I was trying to understand the long line, and came across this reddit thread: https://www.reddit.com/r/math/comments/apfzi/in_topology_the_long_line_or_alexandroff_line_is/, in which there's a comment that says:
You're probably having difficulty because you need to understand ordinal numbers as well as standard run-of-the-mill topology. Basically, to make the long line, you don't label each interval with a real number, you label them with elements of ω₁, the first uncountable ordinal. The reals have the same cardinality as ω₁, but they are not the same ordinal; in particular, they do not have the same topology.
Although if you did simply did take the Cartesian product of the real line with the unit interval, you could still define a similar linear topology on it by using the lexicographic order. You don't necessarily need to identify the "next" interval or ever actually glue anything, you just need to be able to define open sets using a total ordering.
Could someone shed more light on this last paragraph? If it's possible, I would like to see an alternative definition of the long line that does not mention ordinals at all but instead just uses real numbers.
It’s possible to define the long line without explicitly mentioning $\omega_1$ or using the continuum hypothesis, but it’s not possible to avoid using a well-ordered set that is order-isomorphic to $\omega_1$.
Let $\preceq$ be a well-order on $\Bbb R$, for each $x\in\Bbb R$ let $P_x=\{y\in\Bbb R:y\prec x\}$, and let $X=\{x\in\Bbb R:P_x\text{ is countable}\}$. Give $X\times[0,1)$ the lexicographic order with respect to $\preceq$ on $X$ and the usual order on $[0,1)$; then $X\times[0,1)$ with the linear order topology induced by that lexicographic order is the long line. This doesn’t explicitly mention $\omega_1$, but in fact $\langle X,\preceq\rangle$ is order-isomorphic to $\omega_1$ with its usual order, so $\omega_1$ is actually present implicitly — as in fact it must be in any construction of the long line.