# What uncountable ordinals live in the Long Line?

#### lugita15

It is a relatively simple exercise to prove that a well-ordered set is order-isomorphic to a subset of R (under the usual ordering) if and only if it is countable. You can say that R is "too small" to contain any uncountable well-ordered sets.

So my question is, can you embed bigger well-ordered sets in the long line? For those who don't know, the long line can be constructed by taking the minimal uncountable well-ordered set (i.e. omega_1) and taking its Cartesian product with [0,1) under the dictionary order. So obviously omega_1 itself is emebeddable in the long line, just by taking the left endpoints of all the intervals [0,1). But can you embed bigger uncountable ordinals, and if so how big? I'm guessing that you may be able to embed all well-ordered sets with cardinality less than or equal to aleph_1, the cardinality of the set of countable ordinals.

Any help would be greatly appreciated.

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#### fresh_42

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It seems as if you were not allowed to use AC. This is a quite unusual mathematical concepts, so some of your "easy proofs" would be interesting to see.

#### SSequence

It seems as if you were not allowed to use AC. This is a quite unusual mathematical concepts, so some of your "easy proofs" would be interesting to see.
I think that, under reasonable interpretation, the first sentence in OP is correct (I am not good at parsing formal statements, so I am not completely sure). Consider some arbitrary set $S \subseteq R$. Now assuming well-ordering principle (for simplicity), indeed there is always going to an $S$ that has well-orders of (only) uncountable length.

However, I think the intention might be as follows. Consider some $S \subseteq R$ and consider some specific well-order of $S$ that is of length, say $\alpha$. I think the intention here might be that if we have any two arbitrary ordinal values $x_1<x_2<\alpha$ and we write the real numbers "occupying" the positions $x_1,x_2$ as $r_{x_1},r_{x_2}$ respectively, then the following must be true:
$r_{x_1}<r_{x_2}$
The comparison in the above line is meant to be the usual comparison relation between the real numbers.

If we assume that an arbitrary well-order of $S$ satisfies the condition in previous paragraph, then it seems to me that $\alpha$ should always be countable (unless I am making an easy mistake). I think one can just use the following two properties (from the basic properties of real numbers), with two or three steps in-between, to show that $\alpha<\omega_1$:
----- between any two distinct real numbers there is a rational number
----- rational numbers are countable

Edit:
I think there is yet another alternative way of writing the same statement. It should probably go like: Consider the linear-order formed by endowing $R$ with the usual comparison relation for real numbers. Then one can't "embed" an uncountable ordinal within this linear-order. Or perhaps(?) more casually: "one can't embed an uncountable ordinal in the real number line".

But, as far as I know, the term embed is supposed to have a technical meaning. I have the feeling that it agrees with the way I have used the term in previous paragraph, but I don't know for sure (it would actually be good to know).

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"What uncountable ordinals live in the Long Line?"

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