Why does it hold that $\delta \underline{W(\omega_1)} = \aleph_1?$

Question

Given order topology on a set $W(\omega_1) = \{ \beta \ | \ \beta < \omega_1 \}$. How can one prove that it's density is $\aleph_1$?

Since $W(\omega_1)$ is dense in $\underline{W(\omega_1)}$, so it must hold that $\delta \underline{W(\omega_1)} \le \aleph_1$. It is only needed to prove that $\delta \underline{W(\omega_1)} \ge \aleph_1$. Or in other words, that for any dense set $D$ in $\underline{W(\omega_1)}$ it would hold that $card(D) \ge \aleph_1$.

If I understand it right then for $0 \in W(\omega_1)$ it holds that $\{0\}$ is open and thus $0 \in D$ as well as for any successor-ordinal $\beta \in W(\omega_1)$ it must hold that $\{\beta\}$ is open and we get $\beta \in D$.

Here is where I'm stuck. I'm not sure if the cardinality of $D$ at this point is bigger-than or equal to $\aleph_1$ and if so, then why it is the case. If not, then most probably I'd have to work with limit ordinals up-to $w_1$ somehow.

Here $\underline X$ denotes a topological space, e.g. $\underline X = (X, T)$ with $X$ being a set and $T$ a topology on that set. The $\delta \underline X$ corresponds to the density of this topological space and $\omega \underline X$ to its weight.

PS. If proven then I guess it'd automatically mean that $w\underline{W(\omega_1)} = \aleph_1$ as well.

Answer 1

If $D \subseteq W(\omega_1)$ is countable, there is some $\beta \in W(\omega_1)$ (because a countable union of countable ordinals is again a countable ordinal) so that $\forall \alpha \in D: \alpha \le \beta$. But then $O:=\{\gamma\mid \gamma > \beta\}$ is open and non-empty in $W(\omega_1)$ and $O \cap D = \emptyset$, so that $D$ is not dense in $W(\omega_1)$. A subset of $W(\omega_1)$ that is countable so cannot be dense so a minimal dense subset must have size $\ge \aleph_1$ and as this is the trivial upperbound anyway, $d(W(\omega_1)) = \aleph_1$. This certainly implies $w(W(\omega_1)) = \aleph_1$ as well, by trivial considerations, like $d(X) \le w(X) \le |X|$ for all ordered spaces e.g.

Answer 2

If $D$ is dense in $\omega_1$ then $\forall b\in\omega_1\,(D\cap \{b+1\}\ne \emptyset\,)$ because each $\{b+1\}$ is open. So $D\supseteq \{b+1:b\in\omega_1\}.$ And the function $f(b)=b+1$ is injective on $\omega_1,$ so $|\{b+1:b\in\omega_1\}|=\omega_1.$

The cellularity $c(X)$ of a space $X$ is the least infinite cardinal $k$ such that if $F$ is any family of pair-wise disjoint open subsets of $X$ (a.k.a. a discrete open family) then $|F|\le k.$ The density $d(X)$ is the least infinite cardinal $k'$ such that $X$ has a dense subset $D$ with $|D|\le k'.$ (It has been said that "There are no finite cardinals in point-set topology".) We always have $c(X)\le d(X).$

There are many other types of cardinal properties of a topology. They are generally called topological cardinal functions.