What is the cardinality of the open ordinal space $[0,\Omega)$ if we remove neighborhoods of each limit ordinal? Let $\Omega$ be the first uncountable ordinal with the order topology. For each limit ordinal $\lambda < \Omega$ let $U_\lambda$ be an open neighborhood of $\lambda$. What is the cardinality of $[0,\Omega) \setminus (\bigcup \{ U_\lambda : \lambda < \Omega\text{ is limit} \})$?
 A: A weak result
Let's denote by $\newcommand{\LIM}{\mathrm{Lim}}\LIM$ the set of limit ordinals below $\Omega$.  Note that $\LIM$ is a stationary subset of $\Omega$.  If for each $\lambda \in \LIM$ we have an open neighbourhood $U_\lambda$ of $\lambda$, then there is an $f(\lambda) < \lambda$ such that $( f(\lambda) , \lambda ] \subseteq U_\lambda$.  In this way we set up a function $f: \LIM \to \Omega$ such that $f(\lambda) < \lambda$ for all $\lambda \in \LIM$.
Such a function is called regressive, and by the Pressing Down Lemma it follows that there is a stationary (and hence uncountable) $S \subseteq \LIM$ such that $f \restriction S$ is constant, say $f(\lambda) = \alpha$ for all $\lambda \in S$.  From this it follows that $[ 0 , \Omega ) \setminus \bigcup_{\lambda \in \mathrm{Lim}} U_\lambda \subseteq [ 0 , \alpha]$ is therefore countable.

A stronger result
The above uses some fairly heavy machinery to prove something pretty weak. You can actually do a lot more with less. The following is completely independent of the above
The set $A = [ 0 , \Omega ) \setminus \bigcup_{\lambda \in \mathrm{Lim}} U_\lambda$ is actually finite.


*

*proof. Note that $A$ is closed. A basic property of $[0,\Omega)$ is that any strictly increasing sequence converges to an element of $\LIM$. If $A$ were infinite, we could construct a strictly increasing sequence of elements of $A$. So this sequence converges to an element of $\LIM$, but the limit must also belong to $A$ since $A$ is closed. But this contradicts the fact that $A \cap \LIM = \varnothing$! So $A$ must be finite.

A: Not enough information. All we know about $U_\lambda$ is that it is an open neighborhood of some countable limit ordinal $\lambda$.
If $U_\lambda=[0,\lambda]$ then $\bigcup U_\lambda=[0,\lambda)$ and $|[0,\Omega)\setminus(\bigcup U_\lambda)|=|[0,\Omega)\setminus[0,\lambda)|=|[\lambda,\Omega)|=\aleph_1$.
If $U_\lambda=[0,\Omega)$ then $\bigcup U_\lambda=[0,\Omega)$ and $|[0,\Omega)\setminus(\bigcup U_\lambda)|=|[0,\Omega)\setminus[0,\Omega)|=0$.
Now, if you asked about $|[0,\Omega)\setminus(\bigcup_\lambda U_\lambda)|$, that would be a completely different question.
In your title you ask a third question; the answer to that is, if we remove the limit ordinals from $[0,\Omega)$, we are left with the non-limit ordinals, and there are $\aleph_1$ of those.
