the infimum of $A$ in the argument, but I see you called that $s$ instead. So yes, I meant $p = \inf A$. Your proof is not fallacious, though it is not entirely complete. I'm just saying describing it topologically is unnecessary. $\endgroup$
    – Paul Sinclair
    Commented Sep 9, 2023 at 15:19

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