Real line is separable.
Then, why is $\omega_1$ topological space not separable?
IF this is true, doesn't this settle continuum hypothesis?
Also, does the field of real numbers have anything to do with topological spaces? (Just for curiosity.)
Real line is separable.
Then, why is $\omega_1$ topological space not separable?
IF this is true, doesn't this settle continuum hypothesis?
Also, does the field of real numbers have anything to do with topological spaces? (Just for curiosity.)
The order topology on $\omega_1$ has no direct relationship to the Euclidean topology on $\Bbb R$, so there’s no reason for the separability of one to suggest the separability of the other, let alone to imply it. The order topology on $\omega_1$ is not separable for the simple reason that if $A$ is a countable subset of $\omega_1$, then there is a $\beta\in\omega_1$ such that $\alpha<\beta$ for every $\alpha\in A$. This means that if $\gamma\in\omega_1$ and $\gamma>\beta$, the interval $(\beta,\omega_1)$ is an open neighborhood of $\gamma$ that is disjoint from $A$, and therefore $\gamma$ is not in the closure of $A$. Thus, no countable subset of $\omega_1$ can be dense in $\omega_1$.
None of this involves the continuum hypothesis in any way.
‘Anything to do with’ is a pretty vague description. If you’re asking whether the properties of $\Bbb R$ as an algebraic field play a significant rôle in topology, I’d say not, but that may be partly a reflection of my particular topological interests. Be that as it may, it’s fair to say that the order properties of $\Bbb R$ are much more important topologically.
In some models where the axiom of choice fails we have that $\omega_1$ is a countable union of countable ordinals. This means that indeed $\omega_1$ is separable in such model.
Thus, assuming $\omega_1$ is separable is inconsistent with ZFC, but is fairly consistent with ZF. In fact it settles the question whether or not it is a singular cardinal!
Now we can consider the continuum hypothesis in two formulations in ZFC:
While in ZFC these are equivalent they are no longer equivalent in ZF. The assumption that $\omega_1$ is singular automatically implies that AH is false. It need not imply that CH is false, but I'm not sure that has an answer so far.
I should also point out that given an uncountable set one can consider the discrete topology. It is a locally compact metric space, and by that it is a bit closer to the real line than $\omega_1$ (where both are simply linearly ordered). Uncountable discrete spaces are never separable since every dense set would include all points.
For Further Reading: