Prove there are uncountably many open singletons in $ω_1$ Given a well order $(W,≤)$, where $W$ is uncountable, and $$ω_1:= \{x\in W: \text{only countably many } y \in X \text{ s.t. } y \le x\},$$ prove there are uncountably many open singletons in $ω_1$.
I haven't learn anything about cardinals yet.

Let $W'\subseteq W$ be the set of all elements s.t. there exists an uncountable number of smaller elements. $W'$ is nonempty and has a minimum, let's say $w_1$ (in order to avoid the more natural $\omega_1$ that is already used in the question). Any element less than $w_1$ contains only countably many smaller elements, so there are uncountably many elements that coutain only countably many smaller elements.
If there were only countably many open singletons smaller than $w_1$, that would mean that there are uncountably many elements $x$ smaller than $w_1$ s.t. $\{x\}$ is not open; let's call such $x$ a "limit point". However, for $x$ a limit point, $x+1$ is an open singleton and not a limit poin ($x+1$ is defined to be the smallest element of $W$ larger than $x$). So there are at least as many open singletons as limit points in $[0, w_1)$ and we got a contradiction: so there can't be only countably many open singletons smaller than $w_1$.
