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How can a non-dense, well-ordered set like a long ray be uncountable? If it's a set of an uncountable number of [0,1) line segments laid end to end, shouldn't there be a bijective function between the COUNTABLE set of natural numbers and the n-th segment on the ray? Is its very existence not a contradiction?

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No; why on earth should there be? There is not just an $n$-th segment for every $n\in\Bbb N$; there is an $\alpha$-th segment for every $\alpha\in\omega_1$, and $\omega_1$ is uncountable. –  Brian M. Scott Nov 11 '12 at 22:04
    
That makes sense... it's a little counter-intuitive to think of set theory in a physical sort context. I suppose the problem there is visualization. Thank you, that helped a lot. –  hombre Nov 11 '12 at 22:08
    
You’re welcome; glad it helped. –  Brian M. Scott Nov 11 '12 at 22:36

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up vote 3 down vote accepted

The idea behind the long line is to take $\omega_1$ copies of $[0,1)$. If the [non-negative] real numbers can be thought of as a countable union of $[0,1)$ ordered lexicographically, the idea behind the long line is that we take even more.

The fact that there are uncountable sets which are well-ordered follows from the axioms of ZF, it does not even require the axiom of choice.

The long line, if so, is not like a subset of the real numbers. It is a whole other topological space. In that space there is an uncountable well-ordered set, whereas in the real numbers every set which is well-ordered [in the natural order of the real numbers] is countable.

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