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I'm having some troubles recently understand why every set can be well ordered. I mean suppose it's true, then $\mathbb{R}$ can be well orderd, suppose $a\in \mathbb{R}$ is the first, then we will define a function $f:\mathbb{R} \rightarrow \mathbb{N}$ and say $f(a)=1$. Then we will look at $\mathbb{R}\backslash \{a\}$ (it's a subset of $\mathbb{R}$) then there is $b\in \mathbb{R}$ such that he is the first, and we will define $f(b)=2$ and then we can continue like that in contradiction that $\mathbb{R}$ is uncountable.

Can you please tell me where is my mistake? And maybe give me a "feeling" about why we can suppose this axiom is true? Thank you very much.

Rick
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Elad Elmakias
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  • https://math.stackexchange.com/questions/6501/is-there-a-known-well-ordering-of-the-reals This might be useful to look at. – Benjamin Jun 20 '20 at 09:48
  • okay and about my "contradiction" why is it not true – Elad Elmakias Jun 20 '20 at 10:21
  • What this does is find an increasing sequence in $\mathbb{R}$, but there is no reason why this would exhaust all of $\mathbb{R}$. Put differently: the function $f$ you construct is fine, it just will not be surjective. – Mark Kamsma Jun 20 '20 at 10:30
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    Also, I do not understand the down vote. I think this is a valid question... – Mark Kamsma Jun 20 '20 at 10:32
  • I am not an expert but if you read the Ordinal Number section of https://en.wikipedia.org/wiki/Well-order. The point is you shouldn’t expect the well order to be counted in terms of natural numbers ie $\mathbb{N}$, but should expect other ordinals to help you “count” the order. – Benjamin Jun 20 '20 at 10:42
  • I just realised that my first comment is not quite right, I had the domain and codomain of the function $f$ reversed. The idea is still the same though, this process does not exhaust $\mathbb{R}$. I made this into an answer. – Mark Kamsma Jun 20 '20 at 10:56
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    The reason why the Well Ordering Theorem (WOT) is true is beacuse we choose it to be true, at least in most part of modern Mathematics. What I mean with this is that the statement "every set can be well-ordered" is just an axiom, as it is for example that if $a$ and $b$ are sets, then so is ${a, b}$. In particular, the WOT is equivalent to the Axiom of Choice (AC), meaning that assuming WOT you can prove AC and viceversa. – Rick Jun 20 '20 at 10:58

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The construction of your function $f$ can never exhaust all of $\mathbb{R}$. So it will never be defined at all of $\mathbb{R}$. Maybe it is clearer if you try constructing it the other way around. We construct $g: \mathbb{N} \to \mathbb{R}$ as follows. Let $g(0)$ be the least element in the well-order of $\mathbb{R}$. Then let $g(1)$ be the least element in $\mathbb{R} - \{g(0)\}$, and continue this process. This does define a function, but it is just not surjective.

Note that there are more well-ordered sets than just $\mathbb{N}$. For example, we can put two copies of $\mathbb{N}$ after each other. This is a well-order, but it has "infinite" elements. We usually denote this well-order by $\omega + \omega$. We can already create many different well-orders in this way: $\omega + \omega + \omega$ etc. These are all countable, but there are also uncountable ones. Such as for example any well-order we put on $\mathbb{R}$.

This last sentence uses the axiom of choice of course (or the well-ordering theorem, which is equivalent). We can already construct uncountable well-orders without choice using Hartog's lemma.

Mark Kamsma
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