Is it still possible for their to exist a well-ordering on the reals such that there is always a least next element?

Question

Let $a \leq b$ be a well-ordering on $\Bbb{R}$. And suppose for any $a_n \in \Bbb{R}$ there exists another element $a_{n+1} \in \Bbb{R}$ such that under the given ordering $(a_n, a_{n+1})$ is empty. In other words there always exists a minimal next element. Does this contradict the uncountability of $\Bbb{R}$?

Me: Let $a_0$ be the least element of $\Bbb{R}$ since it is well-ordered. Define $\Bbb{N} \to \Bbb{R}$ to be $n \to a_{n}$ in other words map $n$ to the $n$th element after $a_0$ in the order. Then it's clearly injective. But not necc. surjective as there's not necc. a maximal previous element.

I don't see how I can proceed clearly from here, so maybe it doesn't contradict the uncountability of $\Bbb{R}$?

Answer 1

Being well-ordered does not mean that there is an order isomorphism with $\Bbb N$.

For example consider the order on the set $\Bbb N\cup\{\Bbb N\}$ defined as $x\prec y$ if and only if either $x,y\in\Bbb N$ and $x<y$ or $x\in\Bbb N$ and $y=\Bbb N$. This order is a well-order. It is in fact countable, because we added just a single element to $\Bbb N$. But it is not order isomorphic to $\Bbb N$ since it has a maximal element.

Now we can continue adding one point after another, then reach a stage where each element has a successor, and just add a new point which is larger than everything so far. And continue and continue. At some point, we would have done that more than countably many times. Then we have an uncountable well-order. There are other methods of showing that there are in fact uncountable well-orders, as large as you want, and much much larger too.

As for $\Bbb R$, yes it is uncountable. But assuming the axiom of choice we can prove that there is some well-ordering of $\Bbb R$. What exactly does it look like? We can't quite say, because the axiom of choice is essential for this proof, but we know that it is an uncountable well-order.

Now, if there is a largest element in a well-ordering, we can remove it, if there is still a largest element remove that one as well, and continue until either you have no largest element or no element at all. This defines a decreasing sequence, so it has to be finite. Therefore by removing finitely many elements from an infinite well-order we can guarantee that there is no maximal element.

Answer 2

Any well-ordering of (say) the reals for which there is no largest element has the desired property.

For let $a$ be real, and let $A$ be the set of all $x$ such that $a\lt x$, where $\lt$ is a well-ordering. The set $A$ by assumption is non-empty, so has a smallest element $b$. Then the open "interval" $(a,b)$ is empty.

One can make many well-orderings of the reals such that there is no largest element. A natural well-ordering with this property is induced by a bijection between the reals and the smallest ordinal that has the same cardinality as the reals.

Any well-ordering of an infinite set, as long as there is no largest element, as the "empty interval" property. And the empty interval property always holds for all but the largest element, if there is one. In particular, sets with the property can be of any infinite cardinality.

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The property that for any $a$ there is a $b\gt a$ such that the open interval $(a,b)$ is empty does not imply that the ordering is a well-order.

For call two reals equivalent if they differ by an integer. Let $E$ be the set of equivalence classes. Order $E$ arbitrarily. If $x$ and $y$ are real, put $x \lt y$ if the equivalence class of $x$ is less than the equivalence class of $y$, or if the equivalence classes are the same but $x$ is in the real number sense less than $y$. The ordering we get is not a well-order. However, it has the property that for any $a$ there is a unique $b$ such that $a\lt b$ and $(a,b)$ is empty.