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The condition on a linear order to be a well-order is very restrictive. Informally stated, all well-ordered sets begin the same way when arranged with respect to this order, and any two of them are sort of comparable. But why?

Suppose $\langle A, <\rangle$ is a non-empty well-ordered set. Then it has a minimal element, say $a_0$, because $A\subset A$ is a non-empty subset. We have $a_0 < a$ for all $a\in A$, $a\neq a_0$. But if $A\smallsetminus \{a_0\}$ is non-empty, then it has again a minimal element $a_1$, and $a_0 < a_1 < a$ for all $a\in A$, $a\neq a_0$, $a_1$. Can you see the pattern? If $A$ is big enough, we would have $a_2$ the minimal element of $A\smallsetminus \{a_0,a_1\}$, $a_3$ the minimal element of $A\smallsetminus \{a_0,a_1,a_2\}$, and so on. This process may eventually terminate



or it may become infinite



But even if it is infinite, it may not contain all the elements in $A$: think of $\mathbb{N}\times \mathbb{N}$ with the lexicographical order for instance. So we have to go on. How? Well, one could define the set of elements of $A$ that do not belong to the previous row. This property can be formalized into a logical formula (the set of elements $b$ for which there is no finite sequence $a_0,a_1,\dots,a_n$ such that $a_0 < a_1 < \cdots < a_n < b < a$ for all $a\in A\smallsetminus \{a_0,a_1,\dots,a_n,b\}$, and so on) in order to apply the Axiom of Specification (???). And now: this subset has a minimal element! And when this element is removed, we get another one! So a new thread starts



Provided $A$ is not completely exhausted by this procedure, we would have more and more threads



Even infinite threads!



And after all those infinite threads... one could look at the remaining elements and start again! It would look like... a three-dimensional arrangement?



That's why all the well-ordered sets "begin the same way". And this pattern will be capture by the ordinal numbers: the examples above would be represented by the ordinals $4$, $\omega$, $\omega+4$, $\omega\times 4 + 4$, $\omega^2$ and $\omega^2\times 4$ respectively.

This may seem terribly intricate... but it is just the beginning. If you look carefully, all the sets above are numerable, and indeed all the well-ordered sets we can "describe in a plain language" are numerable. Then, how would a non-numerable well-ordered set look like? What if we could define a well-order in $\mathbb{R}$? Its inner structure would be enormous. For instance, think that there would be a first element (for the minimal element property) whose section is non-numerable, all previous elements having numerable section (just like the first element of the second "thread", which gives rise to non-finite sections). Can you imagine it?

Finally, this discussion is closely related to the more than controversial Axiom of Choice. This axiom seems rather stupid at first; it says that for any set $A$, there is a function that assigns any non-empty subset of $A$ an element that belongs to this subset. It sounds obvious: if the subset is non-empty, we can choose an element in it! The hot point is that we need to have all this choices beforehand, all at once, like some sort of criteria. Could you describe some criteria that assigns every non-empty subset of $\mathbb{R}$ an element in it? (Beware, there are very weird subsets of $\mathbb{R}$!)

In fact, one cannot describe such criteria (try as hard as you want). And if such function-criteria existed, the construction above could be replicated. Suppose we have a function of choice in $\mathbb{R}$:

$$f:\mathcal{P}(\mathbb{R})\longrightarrow \mathbb{R}\qquad f(a)\in a\,\forall a\subset \mathbb{R}$$


Ok, now we define

$$ \begin{array}{rcl} a_0 & = & f(\mathbb{R}) \\ a_1 & = & f(\mathbb{R}\smallsetminus\{a_0\}) \\ a_2 & = & f(\mathbb{R}\smallsetminus\{a_0,a_1\}) \\ a_3 & = & f(\mathbb{R}\smallsetminus\{a_0,a_1,a_2\}) \\ a_4 & = & f(\mathbb{R}\smallsetminus\{a_0,a_1,a_2,a_3\}) \\ & \vdots & \\ \end{array} $$


See? And after one thread, we may start another one... it is as if there was a well-order in $\mathbb{R}$. How long can we proceed as above? Well, forever, and it is always the same principle: if there are elements of $\mathbb{R}$ which can't be put into this well-order, which is $f$ of this subset? So it is equivalent: having all these choices beforehand, all at once, and such huge non-numerable well-order existing. Mind-blowing, isn't it? More on that later