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

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.

• Note that $x\mapsto x+1$ gives an injection of $\omega_1$ into the set of countable successor ordinals, so there are indeed uncountably many isolated points in $\omega_1$. Commented Oct 4, 2021 at 12:50
• Can you edit your post and explain your notation for $\delta$, $w$ and the underbar? Commented Oct 7, 2021 at 4:27
• @PatrickR done.
– Aelx
Commented Oct 7, 2021 at 11:17

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.
• It is not true that $w(X)\le |X|$ for all $X$. E.g. let $F$ be a free ultra-filter on $\Bbb N$ .Then $F\cup \{\emptyset \}$ is a topology on $\Bbb N$ of uncountable weight because $F$ cannot be countably generated. Commented Oct 5, 2021 at 2:13
• @DanielWainfleet for ordered spaces the standard base is already bounded by $|X$. Commented Oct 5, 2021 at 5:13
• Yes. Originally you had "$w(X)\le |X|$ for all $X$" and I know that's not what you meant but the student might not see that you meant "all ordered $X$". Commented Oct 7, 2021 at 21:12
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 $$c(X)\le d(X).$$